Kenneth DeMason, July 16, 2021
Contents
1. Complex Numbers
1.1. The Algebra of Complex Numbers
1.1.5. Inequalities
1.1.5.1.
1.1.5.2.
1.1.5.3.
1.1.5.4.
1.2. The Geometric Representation of Complex Numbers
1.2.1. Geometric Addition and Multiplication
1.2.1.1.
1.2.1.2.
1.2.1.3.
1.2.1.4.
1.2.2. The Binomial Equation
1.2.2.1.
1.2.2.2.
1.2.2.3.
1.2.2.4.
1.2.2.5.
1.2.3. Analytic Geometry
1.2.3.1.
1.2.3.2.
2. Complex Functions
2.1. Introduction to the Concept of Analytic Function
2.1.2. Analytic Functions
2.1.2.1.
2.1.2.2.
2.1.2.3.
2.1.2.4.
2.1.2.5.
2.1.2.6.
2.1.2.7.
2.1.4. Rational Functions
2.1.4.1.
2.1.4.2.
2.1.4.3.
2.1.4.4.
2.1.4.5.
2.2. Elementary Theory of Power Series
2.2.3. Uniform Convergence
2.2.3.1.
2.2.3.2.
2.2.3.3.
2.2.3.4.
2.2.3.5.
2.2.3.6.
2.2.4. Power Series
2.2.4.1.
2.2.4.2.
2.2.4.3.
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Ahlfors Exercises
2.2.4.4.
2.2.4.5.
2.2.4.6.
2.2.4.7.
2.2.4.8.
2.2.4.9.
2.3. The Exponential and Trigonometric Functions
2.3.2. The Trigonometric Functions
2.3.2.1.
2.3.2.2.
2.3.2.3.
2.3.2.4.
2.3.4. The Logarithm
2.3.4.1.
2.3.4.2.
2.3.4.3.
2.3.4.4.
2.3.4.5.
2.3.4.6.
2.3.4.7.
2.3.4.8.
3. Analytic Functions as Mappings
3.1. Elementary Point Set Topology
3.1.2. Metric Spaces
3.1.2.1.
3.1.2.2.
3.1.2.3.
3.1.2.4.
3.1.2.5.
3.1.2.6.
3.1.2.7.
3.2. Conformality
3.2.2. Analytic Functions in Regions
3.2.2.1.
3.2.2.2.
3.2.2.3.
3.3. Linear Transformations
3.3.1. The Linear Group
3.3.1.1.
3.3.1.2.
3.3.1.3.
3.3.1.4.
3.4. Elementary Conformal Mappings
3.4.2. A Survey of Elementary Mappings
3.4.2.1.
3.4.2.2.
3.4.2.3.
3.4.2.4.
3.4.2.5.
3.4.2.6.
3.4.2.7.
4. Complex Integration
4.1. The Fundamental Theorems
4.1.3. Line Integrals as Functions of Arcs
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Ahlfors Exercises
4.1.3.1.
4.1.3.2.
4.1.3.3.
4.1.3.4.
4.1.3.5.
4.1.3.6.
4.1.3.7.
4.1.3.8.
4.2. Cauchy’s Integral Formula
4.2.2. The Integral Formula
4.2.2.1.
4.2.2.2.
4.2.2.3.
4.2.3. Higher Derivatives
4.2.3.1.
4.2.3.2.
4.2.3.3.
4.2.3.4.
4.2.3.5.
4.2.3.6.
4.3. Local Properties of Analytical Functions
4.3.2. Zeros and Poles
4.3.2.1.
4.3.2.2.
4.3.2.3.
4.3.2.4.
4.3.2.5.
4.3.3. The Local Mapping
4.3.3.1.
4.3.3.2.
4.3.3.3.
4.3.3.4.
4.3.4. The Maximum Principle
4.3.4.1.
4.3.4.2.
4.3.4.3.
4.3.4.4.
4.3.4.5.
4.5. The Calculus of Residues
4.5.2. The Argument Principle
4.5.2.1.
4.5.2.2.
4.5.2.3.
4.5.3. Evaluation of Definite Integrals
4.5.3.1.
4.5.3.2.
4.5.3.3.
4.5.3.4.
4.6. Harmonic Functions
4.6.2. The Maximum Principle
4.6.2.1.
4.6.2.2.
5. Series and Product Developments
5.1. Power Series Expansions
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Ahlfors Exercises
5.1.1. Weierstrass’ Theorem
5.1.1.1.
5.1.1.2.
5.1.1.3.
5.1.1.4.
5.1.1.5.
5.1.2. The Taylor Series
5.1.2.1.
5.1.2.2.
5.1.2.3.
5.1.2.4.
5.1.2.5.
5.2. Partial Fractions and Factorization
5.2.1. Partial Fractions
5.2.1.1.
5.2.1.2.
5.2.1.3.
5.2.1.4.
5.2.2. Infinite Products
5.2.2.1.
5.2.2.2.
5.2.2.3.
5.2.2.4.
5.2.3. Canonical Products
5.2.3.1.
5.2.3.2.
5.2.3.3.
5.2.3.4.
5.2.3.5.
5.5. Normal Families
5.5.5. The Classical Definition
5.5.5.1.
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Ahlfors Exercises
1. Complex Numbers
1.1. The Algebra of Complex Numbers.
1.1.5. Inequalities.
1.1.5.1. Prove that
a−b
<1
1 − ab
if |a| < 1 and |b| < 1.
Solution: The inequality is equivalent to showing that
|a − b| < |1 − ab|.
By definition of the norm,
|a − b|2 = (a − b)(a − b) = (a − b)(a − b)
= |a|2 − (ab + ab) + |b|2
|1 − ab|2 = (1 − ab)(1 − ab) = (1 − ab)(1 − ab)
= 1 − (ab + ab) + |a|2 |b|2 .
Thus it suffices to show
|a|2 + |b|2 < 1 + |a|2 |b|2 .
Starting with |a| < 1, we have that |a|2 < 1 and thus |a|2 < (1 − |b|2 )/(1 − |b|2 ). since 0 < |b|2 < 1,
1−|b|2 is positive and therefore |a|2 (1−|b|2 ) < 1−|b|2 . Rearranging this gives us the above inequality.
Note that equality holds when either |a| = 1 or |b| = 1.
1.1.5.2. Prove Cauchy’s inequality by induction.
Solution: Cauchy’s inequality states that
n
X
2
aj bj
j=1
≤
n
X
j=1
|aj |2
n
X
|bj |2 .
j=1
The case n = 1 is always an equality since
1
X
2
aj bj
j=1
= |a1 b1 |2 = |a1 |2 |b1 |2 =
1
X
|aj |2
j=1
1
X
|bj |2 .
j=1
For the case n = 2, first note that
(|a1 ||b2 | − |a2 ||b1 |)2 = |a1 |2 |b2 |2 + |a2 |2 |b1 |2 − 2|a1 |2 |a2 |2 |b1 |2 |b2 |2 ≥ 0.
Adding |a1 |2 |b1 |2 + |a2 |2 |b2 |2 to each side and rearranging gives
(|a1 b1 | + |a2 b2 |)2 = |a1 |2 |b1 |2 + 2|a1 |2 |b1 |2 |a2 |2 |b2 |2 + |a2 |2 |b2 |2
≤ |a1 |2 |b1 |2 + |a1 |2 |b2 |2 + |a2 |2 |b1 |2 + |a2 |2 |b2 |2
= (|a1 |2 + |a2 |2 )(|b1 |2 + |b2 |2 )
Then, by the triangle inequality
2
X
2
aj bj
= |a1 b1 + a2 b2 |2 ≤ (|a1 b1 | + |a2 b2 |)2
j=1
≤ (|a1 |2 + |a2 |2 )(|b1 |2 + |b2 |2 ) =
2
X
j=1
|aj |2
2
X
j=1
|bj |2 .
Ahlfors Exercises
Now suppose it holds for n = k; we wish to show it holds for n = k + 1. By the triangle inequality
and direct application of the inductive hypothesis,
k+1
X
2
aj bj
=
k
X
2
aj bj + ak+1 bk+1
j=1
j=1
≤
k
X
2
aj bj + |ak+1 bk+1 |
v
v
u k
u k
uX
uX
≤ t
|aj |2 t
|bj |2 + |ak+1 bk+1 |
j=1
j=1
2
j=1



k
k
k+1
k+1
X
X
X
X
≤
|aj |2 + |ak+1 |2  
|bj |2 + |bk+1 |2  =
|aj |2
|bj |2
j=1
j=1
j=1
j=1
where we used the same inequality trick to pass from the second to third line as in the n = 2 case.
1.1.5.3. If |ai | < 1, λi ≥ 0 for i = 1, ..., n and λ1 + ... + λn = 1, show that
|λ1 a1 + λ2 a2 + ... + λn an | < 1.
Solution: By the general triangle inequality,
|λ1 a1 + ... + λn an | ≤
n
X
|λj ||aj | <
j=1
n
X
λj = 1.
j=1
1.1.5.4. Show that there are complex numbers z and a satisfying
|z − a| + |z + a| = 2|c|
if and only if |a| ≤ |c|. If this condition is fulfilled, what are the smallest and largest possible values
of |z|?
Solution: By the reverse triangle inequality, if such a and z exist then
2|c| = |z − a| + |z + a| ≥ |z − a − (z + a)| = 2|a|
so that |a| ≤ |c|. For the reverse implication, consider the two circles
|z − a| = |c|
|z + a| = |c|.
Clearly if z lies on both circles, it will satisfy |z − a| + |z + a| = 2|c|. Since the circles have the same
radius, finding a solution is easy. It will lie on the
p line perpendicular to the line connecting a and
−a and passing through 0. One such solution is |c|2 − |a|2 /|a|ia.
As a remark, the solution set to this is an ellipse (which may degenerate to a line if |a| = |c|).
The focii are at ±a. This gives insight into determining the maximum and minimum values of |z|,
they are just the major and minor radii.
1.2. The Geometric Representation of Complex Numbers.
1.2.1. Geometric Addition and Multiplication.
1.2.1.1. Find the symmetric points of a with respect to the lines which bisect the angles between
the coordinate axes.
Solution: Let L1 and L2 denote the real axis rotated counterclockwise by π/4 and 3π/4 respectively.
The symmetric point of a about L1 can be computed by rotating the entire frame by −π/4, reflecting
a across the real axis (i.e., taking the complex conjugate of this new point), then rotating back by
π/4. Notice that the magnitude does not change, so we just need to find the argument. An important property about the argument is that arg(z) = − arg(z). So, after the first rotation by −π/4
the argument is arg(a) − π/4. Taking the complex conjugate gives −(arg(a) − π/4) = π/4 − arg(a).
Finally, rotating back by π/4 gives a final argument of π/2 − arg(a). The same computation works
for L2 , just changing π/4 with 3π/4 everywhere. The final argument in this case is 3π/2 − arg(a).
Ahlfors Exercises
Call these symmetric points a1 and a2 . Then we have
arg(a1 ) = arg(π/2) − arg(a) = arg(i) + arg(a) = arg(ia)
arg(a2 ) = arg(3π/2) − arg(a) = arg(−i) + arg(a) = arg(−ia)
Since |ia| = |a| = |a1 | and similarly for a2 , it follows that a1 = ia and a2 = −ia.
1.2.1.2. Prove that the points a1 , a2 , a3 are vertices of an equilateral triangle if and only if a21 + a22 +
a23 = a1 a2 + a2 a3 + a3 a1 .
Solution: At a first glance, this problem has too much generality if we do not fix the triangle
ins some particular location. Let us show that the equality given is actually translation invariant
– this is a helpful trick for many problems, for example one often uses scale invariance to deduce
information about geometric inequalities.
Suppose that all the vertices a1 , a2 , a3 are translated by a single complex number z. Then,
(a1 − z)2 + (a2 − z)2 + (a3 − z)2 = a21 + a22 + a23 − a1 z − z(a1 − z) − a2 z − z(a2 − z) − a3 z − z(a3 − z)
= a1 a2 + a2 a3 + a3 a1 − a1 z − z(a1 − z) − a2 z − z(a2 − z) − a3 z − z(a3 − z)
= [a1 a2 − a1 z − z(a1 − z)] + [a2 a3 − a2 z − z(a2 − z)] + [a3 a1 − a3 z − z(a3 − z)]
= (a1 − z)(a2 − z) + (a2 − z)(a3 − z) + (a3 − z)(a1 − z).
So, the equality is translation invariant. Now we assume that a1 = 0, which will kill two of the terms
in the right hand side! Thus, we want to show that 0, a2 , a3 are vertices of an equilateral triangle if
and only if a22 + a23 = a2 a3 . We can reduce one more level of generality by rotating and dilating the
frame. That is, multiply each point by a2 /|a2 |2 (which we can assume is nonzero, since otherwise
a3 = 0 automatically). The resulting triangle has vertices at 0, 1, and ã3 = a3 a2 /|a2 |2 . We finally
wish to show that this forms an equilateral triangle
if and only if 1 + ã23 = ã3 . Note that we have an
√
equilateral triangle if and only if ã3 = 1/2 ± 3/2i. These are exactly the two roots of the quadratic
equation we have above.
1.2.1.3. Suppose that a and b are two vertices of a square. Find the two other vertices in all possible
cases.
Solution: Note that a square has four-fold symmetry, so that multiplication by i permutes the
vertices. Hence, all the vertices are a, ia, −a, −ia. One of these will be b.
1.2.1.4. Find the center and the radius of the circle which circumscribes the triangle with vertices
a1 , a2 , a3 . Express the result in symmetric form.
Solution: Let c denote the center of the circumscribed circle and r its radius. Since the circle
passes through all the vertices ai ,
|c|2 + |ai |2 − cai − ai c = (c − ai )(c − ai ) = |c − ai |2 = r2 .
Rewriting each of these gives
a1 c + a1 c + [r2 − |c|2 ] = |a1 |2
a2 c + a2 c + [r2 − |c|2 ] = |a2 |2
a3 c + a3 c + [r2 − |c|2 ] = |a2 |2
Now translate a1 and a2 by a3 so that we get a new triangle with vertices 0, a1 − a3 , a2 − a3 . The
circumcenter will now be c − a3 but the radius XXX
1.2.2. The Binomial Equation.
Ahlfors Exercises
1.2.2.1. Express cos 3ϕ, cos 4ϕ, and sin 5ϕ in terms of cos ϕ and sin ϕ.
Solution: By de Moivre’s formula,
cos 3ϕ + i sin 3ϕ = (cos ϕ + i sin ϕ)3 = ([cos2 ϕ − sin2 ϕ] + i[2 cos ϕ sin ϕ])(cos ϕ + i sin ϕ)
= [cos3 ϕ − 3 cos ϕ sin2 ϕ] + i[3 cos2 ϕ sin ϕ − sin3 ϕ]
cos 4ϕ + i sin 4ϕ = (cos ϕ + i sin ϕ)4 = (cos ϕ + i sin ϕ)3 (cos ϕ + i sin ϕ)
= [cos4 ϕ − 6 cos2 ϕ sin2 ϕ + sin4 ϕ] + i[4 cos3 ϕ sin ϕ − 4 cos ϕ sin3 ϕ]
cos 5ϕ + i sin 5ϕ = (cos ϕ + i sin ϕ)5 = (cos ϕ + i sin ϕ)4 (cos ϕ + i sin ϕ)
= [cos5 ϕ − 10 cos3 ϕ sin2 ϕ + 5 cos ϕ sin4 ϕ]
i[5 cos4 ϕ sin ϕ − 10 cos2 ϕ sin3 ϕ + sin5 ϕ].
Equating real and imaginary parts gives
cos 3ϕ = cos3 ϕ − 3 cos ϕ sin2 ϕ
cos 4ϕ = cos4 ϕ − 6 cos2 ϕ sin2 ϕ + sin4 ϕ
sin 5ϕ = 5 cos4 ϕ sin ϕ − 10 cos2 ϕ sin3 ϕ + sin5 ϕ.
There is an easy way to compute these in general. Consider the binomial expansion of (x + y)n .
Starting with xn , select every other term and add them together in order. Next, collect the remaining terms and add them together. In each sum x with cos ϕ and y with sin ϕ, then change the signs
so they alternate (starting with +). The first sum is cos nϕ while the latter is sin nϕ.
1.2.2.2. Simplify 1 + cos ϕ + cos 2ϕ + ... + cos nϕ and 1 + sin ϕ + sin 2ϕ + ... + sin nϕ.
Solution: Let z = cos ϕ + i sin ϕ and consider the sum
S = 1 + z + z 2 + ... + z n .
By the standard geometric series formula, this is
(1 − z n+1 )(1 − z)
(1 − cos(n + 1)ϕ − i sin(n + 1)ϕ)(1 − cos ϕ + i sin ϕ)
1 − z n+1
=
=
2
1−z
1 − 2 Re(z) + |z|
2 − 2 cos ϕ
(1 − cos ϕ)(1 − cos(n + 1)ϕ) + sin ϕ sin(n + 1)ϕ sin ϕ(1 − cos(n + 1)ϕ) − (1 − cos ϕ) sin(n + 1)ϕ
+
i
=
2(1 − cos ϕ)
2(1 − cos ϕ)
1 − cos(n + 1)ϕ sin ϕ sin(n + 1)ϕ
sin ϕ(1 − cos(n + 1)ϕ) sin(n + 1)ϕ
=
+
−
+
i
2
2(1 − cos ϕ)
2(1 − cos ϕ)
2
S=
The real and imaginary parts of these are the desired sums via de Moivre’s theorem.
1.2.2.3. Express the fifth and tenth roots of unity in algebraic form.
Solution: Observe that
1 + ω + ... + ω n−1 = 0
when ω is an n-th root of unity. Hence, XXX
1.2.2.4. If ω is given by
ω = cos
2π
2π
+ i sin
,
n
n
prove that
1 + ω h + ω 2h + ... + ω (n−1)h = 0
for any integer h which is not a multiple of n.
Ahlfors Exercises
Solution: It suffices to show that ω h is an n-th root of unity. We have that
arg(ω h ) “ = ” h arg(ω) =
2πh
n
Here we need the assumption that h is not a multiple of n to proceed. Because of this, we can take
2πh/n and reduce it modulo 2π. This will just be one of 2π/n, ..., 2(n − 1)π/n, which are the arguments of the n-th roots of unity not equal to 1. Since the modulus is multiplicative, |ω h | = |ω| = 1,
and ω h is an n-th root of unity.
1.2.2.5. What is the value of
1 − ω h + ω 2h − ... + (−1)n−1 ω (n−1)h ?
Solution: Notice the above sum is
S = 1 + (−ω)h + (−ω)2h + ... + (−ω)(n−1)h =
n−1
X
(−ω h )k .
k=0
If ω h = −1, then the above sum is clearly just n. Note that this can only occur when n is even,
since the n-th roots of unity are vertices of a regular n-gon with one vertex at 1. In particular,
ω h = −1 only when h = n/2 + nm = n(2m + 1)/2 for some integer m. In all other cases, we apply
the geometric series formula to obtain
S=
1 − (−1)n (ω n )h
1 − (−1)n
1 − (−ω h )n
=
=
1 − (−ω h )
1 + ωh
1 + ωh
since ω n = 1. Observe that if n is even, then we have S = 0. This suggests a nice geometric proof.
Indeed, in this case the regular polygon generated is symmetric with respect to reflection about the
origin. In particular, if ω is a root of unity then so too is −ω. Hence, we can apply the standard
result we know.
1.2.3. Analytic Geometry.
1.2.3.1. When does az + bz + c = 0 represent a line?
Solution: Suppose this does represent the equation of a line. Then for solutions u, v it should
be that u − v is always a multiple of a particular complex number, say λw. Then, u − v = λz. By
assumption, au + bu + c = 0 and similarly for v, so subtracting them gives
λ[aw + bw] = 0
for any λ. Thus, aw + bw = 0 for some w. This implies that |a| = |b|. Moreover, since u satisfies
the equation we have au + bu + c = 0. Multiply this by b and the original equation by a, then take
the difference. This yields
abu + bbu + bc − [aau + abu + ac] = 0
Two terms obviously cancel, and two more cancel since |a| = |b|. Hence, we are left with ac + bc = 0.
Clearly, we cannot have a = b = 0 since this implies c = 0, which may not be the case a priori. One
can also check that this is sufficient. XXX
1.2.3.2.
2. Complex Functions
2.1. Introduction to the Concept of Analytic Function.
2.1.2. Analytic Functions.
Ahlfors Exercises
2.1.2.1. If g(w) and f (z) are analytic functions, show that g(f (z)) is also analytic.
Solution: Write g(w) = ug (x, y) + ivg (x, y) and f (z) = uf (x, y) + ivf (x, y). Then ug , vg and uf , vf
all have continuous first partial derivatives and pairwise satisfy the Cauchy-Riemann equations. Let
h(z) = g(f (z)) so that
h(z) = uh (x, y) + ivh (x, y) = ug (uf (x, y), vf (x, y)) + ivg (uf (x, y), vf (x, y)).
By the classical chain rule for multivariable real functions,
∂uh
∂ug ∂uf
∂ug
=
+
∂x
∂x ∂x
∂y
∂uh
∂ug ∂uf
∂ug
=
+
∂y
∂x ∂y
∂y
∂vf
∂x
∂vf
∂y
∂vh
∂vg ∂uf
∂vg
=
+
∂x
∂x ∂x
∂y
∂vh
∂vg ∂uf
∂vg
=
+
∂y
∂x ∂y
∂y
∂vf
∂x
∂vf
∂y
which all exist and are continuous. Next we check they satisfy the Cauchy-Riemann equations
∂ug ∂uf
∂ug ∂vf
∂uh
=
+
∂x
∂x ∂x
∂y ∂x
∂ug ∂vf
∂ug ∂uf
=
−
∂x ∂y
∂y ∂y
∂vg ∂vf
∂vg ∂uf
∂vh
=
+
=
∂y ∂y
∂x ∂y
∂y
∂uh
∂ug ∂uf
∂ug ∂vf
=
+
∂y
∂x ∂y
∂y ∂y
∂ug ∂vf
∂ug ∂uf
=−
+
∂x ∂x
∂y ∂x
∂vg ∂vf
∂vg ∂uf
∂vh
=−
−
=−
∂y ∂x
∂x ∂x
∂x
where, the move from the first to second line in each we use the Cauchy-Riemann equations for f ,
then to move from the second to third line we use the Cauchy-Riemann equations for g. It follows
that h(z) = g(f (z)) is harmonic.
2.1.2.2. Verify Cauchy-Riemann’s equations for the functions z 2 and z 3
Solution: We must first compute the real and imaginary parts of these functions:
z 2 = (x + iy)2 = x2 − y 2 + 2xyi
z 3 = (x + iy)3 = (x2 − y 2 + 2xyi)(x + yi) = [x3 − 3xy 2 ] + [3x2 y − y 3 ]i
First, for z 2 , let u(x, y) = x2 − y 2 and v(x, y) = 2xy so that z 2 = u + iv. Then,
∂u
∂v
= 2x =
∂x
∂y
∂u
∂v
= −2y = −
∂y
∂x
so that the Cauchy-Riemann equations hold. Next let u(x, y) = x3 − 3xy 2 and v(x, y) = 3x2 y − y 3
so that z 3 = u + iv. Then,
∂u
∂v
= 3x2 − 3y 2 =
∂x
∂y
∂v
∂u
= −6xy = −
∂y
∂x
so that the Cauchy-Riemann equations also hold.
2.1.2.3. Find the most general harmonic polynomial of the form ax3 + bx2 y + cxy 2 + dy 3 . Determine
the conjugate harmonic function and the corresponding analytic function by integration and by the
formal method.
Ahlfors Exercises
Solution: Let u(x, y) = ax3 + bx2 y + cxy 2 + dy 3 . Then,
∂u
= 3ax2 + 2bxy + cy 2
∂x
∂2u
= 6ax + 2by
∂x2
∂u
= bx2 + 2cxy + 3dy 2
∂y
∂2u
= 2cx + 6dy
∂y 2
So, for u to be harmonic, it must be that
∂2u ∂2u
+ 2 = 2(3a + c)x + 2(b + 3d)y = 0
∂x2
∂y
evidently, this occurs if and only if c = −3a and b = −3d. Now, to find the conjugate harmonic
polynomial v it must be that ∂v/∂x = −∂u/∂y and ∂v/∂y = ∂u/∂x. Using the first relation, we
find that
Z
Z
v = − bx2 + 2cxy + 3dy 2 dx = 3dx2 + 6axy − 3dy 2 dx = dx3 + 3ax2 y − 3dxy 2 + ϕ(y)
where we have substituted the values of c and b found, and ϕ(y) is some function of y. Differentiating
this with respect to y gives
3ax2 − 6dxy + ϕ0 (y) =
∂v
∂u
=
= 3ax2 + 2bxy + cy 2 = 3ax2 − 6xy − 3ay 2 .
∂y
∂x
It follows that ϕ0 (y) = −3ay 2 so that ϕ(y) = −ay 3 + C. In total,
v = dx3 + 3ax2 y − 3dxy 2 − ay 3 + C
for a real constant C. The corresponding analytic function is
f (z) = [ax3 − 3dx2 y − 3axy 2 + dy 2 ] + [dx3 + 3ax2 y − 3dxy 2 − ay 3 + C]i.
The formal method states that we can construct f (z) by
f (z) = 2u(z/2, z/(2i)) − u(0, 0) + Ci
where C is a real constant. Reducing this gives
z 2 z z z 2
z 3 z 3
f (z) = 2 a
− 3d
− 3a
+d
+ Ci
2
2
2i
2
2i
2i
z z 2
z 3
1
az 3 − 3dz 2
− 3az
+d
=
+ Ci
4
i
i
i
= (a + di) z 3 + Ci.
Expanding out the last expression recovers our formula for f (z) deduced from integration.
2.1.2.4. Show that an analytic function cannot have a constant absolute value without reducing to
a constant.
Solution: First, if |f (z)| = 0 then f (z) = 0, so there is nothing to prove. Otherwise, let |f (z)| = c > 0.
We have that
|f (z)|2
c2
f (z) =
=
f (z)
f (z)
so that f (z) is analytic whenever f (z) is nonzero. But, by assumption f (z) is never zero so that
f (z) is analytic. Writing f (z) = u + iv, the Cauchy-Riemann equations applied to f and f imply
that
∂u
∂v
∂u
∂v
=
=−
∂x
∂y
∂x
∂y
∂u
∂v
∂u
∂v
=−
=
∂y
∂x
∂y
∂x
Ahlfors Exercises
Note the sign change that occurs in the Cauchy-Riemann equations for f . Together, these imply
that all the partial derivatives are zero, and thus u and v are constant functions.
2.1.2.5. Prove rigorously that the functions f (z) and f (z) are simultaneously analytic.
Solution: Suppose first that f (z) is analytic. Let g(z) = f (z). Let uf , vf and ug , vg denote the
real and imaginary parts of f and g respectively. Then, these are related by
ug (x, y) = uf (x, −y)
vg (x, y) = −vf (x, −y).
Since f is analytic, uf and vf have continuous partial derivatives – so too do ug , vg then. Next we
verify the Cauchy-Riemann equations:
∂uf
∂ug
=
∂x
∂x
∂ug
∂uf
=−
∂y
∂y
∂vg
∂vf
=−
∂x
∂x
∂vg
∂vf
=
∂y
∂y
Since f is analytic, uf and vf satisfy the Cauchy-Riemann equations. Thus,
∂ug
∂uf
∂vf
∂vg
=
=
=
∂x
∂x
∂y
∂y
∂uf
∂vf
∂vg
∂ug
=−
=
=−
∂y
∂y
∂x
∂y
Thus, ug and vg satisfy the Cauchy-Riemann equations. It follows that g(z) = f (z) is analytic.
Finally, observe that g(z) = f (z) so that the above proof works to show the converse, with the roles
of f and g reversed.
2.1.2.6. Prove that the functions u(z) and u(z) are simultaneously harmonic.
Solution: Let f (z) = u(z) and g(z) = f (z) = u(z), where u : C → R. Then apply Exercise
2.1.2.5, we see that f and g are simultaneously analytic. This implies that their real parts – u(z)
and u(z) – are harmonic.
2.1.2.7. Show that a harmonic function satisfies the formal differential equation
∂2u
= 0.
∂z∂z
Solution: Recall if f is analytic with f = u + iv, then we formally have
1 ∂f
∂f
∂f
1 ∂f
∂f
∂f
=
−i
=
+i
.
∂z
2 ∂x
∂y
∂z
2 ∂x
∂y
Now consider the function f = u, which has no imaginary part. Then,
∂2u
∂ ∂u
1 ∂ ∂u
∂u
1 ∂ ∂u
∂u
∂ ∂u
∂u
=
=
+i
=
+i
−i
+i
∂z∂z
∂z ∂z
2 ∂z ∂x
∂y
4 ∂x ∂x
∂y
∂y ∂x
∂y
1 ∂2u ∂2u
∂2u
∂2u
=
+ 2 +i
−i
=0
4 ∂x2
∂y
∂x∂y
∂y∂x
We use the fact that u is harmonic and symmetry of the mixed partials to conclude.
2.1.4. Rational Functions.
2.1.4.1. Use the method of the text to develop
z4
z3 − 1
in partial fractions.
and
1
z(z + 1)2 (z + 2)3
Ahlfors Exercises
Solution: XXX
2.1.4.2. If Q is a polynomial with distinct roots α1 , ..., αn , and if P is a polynomial of degree < n,
show that
n
P (z) X
P (αk )
=
.
0
Q(z)
Q (αk )(z − αk )
k=1
Solution: I’m fairly certain we should assume that Q has simple roots α1 , ..., αn , otherwise the right
hand side makes little sense. For example, Q0 (αk ) = 0 at a root with order greater than 1, so
the right hand-side is ill defined in this case. We thus assume that Q(z) has degree n. Also, we
may assume the leading coefficient is 1, by just multiplying both sides of the above equation by the
leading coefficient (were there one). Hence, Q takes the form
n
Y
Q(z) =
(z − αj )
j=1
and its derivative
Q0 (z) =
n Y
X
(z − αj ).
l=1 j6=l
So, at some zero αk we have
Y
Q0 (αk ) =
(αk − αj ).
j6=k
Now consider the quantity
Qn
Q
Q(z)
j=1 (z − αj )
j6=k (z − αj )
Q
=
=Q
Q̃k (z) := 0
Q (αk )(z − αk )
(z − αk ) j6=k (αk − αj )
j6=k (αk − αj )
for any k = 1, ..., n. It is clear that Q̃k (αl ) = δkl .
Next, observe that we can prove the desired equality by instead showing
P (z) =
n
X
k=1
n
X
P (αk )Q(z)
P (αk )Q̃k (z).
=
Q0 (αk )(z − αk )
k=1
Moreover, a polynomial of degree n is determined by its values at n distinct points. Since deg(P ) < n,
we just need to show the above equality at less than n points. In particular, we can show it at z = αk
for each k. For fixed αl we have
n
X
k=1
P (αk )Q̃k (αl ) =
n
X
P (αk )δkl = P (αl ).
k=1
2.1.4.3. What is the general form of a rational function which has absolute value 1 on the circle
|z| = 1? In particular, how are the zeros and poles related to each other?
Solution: Consider the quantity
R(z)R(1/z).
Observe that if |z| = 1, then zz = 1 and so z = 1/z. Hence, on |z| = 1
R(z)R(1/z) = R(z)R(z) = |R(z)|2 = 1.
It follows that
R(z) = 1/R(1/z)
on |z| = 1. But, both sides are rational functions. If two rational functions agree on an infinite set,
then they must be the same everywhere (take the difference, which has infinitely many zeros – but,
as the book shows, every rational function has finitely many zeros). So the above identity actually
Ahlfors Exercises
holds everywhere. This tells us that if w is a zero (resp. zero) of order k of R, then it has a pole
(resp. zero) of order k at 1/w = w/|w|2 . We thus write R(z) as
kj
n Y
z − αj
R(z) = z0 z k0
z − 1/αj
j=1
where the αj are the distinct zeros and poles of R(z) inside the unit disc, with order kj . The leading
coefficient z0 is such that |z0 | = 1 since R(z) = 1/R(1/z).
2.1.4.4. XXX
Solution: XXX
2.1.4.5. If R(z) is a rational function of order n, how large and how small can the order of R0 (z) be?
Solution: We need only count the poles. Note that if R(z) has a pole at w ∈ C of order k, then R0 (z)
has a pole at w ∈ C of order k + 1. We must also account for poles at infinity. Note that the order
of a pole at infinity is determined by computing the order the pole at z = 0 of R1 (z) := R(1/z). By
the chain rule,
d
1
R1 (z) = − 2 R0 (1/z)
dz
z
so that we have
d
R0 (1/z) = −z 2 R1 (z).
dz
To determine the order of a pole at infinity of R0 (z), we look at the order of the pole at z = 0 of
R0 (1/z). If n = deg(P ) > deg(Q) = m then R1 (z) has a pole at 0 of degree k = n − m. Applying
the above analysis for finite poles, d/dzR1 (z) has a pole at z = 0 of order k + 1. The additional factor of z 2 in front decreases the order of a pole by 2, so that R0 (1/z) has a pole of order k − 1 at z = 0.
More explicitly,
p0 + p1 z + ... + pn z n
R(z) =
q0 + q1 z + ... + qm z m
(p1 + 2p2 z + ... + npn z n−1 )(q0 + ... + qm z m ) − (q1 + 2q2 z + ... + mqm z m−1 )(p0 + ... + pn z n )
R0 (z) =
(q0 + q1 z + ... + qm z m )2
[p1 q0 − q1 p0 ] + 2[p2 q0 − p0 q2 ]z + ... + (n − m)[pn qm ]z n+m−1
=
(q0 + q1 z + ... + qm z m )2
2m
z
[p1 q0 − q1 p0 ]z n+m−1 + 2[p2 q0 − p0 q2 ]z n+m + ... + (n − m)[pn qm ]
R0 (1/z) = n+m−1
z
(q0 q m + q1 z m−1 + ... + qm )2
so that R0 (1/z) has a pole at z = 0 of order n − m − 1 = k − 1.
To determine the order of R0 , we simply need to add up the orders of all the poles. Thus,
X
deg(R0 ) =
[kj + 1] + [k∞ − 1]
j
where kj isPthe order of pole wj of R(z) and k∞ ≥ 1 is the order of the infinite pole of R(z). Since
deg(R) = j kj + k∞ , we have
X
deg(R0 ) = deg(R) +
1 − 1 = deg(R) + d − 1
j
where d is the number of distinct finite poles of R(z). If all the poles are distinct, then deg(R0 ) =
2 deg(R) and if all the poles are infinite, then deg(R0 ) = deg(R) − 1. XXX
2.2. Elementary Theory of Power Series.
2.2.3. Uniform Convergence.
Ahlfors Exercises
2.2.3.1. Prove that a convergent sequence is bounded.
Solution: Let {zn }∞
n=1 be a convergent sequence of complex numbers such that zn → z ∈ C. Let
> 0, then there exists an N ∈ N such that if n ≥ N then |zn − z| < . Consequently,
|zn | = |zn − z + z| ≤ |zn − z| + |z| < |z| + for n ≥ N . For concreteness, choose = 1. Now set M = max{|z1 |, ..., |zN |, |z| + 1}. By the above,
we see that |zn | < M for all n.
2.2.3.2. If limn→∞ zn = A, prove that
lim
n→∞
1
(z1 + z2 + ... + zn ) = A.
n
Solution: It suffices to prove the real case, since Re(·) and Im(·) are continuous functions. So,
consider a convergent sequence of real numbers an → a. For > 0 there exists an N such that if
n ≥ N , |an − a| < . Expanding this, we have a − < an < a + . Define sn := 1/n(a1 + ... + an ).
Then, if n = N + k for k ≥ 0,
sn =
1
1
(a1 + ... + aN −1 + aN + ... + aN +k ) < (a1 + ... + aN −1 + (n − N + 1)(a + )) ,
n
n
and we get a similar lower bound. Since
a1 + ... + aN −1
→0
n
(n − N + 1)(a ± )
→a±
n
as n → ∞, we have that
a − < sn < a + for n ≥ N . Of course, this says that for any > 0 there exists an N such that if n ≥ N then
|sn − a| < . So, sn → a.
2.2.3.3. Show that the sum of an absolutely convergent series does not change if the terms are
rearranged.
P∞
P∞
Solution: Let n=1 zn be an absolutely convergent series and n=1P
wn any rearrangement (that is,
there exists a bijective map σ : N → N such that wn = zσ(n) ). Since n zn is absolutely convergent,
the
itself converges to some complex number z. Let sn and tn be the n-th partial sums of
P series P
zn and
wn respectively. Then,
|tn − z| ≤ |tn − sn | + |sn − z|.
The second term evidently tends to zero. We now wish to estimate |tn − sn |. Since tn is a partial
sum of the rearrangement, some of the terms will cancel with those in sn . Let
S(n) = {1, ..., n}∆{σ(1), ..., σ(n)}
and set K = max S(n) and k = min S(n). It follows that the remaining terms in tn − sn have indices
between k and K. Thus,
|tn − sn | ≤
K
X
j=k
|zj | ≤
∞
X
|zj |.
j=k
As n increases, so too does k. By absolute convergence of
zero. Hence also |tn − sn | does.
P
zj , we see that the remainders tend to
Ahlfors Exercises
2.2.3.4. Discuss completely the convergence and uniform convergence of the sequence {nz n }∞
n=1 .
Solution: First if |z| < 1, then we have |nz n | = n|z|n → 0, and so nz n → 0. If |z| ≥ 1, then
we have |nz n | = n|z|n ≥ n → ∞, and thus nz n → ∞. Defining fn (z) = nz n , if the convergence
were uniform then the limit would be continuous. But the limit function (in the extended complex
numbers) is 0 inside the unit disc and ∞ outside it. This is discontinuous, and therefore the limit is
not uniform. We can ask, however, if the limit is uniform in |z| < 1. First, consider convergence on
the set |z| < r where |r| < 1. Then, |nz n | < nrn , where by continuity of xrx and since |r| < 1, it is
clear that the convergence is uniform. On the other hand, if we try to obtain uniform convergence
for |z| < 1, we see it is impossible. For example, along the sequence zn = n−1/n we actually have
fn (zn ) = 1. One can show (for example, via L’hopitals rule) that zn → 1. Yet, uniform convergence
implies that fn (zn ) → f (z).
The intuition for this problem is that multiplication of complex numbers amounts to multiplying their distance to the origin and adding their arguments. So, z n is rotates around the origin as
n increases, and either shrinks (|z| < 1) or grows (|z| > 1). This causes a spiraling effect.
2.2.3.5. Discuss the uniform convergence of the series
∞
X
x
n(1
+
nx2 )
n=1
for real values of x.
P
Solution: Define fn (x) = x/(n(1+nx2 )), so that the series above is n fn (x). Note that limx→∞ fn (x) =
0 for any n. Next,
1 − nx2
n(1 + nx2 ) − nx(2nx)
=
.
fn0 (x) =
n2 (1 + nx2 )2
n(1 + nx2 )2
√
So, fn0 (x) = 0 when 1 − nx2 = 0, or when x = ±1/ n. Substituting these into fn (x) gives
√
√
1
−fn (−1/ n) = fn (1/ n) = √ .
2n n
√
−3/2
Combining this with
,
P∞ the fact that each fn tends to zero, we have |fn | ≤ 1/(2n n). If an = 1/2n
then the series n=1 an converges. Hence by the Weierstrass M -test, the above series converges
uniformly.
2.2.3.6. XXX
Solution:
2.2.4. Power Series.
2.2.4.1. Expand (1 − z)−m , m a positive integer, in powers of z.
Solution: We know that
f 00 (0) 2
z + ...
2!
−m
so we just need to compute the derivatives of f (z) = (1 − z)
and evaluate them at zero. Observe
that
f (z) = f (0) + f 0 (0)z +
f (z) = (1 − z)−m
f 0 (z) = m(1 − z)−m−1
f 00 (z) = m(m + 1)(1 − z)−m−2 ...
By repeated differentiation, we see that in general
f (n) (z) =
(m + n − 1)!
(1 − z)−m−n .
(m − 1)!
Ahlfors Exercises
Thus,
f (n) (0) =
(m + n − 1)!
.
(m − 1)!
The power series expansion is then
f (z) =
∞
∞ X
(m + n − 1)! n X m + n − 1 n
z =
z .
n
(m − 1)!n!
n=0
n=0
Note the similarity to the binomial theorem.
2.2.4.2. Expand (2z + 3)/(z + 1) in powers of z − 1. What is the radius of convergence?
Solution: We can write
1
1
1/2
2z + 3
=2+
=2+
=2+
z+1
z+1
2 − (−(z − 1))
1 − (−(z − 1)/2)
n
∞
∞ X
X
(−1)n (z − 1)n
1
(z − 1)
=2+
=2+
−
2 n=0
2
2n+1
n=0
By the geometric series test, it follows that this series converges when
− (x − 1)
<1
2
so that the radius of convergence is R = 2.
2.2.4.3. Find the radius of convergence of the following power series:
X
X zn X
X 2
X
np z n ,
,
n!z n ,
qn zn ,
z n!
n!
where |q| < 1.
Solution: Recall Hadamard’s formula, which states that
p
1/R = lim sup n |an |
n→∞
P
defines the radius of convergence for the power series an z n . Note that every power series converges
for at least one z, namely z = 0; This occurs if R = 0 and 1/R = ∞. On the other hand,pa power
series can converge everywhere, in which case R = ∞ and 1/R = 0. For the latter, since n | · | is a
continuous function, if an → 0 then 1/R = 0. We now treat each power series separately.
P
i) ( np z n ). Since an = np , we have
p
p
√
n
n
lim sup |an | = lim sup n = 1p = 1.
n→∞
n→∞
So,
P R = 1.
ii) ( z n /n!). Since an = 1/n!, which converges to 0, we have R = ∞. √
P
iii) ( n!z n ). Here an = n!, and√by Stirling’s approximation n! ≈ 2πn(n/e)n . So, by
L’hopitals rule one can show n n! ≈ (2πn)1/(2n) n/e → ∞. It follows that R = 0, so
the power series converges only at z = 0.
P 2
2
iv) ( q n z n ). Now an = q n , where |q| < 1. It is clear to see that |an |1/n = |q|n → 0, so that
RP= ∞.
v) ( z n! ). This one ispslightly odd, since ak = 1 when k = n! for some n, and is zero
otherwise. Let bk = k |ak |, then bk = 1 when k = n! and zero else. Since n! grows, for any
tail {k, k + 1, ...} one of the indices will be a factorial. Hence, the limsup is always 1, and
R = 1.
Ahlfors Exercises
2.2.4.4.
P 2 n If
an z ?
P
an z n has radius of convergence R, what is the radius of convergence of
P
an z 2n ? of
P
P
Solution: Let bn = an/2 if n is even and zero otherwise. Then
an z 2n =
bn z n . So, to find
the radius of convergence we can apply Hadamard’s formula. Note that if n = 2k then,
p
1/2
p
p
n
|bn | = 2k |ak | = k |ak |
.
By definition,
1/R1 = lim sup
1/2
1/2 p
p
p
n
= 1/R1/2
|bn | = lim sup k |ak |
= lim sup k |ak |
n→∞
k→∞
k→∞
√
since x is a monotone continuous function and we can discard the the remaining terms of
since they are just zero.
Now let cn = a2n so that
|bn |
a2n z n . By Hadamard’s formula,
2
p
p
n
n
1/R2 = lim sup |cn | = lim sup |an | = 1/R2
P
cn z n =
p
n
P
n→∞
n→∞
using the same logic as above.
2.2.4.5. If f (z) =
P
an z n , what is
P
n3 an z n ?
Solution: It is clear to see that
0
zf (z) = z
∞
X
nan z
n−1
=
∞
X
nan z n
n=0
n=0
by applying Abel’s theorem for the differentiation of power series. We inductively see that
∞
X
d
df
d
nk a n z n = z
z ... z
dz
dz
dz
n=0
where there are k many d/dz. Hence,
∞
X
d
n an z = z
dz
n=0
3
n
d
d
0
z (zf (z)) = z
zf 0 (z) + z 2 f 00 (z)
dz
dz
= zf 0 (z) + 3z 2 f 00 (z) + z 3 f 000 (z)
P
P
2.2.4.6. If Pan z n and
bn z n have radii of convergence R1 and R2 , show that the radius of conn
vergence of
an bn z is at least R1 R2 .
Solution: We first show that if xn and yn are sequences of nonnegative real numbers then
lim sup xn yn ≤ lim sup xn lim sup yn
n→∞
n→∞
n→∞
Observe that
sup{xk yk } ≤ sup sup{xj yk } ≤ sup{xj sup{yk }} = sup{xj } sup{yk }
k≥n
j≥n k≥n
j≥n
k≥n
j≥n
k≥n
The first inequality comes from taking the sup over a larger set, while the second comes from considering xj yk for fixed j and taking the sup over k ≥ n. Taking the limit as
n → ∞ on bothpsides
p
n
n
shows the desired
inequality.
To
solve
the
problem,
apply
it
with
x
=
|a
|bn |,
n
n | and yn =
p
p
p
n
n
n
noting that |an bn | = |an | |bn | = xn yn .
Ahlfors Exercises
2.2.4.7. If limn→∞ |an |/|an+1 | = R, prove that
P
an z n has radius of convergence R.
Solution: We show that if bn is a sequence of positive real numbers then
p
p
bn+1
bn+1
lim inf
≤ lim inf n bn ≤ lim sup n bn ≤ lim sup
.
n→∞
n→∞
bn
bn
n→∞
n→∞
Set l, m, M, L as the above limits respectively, so we wish to show l ≤ m ≤ M ≤ L. Let > 0, then
there exists an N such that if n ≥ N then supn≥N bn+1 /bn < L + . In particular, for all n ≥ N we
have bn+1 < (L + )bn . If n = N + k with k ≥ 1 then
bn < (L + )bn−1 < (L + )2 bn−2 < ... < (L + )k bn−k = (L + )n−N bN .
Recall that c1/n → 1 for c > 0. Hence, taking n-th roots and the limsup of both sides, we get
1/n
p
bN
= L + .
lim sup n bn ≤ (L + ) lim sup
(L + )N
n→∞
n→∞
This holds for all > 0, so that M ≤ L. Showing l ≤ m is similar, except using l − . The inner
inequality is trivial.
Now, if limn→∞ |an |/|an+1 | = R then limn→∞ |an+1 |/|an | = 1/R. Setting bn = |an | and using
the fact that, for convergent sequences, the limsup and liminf agree, we have
p
p
1
1
≤ lim inf n |an | ≤ lim sup n |an | ≤ .
n→∞
R
R
n→∞
We conclude by applying Hadamard’s formula.
2.2.4.8. For what values of z is
n
∞ X
z
1+z
n=0
convergent?
Solution: As a geometric series, it will converge when |z/(1 + z)| < 1, or equivalently |z| < |1 + z|.
Squaring both sides does not affect the solution set to this inequality, so we get
zz = |z|2 < |1 + z|2 = (1 + z)(1 + z) = 1 + z + z + zz.
Recalling that z + z = 2 Re(z), this is equivalent to
0 < 1 + 2 Re(z)
⇔
Re(z) > −1/2.
2.2.4.9. Same question for
∞
X
zn
.
1 + z 2n
n=0
Solution: Suppose that |z| = 1. Then, |1 + z 2n | ≤ 1 + |z|2n = 2. Consequently,
1
1
zn
≥ .
=
1 + z 2n
|1 + z 2n |
2
So, the terms never go to zero, and thus the series diverges. Also note that
∞
X
∞
X
zn
1
=
2n
−n
1+z
z + zn
n=0
n=0
so that the behavior for |z| > 1 is the same as that of |z| < 1. Now, if |z| > 1 then for any r > 1
there exists an N such that if n ≥ N ,
|1 + z 2n | ≥ |1 − |z|2n | ≥
|z|2n
.
r
Ahlfors Exercises
The first inequality is just the reverse triangle inequality. The latter comes from the limit
lim sup
n→∞
|1 − |z|2n |
=1
|z|2n
and the fact that r > 1. Next, applying the root test
s
s
√
n
n
zn
r
1
n
n |z| r
≤
lim
sup
lim sup
=
lim
sup
=
< 1.
2n
2n
1+z
|z|
|z|
n→∞
n→∞
n→∞ |z|
Since the limit is less than one, the series converges.
2.3. The Exponential and Trigonometric Functions.
2.3.2. The Trigonometric Functions.
2.3.2.1. Find the values of sin i, cos i, tan(1 + i).
Solution: We recall that
sin z =
eiz − e−iz
2i
cos z =
eiz + e−iz
2
So that direct substitution of z = i gives
2
2
ei − e−i
1/e − e
(e2 − 1)i
sin i =
=
=
2i
2i
2e
i2
−i2
1/e + e
1 + e2
e +e
=
=
cos i =
2
2
2e
By the addition formulas, we have
(1 + e2 ) sin(1) + (e2 − 1) cos(1)i
2e
(1 + e2 ) cos(1) + (1 − e2 ) sin(1)i
cos(1 + i) = cos(1) cos(i) − sin(1) sin(i) =
2e
sin(1 + i) = cos(1) sin(i) + sin(1) cos(i) =
Hence,
sin(1 + i)
(1 + e2 ) sin(1) + (e2 − 1) cos(1)i
=
cos(1 + i)
(1 + e2 ) cos(1) + (1 − e2 ) sin(1)i
2
((1 + e ) sin(1) + (e2 − 1) cos(1)i)((1 + e2 ) cos(1) + (e2 − 1) sin(1)i)
=
(1 + 2e2 + e4 ) cos(1)2 + (1 − 2e2 + e4 ) sin(1)2
2
4e sin(1) cos(1) + (e4 − 1)i
=
1 + e4 + 2e2 cos(2)
tan(1 + i) =
If we adopt the definitions of cosh(z) and sinh(z) in the next problem, then observe that
e2 + 1/e2
e4 + 1
=
2
2e2
e2 − 1/e2
e4 − 1
sinh(2) =
=
2
2e2
cosh(2) =
Thus, we can write tan(1 + i) as
tan(1 + i) =
2 sin(1) cos(1) + (e4 − 1)/(2e2 )i
sin(2) + sinh(2)i
=
.
4
2
(1 + e )/(2e ) + cos(2)
cos(2) + cosh(2)
Ahlfors Exercises
2.3.2.2. The hyperbolic cosine and sine are defined by cosh z = (ez + e−z )/2, sinh z = (ez − e−z )/2.
Express them through cos iz, sin iz. Derive the addition formulas, and formulas for cosh 2z, sinh 2z.
Solution: It is easy to see that
cos(iz) =
sin(iz) =
2
z
2
z
ei
ei
2
+ e−i z
ez + e−z
=
= cosh(z)
2
2
2
− e−i z
(ez − e−z )i
=
= sinh(z)i.
2i
2
The addition formulas are thus
cosh(a + b) = cos(ai + bi) = cos(ai) cos(bi) − sin(ai) sin(bi)
= cosh(a) cosh(b) − sinh(a) sinh(b)i2 = cosh(a) cosh(b) + sinh(a) sinh(b)
sinh(a + b) = −i sin(ai + bi) = −i[sin(ai) cos(bi) + sin(bi) cos(ai)]
= [−i sin(ai)] cos(bi) + [−i sin(bi) cos(ai)] = sinh(a) cosh(b) + sinh(b) cosh(a)
and from these we derive the double angle formulae
cosh(2z) = cosh2 (z) + sinh2 (z)
sinh(2z) = 2 sinh(z) cosh(z)
2.3.2.3. Use the addition formulas to separate cos(x + iy), sin(x + iy) in real and imaginary parts.
Solution: We essentially did this in Exercise 2.3.2.1, but without the aid of the hyperbolic trig
functions. Direct application of the addition formulae give
cos(x + iy) = cos(x) cos(iy) − sin(x) sin(iy) = cos(x) cos(iy) − i[−i sin(iy)] sin(x)
= cos(x) cosh(y) − sin(x) sinh(y)i
sin(x + iy) = sin(x) cos(iy) + sin(iy) cos(x) = sin(x) cos(iy) + i[−i sin(iy)] cos(x)
= sin(x) cosh(y) + cos(x) sinh(y)i
2.3.2.4. Show that
| cos z|2 = sinh2 y + cos2 x = cosh2 y − sin2 x =
1
(cosh 2y + cos 2x)
2
| sin z|2 = sinh2 y + sin2 x = cosh2 y − cos2 x =
1
(cosh 2y − cos 2x).
2
and
Solution: If z = x + iy then by Exercise 2.3.2.3 we have
| cos z|2 = Re(cos(z))2 + Im(cos(z))2 = cos2 (x) cosh2 (y) + sin2 (x) sinh2 (y)
= cos2 (x)(1 + sinh2 (y)) + (1 − cos2 (x)) sinh2 (y)
= cos2 (x) + sinh2 (y)
where we have made use of the fact that cosh2 (y) − sinh2 (y) = 1 and cos2 (x) + sin2 (x) = 1. To
convert to the second formulation, we simply apply these identities again
| cos z|2 = [1 − sin2 (x)] + [cosh2 (y) − 1] = cosh2 (y) − sin2 (x).
To get the third formulation add both of these identities and apply the double angle formulae.
For sin z we have
| sin z|2 = sin2 (x) cosh2 (y) + cos2 (x) sinh2 (y)
= sin2 (x)(1 + sinh2 (y)) + (1 − sin2 (x)) sinh2 (y) = sin2 (x) + sinh2 (y).
Deriving the other identities proceeds in virtually the same way as for cos z. Note that this aligns
with Euler’s identity eiz = cos z + i sin z and the modulus of the exponential, |ex+iy | = ex .
2.3.4. The Logarithm.
Ahlfors Exercises
2.3.4.1. XXX
Solution: XXX
2.3.4.2. XXX
Solution:
2.3.4.3. Find the value of ez for z = −πi/2, 3πi/4, 2πi/3.
Solution: We apply Euler’s identity eiz = cos z + i sin z.
−π
−π
+ i sin
= −i
2
2
√
√
3π
2
2
3π
= cos
+ i sin
=−
+
i
4
4
2 √ 2
3
2π
1
2π
+ i sin
=− +
i
= cos
3
3
2
2
e−π/2i = cos
e3π/4i
e2π/3i
2.3.4.4. For what values of z is ez equal to 2, −1, i, −i/2, −1 − i, 1 + 2i?
Solution: For each w in the list above, we want to find
log w = log |w| + i arg w
where arg w ∈ (−π, π]. Then, every value can be obtained from adding integer multiples of 2πi. One
can easily find that
arg −1 = π, arg i = π/2,
arg −i/2 = arg −i = −π/2,
√
√ !
2
2
arg(−1 − i) = arg −
−
i = −3π/4, arg(1 + 2i) = arctan(2)
2
2
arg 2 = 0,
since arguments are not affected by dilations. Hence, all the values of z such that ez is equal to the
above listed numbers are
z(2) = log 2 + 2πni = log 2 + 2πni
z(−1) = log(−1) + 2πni = (2n + 1)πi
z(i) = log(i) + 2πni = (4n + 1)π/2i
z(−i/2) = log(−i/2) + 2πni = − log 2 + (4n − 1)π/2i
z(−1 − i) = log(−1 − i) + 2πni = 1/2 log 2 + (8n − 3)π/4i
z(1 + 2i) = log(1 + 2i) + 2πni = 1/2 log 5 + (2πn + arctan(2))i
2.3.4.5. Find the real and imaginary parts of exp(ez ).
Solution: Write z = a + bi so that
ez = ea ebi = ea (cos b + i sin b).
Then, applying the exponential once more gives
exp(ez ) = ee
a
(cos b+i sin b)
= ee
a
cos b iea sin b
e
= ee
a
cos b
(cos(ea sin b) + i sin(ea sin b)).
Hence,
Re(exp(ez )) = ee
a
cos b
cos(ea sin b)
Im(exp(ez )) = ee
a
cos b
sin(ea sin b)
Ahlfors Exercises
2.3.4.6. Determine all values of 2i , ii , (−1)2 i.
Solution: We recall that ab := eb log a . We simply apply this definition with Euler’s identity several
times. Thus,
2i = ei log 2 = ei(log 2+2πni) = e−2πb ei log 2 = e−2πn (cos log 2 + i sin log 2)
ii = ei log i = ei(2πni+π/2i) = e−π/2 e−2πn
(−1)2i = e2i log(−1) = e2i(2n+1)πi = e−2(2n+1)π
for n ∈ Z.
2.3.4.7. Determine the real and imaginary parts of z z .
Solution: If z = a + bi we have that
z z = ez log z = ez(log |z|+i arg z) = ea log |z|−b arg z+i(arg z+b log |z|)
= ea log |z|−b arg z ei(arg z+b log |z|) = ea log |z|−b arg z (cos(arg z + b log |z|) + i sin(arg z + b log |z|))
Thus, the real and imaginary parts are
Re(z z ) = ea log |z|−b arg z cos(arg z + b log |z|)
Im(z z ) = ea log |z|−b arg z sin(arg z + b log |z|)
Of course, this assumes we use the principal values of arg z.
2.3.4.8. Express arctan w in terms of the logarithm.
Solution: We have that
w = tan z = −i
e2iz − 1
eiz − e−iz
= −i 2iz
.
iz
−iz
e +e
e +1
Rearranging this gives
e2iz =
i−w
w+i
so that taking the logarithm yields
1
arctan w = z =
log
2i
i−w
i+w
.
3. Analytic Functions as Mappings
3.1. Elementary Point Set Topology.
3.1.2. Metric Spaces.
3.1.2.1. If S is a metric space with distance function d(x, y), show that S with the distance function
δ(x, y) = d(x, y)/[1 + d(x, y)] is also a metric space. The latter space is bounded in the sense that
all distances lie under a fixed bound.
Solution: That δ(x, y) ≥ 0, δ(x, y) = 0 if and only if x = y, and δ(x, y) = δ(y, x) follow immediately from the fact that d(x, y) is a metric. Next, observe that
1
δ(x, y) = 1 −
.
1 + d(x, y)
Since d(x, y) ≤ d(x, z) + d(z, y), it follows that
1
1
δ(x, y) = 1 −
≤1−
.
1 + d(x, y)
1 + d(x, z) + d(z, y)
Next, one can show that if x, y ≥ 0 then
1
1
1
+
≤1+
1+x 1+y
1+x+y
Ahlfors Exercises
Applying this with d(x, z) and d(y, z) gives
1
1
1
≤1−
+1−
= δ(x, z) + δ(z, y).
1−
1 + d(x, z) + d(z, y)
1 + d(x, y)
1 + d(z, y)
One can also appeal to monotonicity of the function x 7→ x/(1 + x).
3.1.2.2. Suppose that there are given two distance functions d(x, y) and d1 (x, y) on the same space
S. They are said to be equivalent if they determine the same open sets. Show that d and d1 are
equivalent if to every > 0 there exists a δ > 0 such that d(x, y) < δ implies d1 (x, y) < , and vice
versa. Verify that this condition is fulfilled in the preceding exercise.
Solution: Recall that a set U is open in a topology τ induced by a metric d if for each x ∈ U
there exists an rx > 0 such that Bd (x, rx ) ⊂ U . We wish to show that d1 -balls are open in the dtopology. That is, we wish to verify the above with U = Bd1 (x, r) for any r > 0 and x ∈ S. Assume
for every > 0 there exists a δ > 0 such that d(x, y) < δ implies d1 (x, y) < . Let y ∈ Bd1 (x, r) and
set = min{d1 (x, y), r − d1 (x, y)} and extract δ > 0 according to the above. Now if z ∈ Bd (y, δ)
then d(y, z) < δ and d1 (y, z) < ≤ r − d1 (x, y). By the triangle inequality,
d1 (x, z) ≤ d1 (y, z) + d1 (x, y) < r − d1 (x, y) + d1 (x, y) < r
so that z ∈ Bd1 (x, r). The converse is analogous.
Now, it is clear if d1 is the bounded metric from Exercise 3.1.2.1, d(x, y) ≥ d1 (x, y). So, we need
only show for every > 0 there exists a δ > 0 such that d1 (x, y) < δ implies d(x, y) < . Observe
that d1 (x, y) < δ implies
δ
.
d(x, y) < δ[d(x, y) + 1] ⇔ d(x, y) <
1−δ
So, choose δ such that δ/(1 − δ) < and δ < 1.
3.1.2.3. Show by strict application of the definition that the closure of |z − z0 | < δ is |z − z0 | ≤ δ.
Solution: We show that |z − z0 | ≤ δ is closed by showing |z − z0 | > δ is open. Let z be such
that |z − z0 | > δ and set r = |z − z0 | − δ. The minimum distance between a point w with |w − z| ≤ r
and z0 is attained at w = z + r(z0 − z)/|z0 − z|. Then,
|w − z0 | = |z − z0 + (1 − δ/|z − z0 |)(z0 − z)| = δ.
But, note that |w − z| = r so that in fact if w ∈ B(z, r) then |w − z0 | > δ.
Since |z − z0 | ≤ δ (denoted S) is closed, it follows that Cl(B(z0 , δ)) ⊂ S. Now let U be open and
disjoint from B(z0 , δ). We show that U is disjoint from S. So, let z ∈ S be such that |z − z0 | = δ.
Suppose that z ∈ U . Since U is open, there exists an δ > r > 0 such that B(z, r) ⊂ U . Consider
the point w = z + r/2(z0 − z)/|z − z0 |. We see that
|w − z| = |z − z + r/2(z0 − z)/|z − z0 || = r/2
1
|w − z0 | = |z − z0 + r/2(z0 − z)/δ| = |δ(z − z0 ) + r/2(z0 − z)| = |r/2 − δ| < δ
δ
Hence, w ∈ B(z0 , δ) ∩ B(z, r), a contradiction since B(z0 , δ) ∩ U = Ø. This shows that every closed
set containing B(z0 , δ) must contain S. Hence, S = Cl(B(z0 , δ)).
3.1.2.4. If X is the set of complex numbers whose real and imaginary parts are rational, what is
Int X, Cl X, ∂X?
Solution: Let z ∈ X and consider B(z, δ) for some δ > 0. Let w be such that Im w = Im z
but | Re w − Re z| < δ. There are obviously irrational real numbers satisfying this, so we can always
find a w ∈ B(z, δ) whose real part is irrational. Hence, B(z, δ) is not a subset of X for any δ, and
X is not a neighborhood of any of its points. Thus, Int X = Ø. Now, Cl X = (Int X c )c . By a
Ahlfors Exercises
similar reasoning to the above, for any z ∈ X c and B(z, δ), we can find a w ∈ B(z, δ) whose real and
imaginary parts are rational. So, Int X c = Ø and Cl X = C. By definition, ∂X = Cl X \ Int X = C.
3.1.2.5. It is sometimes typographically simpler to write X 0 for X c . With this notation, how is
(Cl X c )c related to X? Show that (Cl(Cl(Cl(Cl X)c )c )c )c = (Cl(Cl X)c )c
Solution: By definition,
Cl X c =
\
C
X c ⊂C
where C is closed. Note that if X c ⊂ C then C c ⊂ X, and C c is open. So,
!c
[
\
c c
=
C c = Int X.
(Cl X ) =
C
X c ⊂C
X c ⊂C
Next, we show that if X is open
(Cl(Cl X)c )c = X.
It is equivalent to show that
Cl(Cl X)c = Cl X c = X c
since X is open. If Y = X c then the above shows that
(Cl X)c = (Cl Y c )c = Int Y.
By taking closures and noting that Y is closed, we see Cl Int Y = Y = X c . The statement we are
asked to show reduces to the original one above.
3.1.2.6. A set is said to be discrete if all its points are isolated. Show that a discrete set in R or C
is countable.
Solution: We showed in Exercise 3.1.2.4 that the set A of complex numbers with rational real
and imaginary parts is dense in C. Let S be a discrete set in C. Then for any z ∈ S there exists
an r > 0 such that B(z, r) ∩ S = {z}. But by density this ball will contain a point w with rational
real and imaginary parts. We may choose w so that if r0 > |w − z| then B(w, r0 ) ⊂ B(z, r) and
z ∈ B(w, r0 ). It follows that B(w, r0 ) ∩ S = {z}, and thus we can define a map ϕ : S → A such that
ϕ(z) = w. Note that our choice of w is not unique, but we have chosen it so that ϕ is injective.
Since A is countable we are done. The case for R follows from this by way of the subspace topology.
3.1.2.7. Show that the accumulation points of any set form a closed set.
Solution: Let X ⊂ C and denote the set of accumulation points of X by A. Then,
A = {x ∈ Cl X | x is not an isolated point of Cl X}.
Consequently, Ac is defined by
Ac = (Cl X)c ∪ {x ∈ Cl X | x is an isolated point of Cl X}.
The set (Cl X)c is obviously open, so it suffices to show that the isolated points of Cl X is open.
XXX
3.2. Conformality.
3.2.2. Analytic Functions in Regions.
Ahlfors Exercises
3.2.2.1. Give a precise definition of a single-valued branch of
and prove that it is analytic.
√
1+z+
√
1 − z in a suitable region,
√
Solution: We recall that for z we choose the
√ region Ω which is the complement of the negative real axis x ≤ 0, y√= 0. Hence, to define 1 + z we should choose Ω1 as the complement of
x ≤ −1,
√ y = 0 and
√ for 1 − z we should choose Ω2 as the complement of 1 ≤ x, y = 0. In total, to
define 1 + z + 1 − z we choose the region Ω = Ω1 ∩ Ω2 . This is actually the same region used in
defining
arccos z. The branch chosen is that√which has√positive real part. As simple transformations
√
of z in an analytic region, it follows that 1 + z + 1 − z is analytic.
3.2.2.2. Same problem for log log z.
Solution: We look at how the principal branch of log z transforms the region Ω defined by x ≤ 0,
y = 0. The ray x > 0, y = 0 has arg(z) = 0 so that log z = log |z| = log x. So, this gets mapped to
the real axis. All other points get mapped to the strips −π < y < 0 and 0 < y < π. To define the
logarithm again, we must remove those points with x ≤ 0 and y = 0. As above, this corresponds
to the points in Ω with x ≤ 1 and y = 0. We once more choose the branch | Im log log z| < π and
| Im log z| < π. It is analytic by way of a composition of analytic functions.
3.2.2.3. Suppose that f (z) is analytic and satisfies the condition |f (z)2 − 1| < 1 in a region Ω. Show
that either Re f (z) > 0 or Re f (z) < 0 throughout Ω.
Solution: Suppose Re f (z) = 0 at a point z ∈ Ω. Then f (z) = iy 2 for some y ∈ R, and thus
f (z)2 = −y 2 . By the modulus condition |f (z)2 − 1| < 1 we have | − y 2 − 1| < 1 and thus |y 2 + 1| < 1.
This is clearly impossible, so that Re f (z) 6= 0 throughout Ω. But, Re f (z) is continuous and Ω is
connected, so either Re f (z) > 0 or Re f (z) < 0.
3.3. Linear Transformations.
3.3.1. The Linear Group.
3.3.1.1. Prove that the reflection z 7→ z is not a linear transformation.
Solution: Suppose there exist a, b, c, d ∈ C such that
z=
az + b
.
cz + d
First, if z = 0 then z = 0 and so b/d = 0. It follows that b = 0. Next, multiply through by cz + d.
Then
c|z|2 + dz = z(cz + d) = az.
Consider z = r for real r > 0, then cr2 = (a − d)r, or cr = a − d. But this must hold for all r, which
is impossible since a, c, d are constants.
A more intuitive approach may be to prove that z is generally not analytic, but linear transformations are.
3.3.1.2. If
T1 z =
z+2
z+3
T2 z =
z
z+1
find T1 T2 z, T2 T1 z, and T1−1 T2 z.
Solution: Generally consider two linear transformations
T1 z =
a1 z + b1
c1 z + d1
T2 z =
a2 z + b2
.
c2 z + d2
Ahlfors Exercises
Then, the composition T1 T2 z = T1 (T2 z) is given by
a1 (a2 z + b2 ) + b1 (c2 z + d2 )
a1 (a2 z + b2 )/(c2 z + d2 ) + b1
=
c1 (a2 z + b2 )/(c2 z + d2 ) + d1
c1 (a2 z + b2 ) + d1 (c2 z + d2 )
(a1 a2 + b1 c2 )z + (a1 b2 + b1 d2 )
=
(c1 a2 + d1 c2 )z + (c1 b2 + d1 d2 )
T1 (T2 z) =
Observe that if we write T1 and T2 as matrices
a1 b1
T1 =
c1 d 1
T2 =
a2
c2
b2
d2
then T1 T2 is just given by the matrix product
a1 b1
a2 b2
a1 a2 + b1 c2
T1 T2 =
=
=
c1 d1
c2 d2
c1 a2 + d1 c2
With our particular T1 , T2 , we have that
1 2
1
T1 =
T2 =
1 3
1
0
1
T1−1
a1 b2 + b1 d2
.
c1 b2 + d1 d2
=
3
−1
−2
.
1
Thus,
2
1 0
3 2
T1 T2 =
=
3
1 1
4 3
1 0
1 2
1 2
T2 T1 =
=
1 1
1 3
2 5
3 −2
1 0
1 −2
−1
T1 T2 =
=
−1 1
1 1
0 1
1
1
So, the compositions are given by
z+2
3z + 2
T2 T1 z =
T1−1 T2 = z − 2
T1 T2 z =
4z + 3
2z + 5
3.3.1.3. Prove that the most general transformation which leaves the origin fixed and preserves all
distances is either a rotation or a rotation followed by reflection in the real axis.
Solution: Consider a general linear transformation
az + b
Tz =
.
cz + d
If the origin is fixed, then T (0) = 0 and so b/d = 0. Thus, b = 0. Next, if distances are preserved
then
az
aw
aczw + adz − aczw − adw
|ad||z − w|
|z − w| = |T z − T w| =
−
=
=
.
cz + d cw + d
(cz + d)(cw + d)
|cz + d||cw + d|
It follows that |cz + d||cw + d| = |ad| for z 6= w. In particular, this shows that d 6= 0, and if w = 0
then |cz + d| = |a|. Since the right hand side is constant, it must be that c = 0. Substituting this
into |cz + d||cw + d| = |ad|, we see that |d| = |a|. So, also, a 6= 0. Finally we are left with T z = az/d,
where |a/d| = 1. If we set k = a/d then T z = kz where |k| = 1, and it is clear that T is a rotation.
3.3.1.4. Show that any linear transformation which transforms the real axis into itself can be written with real coefficients.
Solution: Suppose that T z = (az + b)/(cz + d). We can assume that a is real, since it is either
0 or we can multiply the numerator and denominator by a.
3.4. Elementary Conformal Mappings.
3.4.2. A Survey of Elementary Mappings. Throughout this section, I ignore the symmetry condition.
I skipped that section.
Ahlfors Exercises
3.4.2.1. Map the common part of the disks |z| < 1 and |z − 1| < 1 on the inside of the unit circle.
Choose the mapping so that the two symmetries are preserved.
Solution: The region is bounded by two circular arcs with common√endpoints, so we take
√ the
approach given in the book. The two common endpoints are a = 1/2 + 3/2i and b = 1/2 − 3/2i.
So, consider first the map
√
√
2z − 1 − 3i
z−a
z − 1/2 − 3/2i
√
√ .
=
z1 =
=
z−b
z − 1/2 + 3/2i
2z − 1 + 3i
We know this maps the region Ω to a sector. To determine the boundary of this sector we make
smart choices of z. First consider z = 1, where we have
√
√
1
3
1 − 3i
√ = e−iπ/3−iπ/3 = e−2π/3i = − −
z1 (1) =
i.
2
2
1 + 3i
Similarly, by choosing z = 0 we get
√
1
3
a
iπ/3−(−iπ/3)
2iπ/3
=e
= +
i.
z1 (0) = = e
b
2
2
To determine the interior of the sector we make one more choice of z. Consider z = 1/2,
√
− 3i
√
= −1.
z1 (1/2) =
3i
It follows that z1 maps Ω to the sector 2π/3 ≤ θ ≤ 4π/3.
Consider now the map z 3/2 = e3/2 log z . Exactly as we define z 1/2 , we may consider the same
branch cut by removing the negative real axis. But if we do that, we will remove a portion of z1 (Ω).
To avoid this, we first rotate z1 (Ω). Let z2 = eiπ/3 z1 . Then z2 rotates the sector 2π/3 ≤ θ ≤ 4π/3
3/2
to the sector π ≤ θ ≤ 5π/3. Now we can apply z3 = z2 , which expands this to the region
3π/2 ≤ θ ≤ 5π/2 ⇔ −π/2 ≤ θ ≤ π/2. This is the right half-plane. We know that the final transformation w = (z3 − 1)/(z3 + 1) maps this half plane to the unit disc |w| < 1. The full composition
is
3/2
w=
3/2
z3 − 1
z
−1
(eiπ/3 z1 )3/2 − 1
iz
−1
i(z − eiπ/3 )3/2 /(z − e−iπ/3 )3/2 − 1
= 23/2
= iπ/3 3/2
= 13/2
=
.
z3 + 1
(e
z1 ) + 1
i(z − eiπ/3 )3/2 /(z − e−iπ/3 )3/2 + 1
z2 + 1
iz1 + 1
I see no good way to simplify this from here.
3.4.2.2. Map the region between |z| = 1 and |z − 1/2| = 1/2 on a half plane.
Solution: The region is bounded by two circles which are tangent at a = 1. The map z1 = 1/(z−a) =
1/(z − 1) sends this region to a parallel strip. We test z1 at different points to determine the boundary curves. First consider z = −1 and z = i which lie on |z| = 1. Then, z1 (−1) = −1/2 and
z1 (i) = 1/(i − 1) = (−1 − i)/2. This shows that one boundary curve is the line x = −1/2 in the
z1 -plane. Since the other boundary curve is parallel to this, we only need to test one point. Consider
z = 0, which lies on the other circle. Then z1 (0) = −1, and so the other boundary curve is x = −1.
We can map this to the upper half plane if the strip was bounded by y = 0 and y = π. But this is
easily done by a rotation, translation, and dilation. First applying the rotation z2 = −iz1 converts
the strip between y = 1/2 and y = 1. Next, the translation z3 = z2 − i/2 converts this to the strip
between y = 0 and y = 1/2. Now, the dilation z4 = 2πz3 converts this to the strip between y = 0
and y = π as desired. The final map w = ez4 transforms this to the upper half plane. In total, the
map is
w = ez4 = e2πz3 = e2π(z2 −i/2) = e2πz2 −iπ = e−2πiz1 −iπ = e−iπ e−2πi/(z−1) = −e−2πi/(z−1) .
Ahlfors Exercises
3.4.2.3. Map the complement of the arc |z| = 1, y ≥ 0 on the outside of the unit circle so that the
points at ∞ correspond to each other.
Solution: Consider the map z1 = (z + 1)/(z − 1). This maps the arc |z| = 1, y ≥ 0 to a straight line.
Choose z = i so that
1+i
(1 + i)(−1 − i)
z1 (i) =
=
= −i.
−1 + i
2
Since z = i belongs to the specified arc, it follows that z1 maps the arc to the ray x = 0, y ≤ 0. This
map sends ∞ to 1. Rotating the domain using z2 = −iz1 takes the ray to the negative real axis, and
1/2
the composition maps ∞ to −i. Using the standard branch cut for√the square root, we use z3 = z2
−π/2i 1/2
−π/4i
)√ = e √ . Now
to map the region to the right half plane. Moreover, ∞ gets sent to −i =√(e
translate and dilate so that e−π/4i is sent to 1 – this is achieved via z4 = 2(z3 + 2/2i) = 2z3 + i.
Note that this keeps the region fixed to the right half plane, and we have sent ∞ to 1 in the
composition. Finally, we use the map w = (z4 + 1)/(z4 − 1) to map this to |w| > 1. It is clear that
∞ 7→ ∞. The full composition is
√
√ 1/2
√
2z3 + i + 1
2z2 + i + 1
2(−iz1 )1/2 + i + 1
z4 + 1
=√
= √ 1/2
=√
w=
z4 − 1
2z3 + i − 1
2(−iz1 )1/2 + i − 1
2z2 + i − 1
p
√ −π/4i 1/2
1/2
(1 − i) (z + 1)/(z − 1) + i + 1
2e
z1 + i + 1
(1 − i)z1 + i + 1
p
=√
=
=
1/2
1/2
(1 − i) (z + 1)/(z − 1) + i − 1
2e−π/4i z1 + i − 1
(1 − i)z1 + i − 1
p
(z + 1)/(z − 1) + i
=p
(z + 1)/(z − 1) − 1
To verify w(∞) = ∞, define w̃ by w̃ = w(1/z). Then,
p
(1 + z)/(1 − z) + i
w̃(z) = p
.
(1 + z)/(1 − z) − 1
Clearly w̃(0) = ∞ so that w(∞) = ∞. Interestingly, our simplified version of w suggests that we
only need three steps – two linear transformations and a square root in between.
3.4.2.4. Map the outside of the parabola y 2 = 2px on the disk |w| < 1 so that z = 0 and z = −p/2
correspond to w = 1 and w = 0. (Lindelöf.)
Solution: We work backwards. XXX
3.4.2.5. Map the inside of the right-hand branch of the hyperbola x2 − y 2 = a2 on the disk |w| < 1
so that the focus corresponds to w = 0 and the vertex to w = −1. (Lindelöf.)
2
2
2
Solution: We are immediately guided towards using z 2 in some way, since Re(z
√ ) = x − y . So,
2
2
the specified region is mapped to x > 0 by z1 = z − a . The focus is at 2a while the vertex
is at a. Under z1 = z 2 − a2 these get mapped to a2 and 0, respectively. Recall that the map
w = (z − 1)/(z + 1) sends x > 0 to |w| < 1. Furthermore, w(1) = 0 and w(0) = −1. So, if we squish
x > 0 so that the boundary stays fixed but a2 is sent to 1, then composing it with the above finishes
the problem. Hence we set z2 = z1 /a2 and w = (z2 − 1)/(z2 + 1). The full composition is
w=
z2 − 1
z1 /a2 − 1
z1 − a2
z 2 − 2a2
=
=
=
= 1 − 2a2 z −2 .
2
2
z2 + 1
z1 /a + 1
z1 + a
z2
3.4.2.6. Map the inside of the lemniscate |z 2 − a2 | = ρ2 (ρ > a) on the disk |w| < 1 so that symmetries are preserved. (Lindelöf.)
Solution: If w = z 2 then the lemniscate is transformed into |w − a2 | = ρ2 , a circle with center
a2 and radius ρ2 . The interior is mapped to the interior. Composing with a translation and a
Ahlfors Exercises
dilation gives the required map. In total
w=
z2
a2
1 2
(z − a2 ) = 2 − 2 .
2
ρ
ρ
ρ
3.4.2.7. Map the outside of the ellipse (x/a)2 + (y/b)2 = 1 onto |w| < 1 with preservation of
symmetries.
4. Complex Integration
4.1. The Fundamental Theorems.
4.1.3. Line Integrals as Functions of Arcs.
4.1.3.1. Compute
Z
x dz
γ
where γ is the directed line segment from 0 to 1 + i.
Solution: Let z(t) = (1 + i)t = t + it. Then z(t) parameterizes γ where z(0) = 0 and z(1) = 1 + i.
Thus,
Z
Z 1
Z 1
1
1 1
t
xdz =
= + i.
x(t)z 0 (t) dt =
t(1 + i) dt = (1 + i)
2 0
2 2
γ
0
0
As an aside, recall that if F (z) is analytic and F (z) = U (x, y) + iV (x, y) then
∂U
∂V
∂V
∂U
F 0 (z) =
+
i=
−i
.
∂x
∂x
∂y
∂y
Since F is analytic, U and V are harmonic functions. Thus,
∂2U
∂2U
∂
∂
+
=
Re(F 0 (z)) −
Im(F 0 (z))
∂x2
∂y 2
∂x
∂y
∂2V
∂2V
∂
∂
0=
+
=
Im(F 0 (z)) +
Re(F 0 (z))
2
∂x
∂y 2
∂x
∂y
Let f (z) = u(x, y) + iv(x, y) where f (z) = F 0 (z). Then, the above can be written as
∂u ∂v
∂v
∂u
−
=0
+
= 0.
∂x ∂y
∂x ∂y
In this problem, we have f (z) = x so that u(x, y) = x and v(x, y) = 0. The first equation fails, so√that
f is not the derivative of an analytic function. Indeed, considering the circle z(t) = 1/2 + 1/ 2eit
gives a different value of the integral, so it is path dependent.
0=
4.1.3.2. Compute
Z
x dz
|z|=r
for the positive sense of the circle, in two ways: first, by use of a parameter, and second, by observing
that x = 1/2(z + z) = 1/2(z + r2 /z) on the circle.
Solution: As seen in the previous problem, f (z) = x is not the derivative of an analytic function so that this contour integral is not necessarily zero. Letting z(t) = reit we have z 0 (t) = ireit
and thus
Z
Z 2π
Z 2π
Z 2π
x dz =
x(t)z 0 (t) dt =
(r cos t)(−r sin t + ir cos t) dt =
−r2 cos t sin t + ir2 cos2 t dt
|z|=r
0
Z
0
2π
2
0
2
2
Z
2π
2
r
ir
ir
ir
sin(2t) +
+
cos(2t) dt =
= πr2 i
2
2
2
2
0
0
Now recall that if C is the circle with center a and radius ρ then
Z
1
dz = 2πi.
C z−a
=
−
Ahlfors Exercises
Consequently,
Z
Z
Z
Z
Z
r2
1
1
r2
r2
1
1
z+
dz =
z dz +
dz =
dz = πr2 i
x dz =
2
z
2
2
z
2
z
|z|=r
|z|=r
|z|=r
|z|=r
|z|=r
where we emphasize that the first integral vanishes since f (z) = z is analytic.
4.1.3.3. Compute
Z
|z|=2
dz
−1
z2
for the positive sense of the circle.
Solution: Here’s one way to do the problem, which leads to some insight. We first split the integral as
Z
Z
Z
Z
dz
1
1
dz
1
dz
1
=
−
dz
=
−
.
2
2(z − 1) 2(z + 1)
2 |z|=2 z − 1 2 |z|=2 z + 1
|z|=2 z − 1
|z|=2
Let a ∈ C and suppose that γ is any simple closed curve which bounds a compact region containing
a. We wish to compute
Z
dz
.
z
−a
γ
We can assume WLOG that a = 0 by a change of variables and that γ is positively oriented. By this,
note that γ is diffeomorphic to S 1 , and so induces an orientation on S 1 by requiring this diffeomorphism be orientation preserving. If the induced orientation on S 1 is positive (counterclockwise), we
say that γ is positively oriented. Recall that F (z) = log z is analytic except along x ≤ 0, and that
F (z) = 1/z (due to our choice of branch cut). Since γ is a closed loop around z = 0, it intersects the
positive real axis somewhere. Denote this point by b. Let K denote the compact region γ bounds.
Since Int K is open and z ∈ Int K, there exists a ball of some radius r about z.
Let γ1 be a straight line from b to r, γ2 the negatively oriented circle |z| = r, and γ3 = −γ1 . Let
γ̃ = γ + γ1 + γ2 + γ3 . Then γ is a simple closed curve which does not bound 0. It follows, since
1/z is the derivative of an analytic function in this region, that the integral vanishes. Also, since
γ3 = −γ1 we have
Z
Z
Z
Z
Z
Z
Z
dz
dz
dz
dz
dz
dz
dz
0=
=
+
+
−
=
+
.
z
z
z
z
z
z
γ̃
γ
γ1
γ2
γ1
γ
γ2 z
Consequently,
Z
γ
dz
=−
z
Z
γ2
dz
= 2πi
z
since γ2 is a negatively oriented circle centered at 0.
Returning to our problem, since ±1 are in the region bounded by |z| = 2, we may apply the
above and conclude
Z
dz
= πi − πi = 0.
2−1
z
|z|=2
This suggests that maybe f (z) = 1/(z 2 − 1) is the derivative of an analytic function. Indeed, we
know that f (z) = F 0 (z) where F (z) = − arctanh z. However, I think the branch points of F (z) are
±1...
One can also deduce that the real and imaginary parts are u(x, y) = (x2 − y 2 − 1)/((x2 − y 2 −
1)2 + (2xy)2 ) and v(x, y) = −2xy/((x2 − y 2 − 1)2 + (2xy)2 ). Letting z(t) = 2eit = 2 cos t + 2i sin t,
one can painfully compute the integral.
Ahlfors Exercises
4.1.3.4. Compute
Z
|z − 1| |dz|.
|z|=1
Solution: Parametrize |z| = 1 by z(t) = eit = cos t + i sin t. Then if f (z) = |z − 1|, we have
q
p
√
f (z(t)) = (cos t − 1)2 + sin2 t = cos2 t − 2 cos t + 1 + sin2 t = 2 − 2 cos t = 2| sin(t/2)|.
Moreover,
|z 0 (t)| =
so that
Z
Z
|z − 1| |dz| =
|z|=1
p
(− sin t)2 + (cos t)2 = 1
2π
Z
0
2π
2π
= −4(−1 − 1) = 8.
sin(t/2) dt = −4 cos(t/2)
2| sin(t/2)| dt = 2
0
0
4.1.3.5. Suppose that f (z) is analytic on a closed curve γ (i.e., f is analytic in a region that contains
γ). Show that
Z
f (z)f 0 (z) dz
γ
0
is purely imaginary. (The continuity of f (z) is taken for granted.)
Solution: Let f (z) = u(x, y) + iv(x, y). Note that
Z
Z b
f (z) dz =
(u(x(t), y(t)) + iv(x(t), y(t)))(x0 (t) + iy 0 (t)) dt
γ
a
Z
=
b
[u(x(t), y(t))x0 (t) − v(x(t), y(t))y 0 (t)] + [u(x(t), y(t))y 0 (t) + v(x(t), y(t))x0 (t)]i dt
a
where z(t), a ≤ t ≤ b is a suitable parametrization of γ. If we write dz = dx + idy, then formally
f (z)dz = (u + iv)(dx + idy) = [udx − vdy] + [udy + vdx]i
R
R
which agrees with the above rigorous computation. We therefore have that Re( γ f (z)dz) = γ Re(f (z)dz).
Then,
Re(f (z)f 0 (z)dz) = Re((u − iv)(ux + ivx )(dx + idy)) = Re(([uux + vvx ] − [uuy + vvy ]i)(dx + idy))
= Re([[uux + vvx ]dx + [uuy + vvy ]dy] + [[uux + vvx ]dy − [uuy + vvy ]dx]i)
1
= [uux + vvx ]dx + [uuy + vvy ]dy = d(u2 + v 2 )
2
where we have used the Cauchy-Riemann equations throughout. It follows that we have an exact
differential, and therefore the integral is zero. This shows that the overall integral is purely imaginary, since its real part vanishes.
4.1.3.6. Assume that f (z) is analytic and satisfies the inequality |f (z) − 1| < 1 in a region Ω. Show
that
Z 0
f (z)
dz = 0
γ f (z)
for every closed curve in Ω. (The continuity of f 0 (z) is taken for granted.)
Solution: If w = f (z) then in the w-plane, the image of f lies in the disc |w − 1| < 1. This is
a circle of radius 1 centered at 1. Importantly, Re f (z) > 0, and −π < arg f (z) < π. Note that log z
is well-defined and analytic on the complex plane minus x ≤ 0. Hence, the composition log f (z) is
well-defined and analytic on Ω. Since d/dz log f (z) = f 0 (z)/f (z), we see that the integrand is the
derivative of an analytic function. Hence the integral over any closed curve in that region vanishes.
Ahlfors Exercises
4.1.3.7. If P (z) is a polynomial and C denotes the circle |z −a| = R, what is the value of
Answer : −2πiR2 P 0 (a).
R
C
P (z)dz?
Solution: By definition,
Z
Z
P (z) dz :=
P (z) dz.
C
C
We can always write P (z) as a power series
P (z) =
n
X
ak (z − a)k .
k=0
Parametrize C by z(t) = a + Reit with 0 ≤ t ≤ 2π. Then,
!
Z 2π
Z
Z
n
n
X
X
it
P (z) dz =
iRe
ak Rk eikt dt = i
ak Rk+1
C
0
k=0
k=0
2π
cos((1 − k)t) − i sin((1 − k)t) dt.
0
All of these evaluate to zero, except for the k = 1 term where we get
Z
P (z) dz = 2πia1 R2 .
C
It is easy to find that a1 = P 0 (a)/1! = P 0 (a). Hence,
Z
P (z) dz = −2πiR2 P 0 (a).
C
4.1.3.8. Describe a set of circumstances under which the formula
Z
log z dz = 0
γ
is meaningful and true.
Solution: Consider F (z) = z(log z − 1). This map is single-valued and analytic in on the complex plane minus the negative real axis x ≤ 0. Since the choice of branch cut affects the values in the
codomain, it does notRput a restriction on γ. Hence any closed curve γ not intersecting the negative
real axis is such that γ log z dz = 0.
4.2. Cauchy’s Integral Formula.
4.2.2. The Integral Formula.
4.2.2.1. Compute
Z
|z|=1
ez
dz.
z
Solution: The map f (z) = ez is analytic everywhere, and in particular on |z| = 1. By the representation formula,
Z
Z
1
f (z)
1
ez
f (0) =
dz =
dz.
2πi |z|=1 z − 0
2πi |z|=1 z
Hence, the integral is 2πie0 = 2πi.
4.2.2.2. Compute
Z
|z|=2
dz
+1
z2
by decomposition of the integrand in partial fractions.
Solution: We have that
Z
Z
Z
Z
dz
1
1
1
dz
1
dz
=
−
dz =
−
.
2+1
z
2i(z
−
i)
2i(z
+
i)
2i
z
−
i
2i
z
+i
|z|=2
|z|=2
|z|=2
|z|=2
Ahlfors Exercises
By an application of Cauchy’s representation formula with the constant function f (z) = 1 at z = ±i,
we arrive at
Z
1
dz
=
(2πi − 2πi) = 0.
2+1
z
2i
|z|=2
It is important that ±i lie in the bounded region determined by |z| = 2.
4.2.2.3. Compute
Z
|z|=ρ
|dz|
|z − a|2
under the condition |a| =
6 ρ. Hint: make use of the equations zz = ρ2 and |dz| = −iρdz/z.
Solution: Let’s first show the second hint in the equation. Consider an arbitrary integrable function
f (z) and parametrize |z| = ρ by z(t) = ρeit = ρ cos t + iρ sin t. Then,
Z
Z 2π
Z 2π
0
f (z) |dz| =
f (z(t))|z (t)| dt =
ρf (eit ) dt.
|z|=ρ
0
0
On the other hand,
Z
Z 2π
Z 2π
Z 2π
f (z)
f (z(t)) 0
f (eit ) it
−iρ
dz =
−iρ
z (t) dt =
−i2 ρ2
e
dt
=
ρf (eit ) dt.
it
z
z(t)
ρe
|z|=ρ
0
0
0
Since these integrals are equal, we are justified in formally writing |dz| = −iρdz/z.
Assume first that a = 0. Then,
Z
Z
Z
|dz|
−iρdz/z
i
dz
i
2π
=
=−
= − (2πi) =
.
2
2
|z|
ρ
ρ
z
ρ
ρ
|z|=ρ
|z|=ρ
|z|=ρ
Consider the change of variables z 7→ eiθ z, θ ∈ [−π, π). The differential element transforms by
|eiθ dz| = |eiθ ||dz| = |dz|. Next,
|z − a|2 = |eiθ z − a|2 = |eiθ |2 |z − ae−iθ |2 .
By choosing θ = arg a, we may assume a is a nonnegative real number. Or,
Z
Z
|dz|
|dz|
=
.
2
|z
−
a|
|z
−
|a||2
|z|=ρ
|z|=ρ
If z lies on |z| = ρ, then we have z = ρ2 /z. Together with the above,
Z
Z
Z
Z
|dz|
|dz|
−iρdz/z
dz
=
=
=
−iρ
2
2
2
|z|=ρ |z − |a||
|z|=ρ (z − |a|)(z − |a|)
|z|=ρ (z − |a|)(ρ /z − |a|)
|z|=ρ (z − |a|)(ρ − |a|z)
!
Z
Z
iρ
dz
dz
= 2
+
|a|
2
2
|a| − ρ
|z|=ρ z − |a|
|z|=ρ ρ − |a|z
!
Z
Z
iρ
dz
dz
= 2
−
.
2
|a| − ρ2
|z|=ρ z − |a|
|z|=ρ z − ρ /|a|
To evaluate the two integrals, we must determine if |a| and ρ2 /|a| lie the bounded region determined
by |z| = ρ or not. That is, if |a| < ρ and/or ρ2 /|a| < ρ. The latter inequality reduces to ρ < |a|.
So, there are two cases: |a| < ρ and |a| > ρ. In the former, the first integral is 2πi while the second
integral is zero (since ρ2 /|a| is in the unbounded region). In the latter case, the roles are reversed.
So,
(
Z
2πρ/(ρ2 − |a|2 )
|a| < ρ
|dz|
=
.
2
2
2
2πρ/(|a| − ρ )
|a| > ρ
|z|=ρ |z − a|
Note that we can include the case |a| = 0 (in fact, we could have included this throughout if we
consider the extend complex numbers).
4.2.3. Higher Derivatives.
Ahlfors Exercises
4.2.3.1. Compute
Z
|z|=1
ez z −n dz,
Z
z n (1 − z)m dz,
|z|=2
Z
|z − a|−4 |dz|
|z|=ρ
where |a| =
6 ρ.
Solution: Let the integrals be denoted by I1 , I2 , and I3 respectively.
i) If n ≤ 0 then ez z −n is analytic, and hence I1 = 0. If n > 0, then by Cauchy’s integral
formula:
Z
2πi
2πi
ez
dz =
=
ez
.
I1 =
n
z
(n
−
1)!
(n
− 1)!
|z|=1
z=0
ii) If n, m ≥ 0 then z n (1 − z)m is analytic, and thus I2 = 0. Suppose n < 0 but m ≥ 0. Then,
Z
(1 − z)m
2πi
d|n|−1
I2 =
dz =
(1 − z)m
|n|
|n|−1
(|n|
−
1)!
z
dz
|z|=2
z=0
2πim!(−1)|n|−1
m
|n|−1
=
= 2πi(−1)
.
|n| − 1
(|n| − 1)!(m − (|n| − 1))!
If |n| − 1 > m, we interpret the above to be zero. Now suppose m < 0 but n ≥ 0. Then,
Z
Z
zn
zn
2πi(−1)m d|m|−1 n
m
I2 =
dz
=
(−1)
dz
=
z
|m|
|m|
(|m| − 1)! dz |m|−1
|z|=2 (1 − z)
|z|=2 (z − 1)
z=1
m
2πin!(−1)
n
= 2πi(−1)m
=
|m| − 1
(|m| − 1)!(n − (|m| − 1))!
which is also zero if |m| − 1 > n. Assume now that both n ≤ 0 and m ≤ 0. Then,
Z
dz
m
I2 = (−1)
.
|n| (z − 1)|m|
z
|z|=2
This is probably going to be hellish to do... I think we’re probably supposed to assume
n, m ≥ 0.
iii) This one also seems a little challenging, but is very similar to 4.2.2.3. We still have |dz| =
−iρdz/z and if z = ρ/z. We can also go ahead and assume that a is a nonnegative real
number, as before. So,
Z
Z
dz
|dz|
I3 =
=
−iρ
4
2 (ρ2 /z − a)2 z
|z
−
a|
(z
−
a)
|z|=ρ
|z|=ρ
Z
Z
z dz
iρ
z dz
= −iρ
=
−
.
2 (ρ2 − az)2
2
2 (z − ρ2 /a)2
(z
−
a)
a
(z
−
a)
|z|=ρ
|z|=ρ
Note that if |a| < ρ, then 1/(z − ρ2 /a)2 is analytic inside |z| = ρ, and if |a| > ρ then
1/(z − a)2 is analytic inside |z| = ρ. We assume |a| < ρ first. Then,
Z
iρ
z/(z − ρ2 /a)2 dz
iρ 2πi d
z
I3 = − 2
=− 2
.
a |z|=ρ
(z − a)2
a 1! dz (z − ρ2 /a)2 z=a
The derivative is
d
z
(z − ρ2 /a)2 − 2z(z − ρ2 /a)
−z 2 + ρ4 /a2
ρ4 a2 − a4 z 2
=
=
=
.
2
2
2
4
2
4
dz (z − ρ /a)
(z − ρ /a)
(z − ρ /a)
(az − ρ2 )4
Evaluating this at z = a gives
d
z
dz (z − ρ2 /a)2
=
z=a
a2 (ρ4 − a4 )
(a2 − ρ2 )4
so that I3 is
I3 =
2πρ(ρ2 + a2 )
.
(ρ2 − a2 )3
Ahlfors Exercises
We now assume |a| > ρ. Then,
Z
2πρ d
z/(z − a)2 dz
z
iρ
= 2
I3 = − 2
a |z|=ρ (z − ρ2 /a)2
a dz (z − a)2
The derivative is
d
a2 − z 2
z
=
2
dz (z − a)
(z − a)4
⇒
d
dz
z
(z − a)2
=
z=ρ2 /a
.
z=ρ2 /a
a2 (a4 − ρ4 )
a2 (a2 + ρ2 )
=
.
2
2
4
(ρ − a )
(a2 − ρ2 )3
Then, I3 is simply
2πρ(a2 + ρ2 )
.
(a2 − ρ2 )3
Note that in the two cases, all that changes is a sign.
4.2.3.2. Prove that a function which is analytic in the whole plane and satisfies an inequality
|f (z)| ≤ |z|n for some n and all sufficiently large |z| reduces to a polynomial.
I3 =
Solution: We have existence of an R > 0 such that |f (z)| ≤ |z|n for all |z| > R. Let r > R,
then by Cauchy’s integral formula,
Z
Z
(n + 1)!
|f (z)| |dz|
(n + 1)!
rn |dz|
(n + 1)!rn+1
(n+1)
|f
(a)| ≤
≤
=
.
n+2
n+2
2π
2π
(r − |a|)n+2
|z|=r |z − a|
|z|=r (r − |a|)
for any a with |a| < r. Of course, for any a we can find an r such that |a| < r and R < r, so as to
arrive at the above estimate. By taking r → ∞ it follows that |f (n+1) (a)| ≤ 0, and so f (n+1) ≡ 0.
Hence f is a polynomial of degree at most n.
4.2.3.3. If f (z) is analytic and |f (z)| ≤ M for |z| ≤ R, find an upper bound for |f (n) (z)| in
|z| ≤ ρ < R.
Solution: By Cauchy’s integral formula,
|f (n) (a)| ≤
n!
2π
Z
|z|=R
|f (z)| |dz|
n!M R
≤
.
|z − a|n+1
(R − |a|)n+1
The right hand side increases with increasing |a|. Hence, it is maximized when |a| = ρ, and for any
a with |z| ≤ ρ,
n!M R
|f (n) (z)| ≤
.
(R − ρ)n+1
4.2.3.4. If f (z) is analytic for |z| < 1 and |f (z)| ≤ 1/(1 − |z|), find the best estimate of |f (n) (0)|
that Cauchy’s inequality will yield.
Solution: Fix n ≥ 0 and let R = n/(n + 1) so that f (z) is analytic on |z| = R. Then by Cauchy’s
inequality,
n
n!
n!
n!(n + 1)n+1
1
=
=
n!
1
+
.
|f (n) (0)| ≤ n max |f (z)| = n
R |z|=R
R (1 − R)
nn
n
The optimal radius R was obtained by minimizing the function g(r) = 1/(rn (1 − r)) on (0, 1). Evaluation of g(r) at R gives the quantity used in the estimate, (n + 1)n+1 /nn . Interestingly, this is
approximately e([n + 1] + n)/2 = (n + 1/2)e.
4.2.3.5. Show that the successive derivatives of an analytic function at a point can never satisfy
|f (n) (z)| > n!nn . Formulate a sharper theorem of the same kind.
Solution: Recall that analytic functions admit a power series
f (z) =
∞
X
f (n) (a)(z − a)n
.
n!
n=0
Ahlfors Exercises
Let an = f (n) (a)/n!, so that for any a we have |an | > nn . By Hadamard’s formula, the radius of
convergence is given by
1
= lim sup |an |1/n = lim sup n = ∞
R
n→∞
n→∞
so that the power series defined in the right hand side is valid only at a. Yet, as an analytic function
it should be analytic in some radius of convergence for R > 0. Of course, if g(n) is any increasing
function (instead of nn ) such that g(n) → ∞, then we arrive at the same conclusion.
Here’s another way to go about the problem, which seems to be more of what Ahlfors intended
by putting this problem here. Cauchy’s estimate tells us that if R > 0,
n!M
|f (n) (z)| ≤ n
R
where M = max|z|=R |f (z)|. Note that M is finite since f (z) is continuous and |z| = R is compact.
If also |f (n) (z)| > n!nn , then nn ≤ M/Rn for all n ≥ 0. For n large enough this inequality fails,
since we can rearrange it to (nR)n ≤ M , and the left hand side is unbounded for every R > 0.
4.2.3.6. A more general form of Lemma 3 reads as follows:
Let the function ϕ(z, t) be continuous as a function of both variables when z lies in a region Ω and
a ≤ t ≤ b. Suppose further that ϕ(z, t) is analytic as a function of z ∈ Ω for any fixed t. Then
Z β
F (z) =
ϕ(z, t) dt
α
is analytic in z and
Z
β
∂ϕ(z, t)
dt
(?).
∂z
α
To prove this represent ϕ(z, t) as a Cauchy integral
Z
ϕ(ζ, t)
1
ϕ(z, t) =
dζ.
2πi C ζ − z
Fill in the necessary details to obtain
!
Z
Z β
1
dζ
F (z) =
ϕ(ζ, t) dt
.
2πi α
ζ −z
C
F 0 (z) =
and use Lemma 3 to prove (?).
Solution: Recall that Cauchy’s integral formula tells us that
Z
f (ζ) dζ
1
f (z) =
2πi C ζ − z
for z in the bounded region determined by C. Applying this with F (z) gives
!
!
Z
Z
Z β
Z
Z β
1
F (ζ) dζ
1
dζ
1
dζ
F (z) =
=
ϕ(ζ, t) dt
=
ϕ(ζ, t) dt
2πi C ζ − z
2πi C
ζ −z
2πi α
ζ −z
α
C
Rβ
(one may also view it as a consequence of Fubini’s theorem). Defining Φ(ζ) = 1/2πi α ϕ(ζ, t) dt,
Lemma 3 tells us that
!
Z
Z
Z β
Φ(ζ) dζ
dζ
1
0
F (z) =
=
ϕ(ζ, t) dt
.
2
2πi α
(ζ − z)2
C
C (ζ − z)
Now apply Lemma 3 once more to ϕ(z, t) directly:
Z
∂ϕ(z, t)
1
ϕ(ζ, t) dζ
=
.
∂z
2πi C (ζ − z)2
Integrating this over α ≤ t ≤ β and using Fubini’s theorem to switch integrals gives the result.
4.3. Local Properties of Analytical Functions.
Ahlfors Exercises
4.3.2. Zeros and Poles.
4.3.2.1. If f (z) and g(z) have the algebraic orders h and k at z = a, show that f g has the order
h + k, f /g the order h − k, and f + g an order which does not exceed max(h, k).
Solution: By definition, f (z) has an algebraic order h at z = a if
a) limz→a |z − a|α |f (z)| = 0 for all real α > h, and
b) limz→a |z − a|α |f (z)| = ∞ for all real α < h.
Consider now the function f g. Suppose α > h + k. Then we can write α as α1 + α2 with α1 > h
and α2 > k. Indeed, if = α − h − k > 0, then we can consider α1 := h + /2 and α2 := k + /2.
Consequently,
lim |z − a|α |f (z)g(z)| = lim |z − a|α1 |f (z)| lim |z − a|α2 |g(z)| = 0.
z→a
z→a
z→a
Similarly, if α < h + k we can write α = α1 + α2 with α1 < h and α2 < k. Then, by the same logic
we can split the limit to a product of limits, each evaluating to ∞.
Next, observe that constant, nonzero functions have algebraic order 0. By applying the above
with g and 1/g, we see that the orders must satisfy k + k̃ = 0, or that the order of 1/g is k̃ = −k.
So, the order of f /g is simply the order of f (1/g), which is h + (−k) = h − k.
Finally, suppose that f + g has order greater than max(h, k). Then there is a nondegenerate interval
between max(h, k) and the order of f + g – let α be in the interior of this interval. Then, by the
triangle inequality and applying the definitions of the orders:
∞ = lim |z − a|α |f (z) + g(z)| ≤ lim |z − a|α |f (z)| + lim |z − a|α |g(z)| = 0.
z→a
z→a
z→a
Clearly this is a contradiction, so that the order of f + g is majorized by max(f, g).
4.3.2.2. Show that a function which is analytic in the whole plane and has a nonessential singularity
at ∞ reduces to a polynomial.
Solution: If limz→∞ f (z) ∈ C, then f (z) is bounded in a region containing ∞. The complement
of this region is compact, and hence f (z) is bounded there too. Consequently, f (z) is a bounded
holomorphic function, and by Louisville’s theorem f (z) reduces to a constant (thus, is a polynomial). Suppose instead that limz→∞ f (z) = ∞. We recall that in Exercise 4.2.3.2, we show that if
|f (z)| grows like |z|n for |z| large, then f is a polynomial. So, we aim to prove an estimate like this.
Consider the function g(z) = 1/f (1/z). Since f (z) → ∞ as z → ∞, we see that g(z) → 0 as z → 0.
Hence, g has a removable singularity at 0. Let m be the algebraic order of this zero, so that h(z) is
a holomorphic function satisfying g(z) = z m h(z) and h(0) 6= 0. Because of this, 1/h is continuous
in some ball |z| < R, and is therefore bounded by M < ∞ in this ball. Consequently,
1
1
M
1
1
=
= m
≤ m.
f
z
|g(z)|
|z| h(z)
|z|
Considering z 7→ 1/z, we see that |f (z)| ≤ M |z|m for |z| ≥ 1/R. This is exactly the estimate we want.
4.3.2.3. Show that the functions ez , sin z and cos z have essential singularities at ∞.
Solution: All these functions are analytic, so that if they had a nonessential singularity at ∞,
then by Exercise 4.3.2.2 they would be polynomials. But, this is not the case (the latter two have
infinitely many zeros, and ez is not zero and not constant).
One can also prove this directly. Consider first f (z) = ez . To describe the nature of f (z) at infinity,
we instead analyze the nature of g(z) = f (1/z) = e1/z at z = 0. For any real α, |z|α |e1/z | = |z|α .
XXX. It is easily seen that this limit blows up for α < 0 and goes to zero for α > 0. But, g(z) tends
to zero as z → 0, so that g has no algebraic order. Hence, ez has no algebraic order; ∞ is thus an
Ahlfors Exercises
essential singularity.
4.3.2.4. Show that any function which is meromorphic in the extended plane is rational.
Solution: Note that rational functions have finitely many poles, so we aim to show this first. Suppose
that f is meromorphic in the extended plane and has poles a1 , a2 , a3 ... Then, since the extended
complex plane is compact, there exists a convergent subsequence (still denoted a1 , a2 , ...). This implies that the poles are not isolated, which contradicts f being meromoprhic. So, there are finitely
many poles – say N . Let hn be the order of the pole at an . Consider the function
h(z) = f (z)
N
Y
(z − an )hn .
n=1
QN
We wish to show that h is a polynomial so that f (z) = h(z)/ n=1 (z − an )hn is a rational function.
By construction, h is an analytic function and therefore only has a pole possibly at ∞. It follows
that h is a polynomial, and we are done.
4.3.2.5. Prove that an isolated singularity of f (z) is removable as soon as either Re f (z) or Im f (z)
is bounded above or below. Hint: Apply a fractional linear transformation.
Solution: Note that the fractional linear transformation T (z) = (z − 1)/(z + 1) maps {Re > 0}
to the unit disc. Consequently, the map Tc (z) = (z − 1 − c)/(z + 1 − c) maps {Re > c} onto the unit
disc. If Im > c, Re < c, Im < c, we can always rotate these domains onto Re > c. In particular, we
can always map these to the unit disc via
in z − 1 − c
Tc,n (z) = n
i z+1−c
for n = 0, 1, 2, 3 (corresponding to the four cases, in order). Hence, without loss of generality we
may assume Re > 0. Consider now the composite map
g(z) =
f (z) − 1
.
f (z) + 1
If g has a removable singularity at zero, then
lim zg(z) = 0
z→0
and consequently,
lim z[f (z) − 1] = lim zf (z) = 0
z→0
z→0
so that
4.3.3. The Local Mapping.
4.3.3.1. Determine explicitly the largest disk about the origin whose image under the mapping
w = z 2 + z is one to one.
Solution: If f (z) = z 2 + z then f (0) = 0 and f 0 (z) = 2z + 1 so that f 0 (0) 6= 0. So, z = 0 is a
simple root and we can apply Theorem 11 to deduce existence of a disc. Note that if f 0 (z) = 0, then
f cannot be injective. The only place where f 0 (z) = 0 is at z = −1/2, so that f is injective in a disc
of radius at most 1/2. Suppose now that z1 6= z2 are such that |z1 |, |z2 | < 1/2. If f (z1 ) = f (z2 ),
then
(z1 − z2 )(z1 + z2 ) = z12 − z22 = z2 − z1 .
Consequently, z1 + z2 = −1. By the triangle inequality,
1 = |z1 + z2 | ≤ |z1 | + |z2 | <
a contradiction. Hence f is injective on |z| < 1/2.
1 1
+ = 1,
2 2
Ahlfors Exercises
4.3.3.2. Same problem for w = ez .
Solution: Note that f (z) = ez has nonzero derivative everywhere, so we cannot use the same
logic as above to bound the injectivity radius. Suppose that ez1 = ez2 . It follows immediately (by
taking absolute values) that Re z1 = Re z2 . Dividing through by this, we have
cos(Im z1 ) + i sin(Im z1 ) = cos(Im z2 ) + i sin(Im z2 ).
By equating real and imaginary parts, one discovers that Im z1 − Im z2 = 2πn for some integer n.
Hence, the largest disc is one whose imaginary parts differ by less than 2π. In other words, a disc
of radius π.
4.3.3.3. Apply the representation f (z) = w0 + ζ(z)n to cos z with z0 = 0. Determine ζ(z) explicitly.
Solution: If z0 = 0 then w0 = 1. Now, f 0 (z) = − sin z so that f 0 (0) = 0, and f 00 (z) = − cos z
so that f 00 (0) = 1. Hence, n = 2.
z 2
f (z) − 1 = cos z − 1 = −2 sin
.
2
√
It follows that ζ(z) = 2i sin(z/2).
4.3.3.4. If f (z) is analytic at the origin and f 0 (0) 6= 0, prove the existence of an analytic g(z) such
that f (z n ) = f (0) + g(z)n in a neighborhood of 0.
Solution: Let h(z) = f (z n ), which is analytic as the composition of analytic functions. Now observe
that h0 (z) = nz n−1 f 0 (z n ). As we differentiate this, the product rule generates more terms but there
is only one which contains f 0 (z n ); the one obtained by differentiating the polynomial coefficient of
f 0 (z n ). All other terms contain z to some nonzero power multiplied by a higher order derivative
of f . All of these evaluate to zero, so we write h(k) (z) = n!/(n − k)!f 0 (z n ) + hk (z), where hk (z) is
analytic and hk (0) = 0. Accordingly, h(n) (0) = n!f 0 (0) + hn (0) = n!f 0 (0) 6= 0. So, h(z) − f (0) has a
zero of order n, and we may apply Theorem 11 to locally write
f (z n ) − f (0) = h(z) − f (0) = (z − z0 )n g̃(z).
Then,
|g̃(z0 )| for |z − z0 | < we can choose a single-valued branch
p if is such that |g̃(z0 ) − g̃(z)| <p
of n g̃(z). By defining g(z) = (z − z0 ) n g̃(z), which is analytic, we have
f (z n ) − f (0) = h(z) − f (0) = (z − z0 )n g̃(z) = g(z)n
as desired.
4.3.4. The Maximum Principle.
4.3.4.1. Show by use of (36), or directly, that |f (z)| ≤ 1 for |z| ≤ 1 implies
1
|f 0 (z)|
≤
.
(1 − |f (z)|2 )
1 − |z|2
Solution: First, (36) tells us that
M (f (z) − w0 )
R(z − z0 )
≤
M 2 − w0 f (z)
R2 − z0 z
when |z0 | < R and |w0 | < M , w0 = f (z0 ). In our particular problem, R = M = 1, and we can
rewrite it as
[f (z) − w0 ]/[z − z0 ]
1
≤
.
1 − w0 f (z)
1 − z0 z
Now, taking z → z0 within |z| < 1 gives
|f 0 (z0 )|
|f 0 (z0 )|
1
1
=
≤
=
.
2
1 − |f (z0 )|
|1 − w0 f (z0 )|
|1 − z0 z0 |
1 − |z0 |2
Ahlfors Exercises
But, the choice of z0 in |z| < 1 was arbitrary, so long as |f (z0 )| < 1. Consequently,
1
|f 0 (z)|
≤
2
1 − |f (z)|
1 − |z|2
on |z| < 1.
4.3.4.2. If f (z) is analytic and Im f (z) ≥ 0 for Im z > 0, show that
|f (z) − f (z0 )|
|f (z) − f (z0 )|
and
≤
|z − z0 |
|z − z0 |
|f 0 (z)|
1
≤
Im f (z)
y
where z = x + iy.
Solution: Consider the linear transformations
z − z0
Tz =
z − z0
Sw =
w − f (z0 )
w − f (z0 )
.
As done in the book, if we can show that Sf (T −1 ζ) satisfies the hypotheses of the Schwarz lemma,
then we have |Sf (T −1 ζ)| ≤ |ζ|, or by setting z = T −1 ζ, |Sf (z)| ≤ |T z|. Define F (ζ) := Sf (T −1 ζ).
We first want to show that F (0) = 0. Since T z0 = 0, we have T −1 (0) = z0 . Moreover, it is easily
seen that Sf (z0 ) = 0, so that F (0) = 0. Now, T z maps Im z > 0 to |ζ| < 1 and Sw maps Im w ≥ 0
to |Sw| ≤ 1. Since f (z) maps Im z > 0 to Im f (z) ≥ 0, it follows that |F (ζ)| ≤ 1 on |ζ| < 1. As a
composition of analytic functions, F is analytic. Thus the Schwarz lemma applies.
The second inequality follows from the first. If we rearrange it to
f (z) − f (z0 )
f (z) − f (z0 )
≤
z − z0
z − z0
then taking z → z0 (from above?) yields
|f 0 (z0 )| ≤
2 Im f (z0 )
f (z0 ) − f (z0 )
Im f (z0 )
=
=
.
z0 − z0
2 Im z0
y0
For arbitrary z0 , this results in the desired inequality.
4.3.4.3. In 4.3.4.1 and 4.3.4.2, prove that equality implies that f (z) is a linear transformation.
Solution: Observe that both proofs use the same method; apply the Schwarz lemma with F (ζ) =
Sf (T −1 (ζ)) for some linear transformations S and T . If equality holds, then we have that Sf (T −1 (ζ)) =
cζ for some c ∈ C. If z = T −1 (ζ), then Sf (z) = cT (z). Then, by composing with S −1 we have
f (z) = S −1 (cT z). As a composition of linear transformations, f is one too.
4.3.4.4. Derive corresponding inequalities if f (z) maps |z| < 1 into the upper half plane.
Solution: Consider the linear transformation Sw = (w − f (0))/(w − f (0)). As observed, Sw maps
the upper half plane onto |Sw| < 1. Hence, the composition F (z) = Sf (z) satisfies |F (z)| < 1 for
|z| < 1. Moreover, Sf (0) = 0 so that F (0) = 0 and F is analytic on |z| < 1. By the Schwarz lemma,
f (z) − f (0)
f (z) − f (0)
= |Sf (z)| ≤ |z|.
Rearranging this yields
f (z) − f (0)
≤ |f (z) − f (0)|
z
Ahlfors Exercises
so that taking z → 0 gives
|f 0 (0)| ≤ |f (0) − f (0)| = 2| Im f (0)| = 2 Im f (0).
4.3.4.5. Prove by use of Schwarz’s lemma that every one-to-one conformal mapping of a disk onto
another (or a half plane) is given by a linear transformation.
Solution: First, the problem reduces to considering only those biholomorphic maps f from the
unit disc to itself. Suppose f : D1 → D2 (where D1 and D2 are discs or half planes) and f is
injective. Then, f is biholomorphic onto D2 . Next, we can consider two linear transformations T
and S which map the unit disc D to D1 and D2 to D. We may choose these to also be biholomorphic. Thus, the composition S ◦ f ◦ T : D → D is biholomorphic. If we showed all such maps are
linear transformations, then by considering appropriate inverses of S and T we can write f as the
composition of three linear transformations.
So, we only consider biholomorphic f : D → D. Let S be given by Sw = [w − f (0)]/[1 − f (0)w]
and define F (z) = Sf (z). Note that |Sw| < 1 for |w| < 1 (since also |f (0)| < 1 by hypothesis), see
1.1.5.1 for details. It follows that F : D → D is such that F (0) = 0. Consequently, we can apply the
Schwarz lemma to F and obtain |F (z)| ≤ |z|. On the other hand, since F : D → D is biholomorphic,
we can apply the Schwarz lemma to F −1 and obtain |F −1 (ζ)| ≤ |ζ|. Letting ζ = F (z), which is
valid since F maps into D, we get
|z| = |F −1 (F (z))| ≤ |F (z)| ≤ |z|.
Hence, |F (z)| = |z| and F (z) = cz for some c ∈ C. It follows that f (z) = S −1 (cz), and thus f is a
linear transformation.
4.5. The Calculus of Residues.
4.5.2. The Argument Principle.
4.5.2.1. How many roots does the equation z 7 − 2z 5 + 6z 3 − z + 1 = 0 have in the disk |z| < 1?
Hint: Look for the biggest term when |z| = 1 and apply Rouché’s theorem.
Solution: Let f (z) = 6z 3 and g(z) = z 7 − 2z 5 + 6z 3 − z + 1. Then on |z| = 1,
|f (z) − g(z)| = |z 7 − 2z 5 − z + 1| ≤ |z|7 + 2|z|5 + |z| + 1 = 5 < 6 = |f (z)|.
By Rouché’s theorem, f (z) and g(z) have the same number of zeros in |z| < 1. Clearly f (z) = 6z 3
has three (a zero of order three at the origin), so that g(z) has three zeros in |z| < 1.
4.5.2.2. How many roots of the equation z 4 − 6z + 3 = 0 have their modulus between 1 and 2?
Solution: We simply need to apply Rouché’s theorem twice with |z| = 1 and |z| = 2. To find
the number of zeros in |z| < 1, let f (z) = 6z and g(z) = z 4 − 6z + 3. Then on |z| = 1,
|f (z) − g(z)| = |z 4 + 3| ≤ |z|4 + 3 = 4 < 6 = |f (z)|.
Thus, f (z) and g(z) have the same number of zeros in |z| < 1. Clearly f (z) has one so that g(z)
has one zero. To find the number of zeros in |z| < 2, choose f (z) = z 4 instead. Then on |z| = 2,
|f (z) − g(z)| = |6z + 3| ≤ 6|z| + 3 = 15 ≤ 16 = |z|4 = |f (z)|.
Thus, g(z) has four zeros in |z| < 2. We’ve already counted one of these, so that g(z) has three zeros
in 1 ≤ |z| < 2.
Ahlfors Exercises
4.5.2.3. How many roots of the equation z 4 + 8z 3 + 3z 2 + 8z + 3 = 0 lie in the right half plane?
Hint: Sketch the image of the imaginary axis and apply the argument principle to a large half disk.
Solution: We parametrize the imaginary axis by z(t) = it for t ∈ R. Then, the image of the
imaginary axis is given by
z(t) = (it)4 + 8(it)3 + 3(it)2 + 8(it) + 3 = t4 − 8t3 i − 3t2 + 8ti + 3.
Note that Re f (it) = t4 − 3t2 + 3 = (t2 − 3/2)2 + 3/4 > 0. It follows that travelling along the
imaginary axis produces 0 turns. Considering this, we take a semicircle of radius R with diameter
along the imaginary axis, whose circular arc extends into the right half-plane. We need only analyze
the change in argument along the circular arc – that is, for θ = −π/2 to θ = π/2. Finally, observe
that for large R the z 4 term dominates, and therefore f (Reiθ ) ≈ R4 e4iθ . As θ ranges from −π/2 to
π/2, we see that f (Reiθ ) ranges from −2π to 2π, or over an interval of length 4π. This tells us that
f winds twice, and therefore there are two roots.
4.5.3. Evaluation of Definite Integrals.
4.5.3.1. Find the poles and residues of the following functions:
a) 1/(z 2 + 5z + 6),
b) 1/(z 2 − 1)2 ,
e) 1/ sin2 z,
f) 1/(z m (1 − z)n )
c) 1/ sin z,
d) cot z,
where n, m are positive integers.
Solution:
a) We can factor z 2 + 5z + 6 as (z + 3)(z + 2), so that there are two poles at z = −3 and z = −2.
The corresponding residues are
1
1
1
1
=
= −1,
Resz=−2 2
=
= 1.
Resz=−3 2
z + 5z + 6
−3 + 2
z + 5z + 6
−2 + 3
b) Clearly the poles of 1/(z 2 − 1)2 will be at z = ±1. Both of these poles are of order 2, so we
apply the formula
Resz=a f (z) =
1
dn−1
lim n−1 ((z − a)n f (z))
(n − 1)! z→a dz
where f (z) has a pole of order n at z = a. In this case,
1
d
1
(z − 1)2
2
Resz=1
=
lim
=− ,
= − lim
2
2
2
2
3
z→1 dz
z→1 (z + 1)
(z − 1) (z + 1)
(z − 1) (z + 1)
4
d
(z + 1)2
2
1
1
= lim
= − lim
= .
Resz=−1
2
2
2
2
3
z→−1 (z − 1)
z→−1 dz
(z − 1) (z + 1)
(z − 1) (z + 1)
4
c) The zeros of sin z occur at z = nπ for integers n. Since d/dz sin z = cos z is nonzero at these
points, the poles are simple. Thus,
1
z − nπ
z
z
Resz=nπ
= lim
= lim
= (−1)n lim
= (−1)n .
z→nπ
z→0
z→0
sin z
sin z
sin(z + nπ)
sin z
d) Since cot z = cos z/ sin z, the poles are still z = nπ for integers n. Since cos z is nonzero at
these points, and the order of the zero is 1 for sin z, we see that they are still simple poles.
By the same logic as above,
z
Resz=nπ (z − nπ) cot z = lim z cot(z − nπ) = lim z cot z = [ lim cos z] lim
= 1.
z→0
z→0
z→0
z→0 sin z
e) The poles are still at z = nπ for integers n, but since we have sin2 z the order of each pole
is 2. Let us compute the residue at z = 0. Observe that
Z
dz
=0
2
|z|=1 sin z
Ahlfors Exercises
since 1/ sin2 z takes the same values on opposite sides of the circle. But, by the residue
theorem
Z
X
dz
1
1
=
Resz=aj
2πi |z|=1 sin2 z
sin2 z
j
where aj are all the isolated singularities in |z| < 1. There is only one, namely z = 0.
Thus, we find Resz=0 1/ sin2 z = 0. The other poles are equal by periodicity (e.g., the same
technique used in part d)).
f) There are two poles at z = 0 and z = 1 with orders m and n, respectively. Hence,
Resz=0
Resz=1
1
zm
1
dm−1
=
lim m−1 m
n
− z)
(m − 1)! z→0 d
z (1 − z)n
n(n + 1)(n + 2)...(n + m − 2)
n+m−2
=
=
m−1
(m − 1)!
n
n−1
1
(z − 1)
d
1
=
lim
z m (1 − z)n
(n − 1)! z→1 dn−1 z m (1 − z)n
(−1)n−1 m(m + 1)...(m + n − 2)
m+n−2
=
=
−
n−1
(−1)n (n − 1)!
z m (1
Interestingly, the two residues are equal up to a sign.
4.5.3.2. Show that in Sec. 5.3, Example 3, the integral may be extended over a right-angled isosceles
triangle. (Suggested by a student.)
Solution: Details for the semicircle method. Let ρ > 0, let γ be the semicircle of radius ρ whose
diameter lies on the real axis and arc lies in the upper half plane, and let γ 0 be the corresponding
arc. Then,
Z
Z
Z
ρ
R(x)eix dx =
−ρ
R(z)eiz dz −
γ
R(z)eiz dz.
γ0
We wish to show the final integral tends to zero in absolute value as ρ → ∞. Note that eiz =
ei(x+iy) = e−y+ix , so that |eiz | = e−y . Since 0 ≤ y ≤ ρ, we have that |eiz | ≤ 1. Suppose our rational
function is given by
(z − a1 )...(z − an )
R(z) = c
(z − b1 )...(z − bm )
with m ≥ n + 2. For any 1 ≤ i ≤ n, along γ 0 we have that
|z − ai | ≤ |z| + |ai | = ρ + |ai |.
0
For any 1 ≤ j ≤ m, along γ we have
|z − bj | ≥ ||z| − |bj || = |ρ − |bj || = ρ − |bj |
when ρ is large enough (e.g. so that all the ai and bj are in |z| < ρ). Consequently, along γ 0 we have
|R(z)| ≤ |c|
(ρ + |a1 |)...(ρ + |an |)
.
(ρ − |b1 |)...(ρ − |bm |)
As soon as m ≥ n + 1 we have that |R(z)| → 0 as ρ → ∞. Hence, the integral over γ 0 tends to zero.
We need to have m ≥ n + 2 because the |dz| (coming from the absolute value) contributes a factor
of πρ.
When using a right-hand isosceles triangle, we situate the hypotenuse so it lies on the real axis,
and so that its two legs lie in the upper half plane. Suppose the hypotenuse is the interval [−ρ, ρ];
denote the two legs by γ 0 . Then,
Z ρ
Z
Z
R(x)eix dx =
R(z)eiz dz −
R(z)eiz dz.
0
γ
γ0
We now need to estimate the last integral. To estimate |R(z)eiz | along γ 0 we use the maximum
principle. Indeed, the two legs can be bounded by a semicircular arc. It follows that the modulus
Ahlfors Exercises
of R(z)eiz along γ 0 is bounded by that along the arc. But, as above, we showed that this tends to
zero. I’m not sure if this is what Ahlfors intended for this problem – he might’ve wanted a direct
computation along the two legs, but I think that would be tedious.
4.5.3.3. Evaluate the following integrals by the method of residues:
Z π/2
Z ∞
Z ∞
dx
x2 dx
x2 − x + 2
,
c)
dx,
a)
,
|a|
>
1,
b)
4
2
x4 + 5x2 + 6
a + sin2 x
0
0
−∞ x + 10x + 9
Z ∞
Z ∞
Z ∞
x2 dx
cos x
x sin x
d)
,
a
real,
e)
dx,
a
real,
f)
dx, a real,
2
2
3
2
2
(x + a )
x +a
x2 + a2
0
0
0
Z ∞ 1/3
Z ∞
Z ∞
x
log x
log(1 + x2 )
g)
dx,
h)
dx,
i)
dx, 0 < α < 2.
2
2
1+x
1+x
x1+α
0
0
0
Solution:
a) First, recall the trig identity
sin2 (x) =
1 − cos(2x)
2
so the integral becomes
Z π/2
Z π
Z π/2
dx
dx
dx
=
=
2
a
+
(1
−
cos(2x))/2
2a
+
1
− cos x
a
+
sin
x
0
0
0
after a simple substitution. The above integral is similar to Example 1 in the text. Letting
b = −(2a + 1), we see that
2 < |2a| ≤ |2a + 1| + 1 = |b| + 1
so that |b| > 1. Now, suppose b > 1. This is exactly the integral in the text, so
Z π
π
dx
π
π
=p
.
=√
= √ √
2
2
b
+
cos
x
2
a
a+1
b −1
(2a + 1) − 1
0
If b < −1, we can use essentially the same
the residue,
√ method in the text. When computing
√
we choose instead however β = −b − b2 − 1. The residue then is −1/(2 b2 − 1, and the
integral becomes
Z π
dx
π
π
= −√
.
=− p p
2
b −1
2 |a| |a + 1|
0 b + cos x
In total,
Z
0
The integral is thus
Z π/2
0
π
dx
− sgn(a)π
= p p
.
b + cos x
2 |a| |a + 1|
dx
=−
a + sin2 x
Z
0
π
dx
sgn(a)π
= p p
.
b + cos x
2 |a| |a + 1|
b) We are integrating a rational function over −∞ to ∞, and the numerator and denominators
have the appropriate degrees. Therefore we can apply the same method as in Example 2 of
the text. Factoring the denominator gives
z 4 + 10z 2 + 9 = (z 2 + 9)(z 2 + 1) = (z + 3i)(z − 3i)(z + i)(z − i).
Note that all of these are poles since substituting z = ±3i, ±i into z 2 − z + 2 is clearly
nonzero. The necessary residues are at z = 3i, i since these have positive imaginary part.
Then,
Resz=3i
Resz=i
−9 − 3i + 2
7 + 3i
z2 − z + 2
=
=
(z + 3i)(z − 3i)(z + i)(z − i)
6i(−9 + 1)
48i
z2 − z + 2
−1 − i + 2
1−i
3 − 3i
=
=
=
.
(z + 3i)(z − 3i)(z + i)(z − i)
(−1 + 9)(2i)
16i
48i
Ahlfors Exercises
Appealing to the residue theorem gives
Z ∞
z2 − z + 2
z2 − z + 2
5π
x2 − x + 2
dx = 2πi Resz=3i 4
+ Resz=i 4
=
.
4
2
2
2
z + 10z + 9
z + 10z + 9
12
−∞ x + 10x + 9
c) As with the previous rational integrands, we can apply the same theory. First note that if
a = 0, the integral is infinite. Indeed, in this case the pole lies on the real axis. So, assume
a 6= 0. The denominator factors over C as
(z 2 + a2 )3 = (z − ai)3 (z + ai)3
so that z = ±ai are poles of order 3. WLOG assume that a > 0, hence the only pole in the
upper half plane is z = ai. The residue at this point is
1
z 2 (z − ai)3
1
z2
d2
d2
z2
=
=
lim
lim
Resz=ai
(z − ai)3 (z + ai)3
2! z→ai dz 2 (z − ai)3 (z + ai)3
2 z→ai dz 2 (z + ai)3
1
12z
12z 2
1
2
= lim
−
+
=
.
2 z→ai (z + ai)3
(z + ai)4
(z + ai)5
16a3 i
Then,
Z
0
∞
x2 dx
1
=
2
2
3
(x + a )
2
d) Observe that
Z ∞
0
∞
Z
0
cos x
1
dx =
2
2
x +a
2
2
x2 dx
= πi Resz=ai
2
(x + a2 )3
∞
Z
−∞
1
cos x
dx = Re
2
2
x +a
2
z2
2
(z + a2 )3
Z
∞
−∞
=
π
.
16a3
eix
dx
.
x2 + a2
2
Let R(z) = 1/(z + a ), which has a zero of order 2 at ∞. So, we can apply the method
used in Example 3 to conclude
Z
X
1 ∞
R(x)eix dx = πi
Res R(z)eiz .
2 −∞
y>0
The poles of R(z) occur at z = ±ai, so we only compute the residue at z = ai. Doing this
yields
e−a
eiz
eiz
=
.
=
lim
Resz=ai 2
z→ai (z + ai)
z + a2
2ai
Hence,
Z ∞
cos x
πe−a
.
dx
=
x2 + a2
2a
0
Note that the integral diverges for a = 0.
e) Similar to the previous example, we have
Z ∞
Z
Z ∞
xeix
x sin x
1 ∞ x sin x
1
dx =
dx = Im
dx .
2
2
x2 + a2
2 −∞ x2 + a2
2
−∞ x + a
0
Letting R(z) = z/(z 2 + a2 ), we see there is only a simple zero at infinity. A priori, we can
only compute the principal value via
Z ∞
X
X
pv
R(x)eix dx = 2πi
Res R(z)eiz + πi
Res R(z)eiz .
−∞
y>0
y=0
The poles of R(z) are at z = ±ai, and the residue at z = ai is
Resz=ai
zeiz
e−a
zeiz
=
lim
=
.
z→ai (z + ai)
z 2 + a2
2
Hence,
Z
∞
pv
−∞
R(x)eix dx = πe−a i.
Ahlfors Exercises
It follows that
∞
πe−a
x sin x
1
−a
dx
=
Im
πe
i
=
.
x2 + a2
2
2
0
f) Our integrand is of the form xα R(x) with 0 < α < 1 and R(x) = 1/(1 + x2 ) a rational
function. Therefore we follow the procedure in the text, without making the initial substitution. Considering z 1/3 , we exclude the negative imaginary axis and form a branch cut –
the argument should lie between −π/2(1/3) = −π/6 and 3π/2(1/3) = π/2. We now show
that the integral along semicircular arcs vanishes. Consider z(t) = Reit , with 0 ≤ t ≤ π.
Then,
Z
Z
|z|1/3
πR4/3
z 1/3
≤
dz
|dz|
=
2
2
|R2 − 1|
γR ||z| − 1|
γR 1 + z
where γR is the image of z(t). The right hand side tends to zero as R goes to infinity or 0.
Using the contour described in the text, we obtain
1/3 Z ∞
Z ∞ 1/3
Z ∞
X
z
x1/3
x
(−x)1/3
Res
2πi
=
dx
=
dx
+
dx.
2
1 + z2
1 + x2
1 + x2
−∞ 1 + x
0
0
y>0
Z
pv
We note that (−1)1/3 = (eiπ )1/3 = eiπ/3 , so that
1/3 Z ∞ 1/3
2πi X
z
x
dx =
Res
.
2
πi/3
1+x
1 + z2
1+e
0
y>0
The poles of z 1/3 /(1 + z 2 ) are at z = ±i, so we need only compute the residue at z = i. It is
1/3 i1/3
eiπ/6
z
Resz=i
=
=
.
1 + z2
2i
2i
Then, the integral is
iπ/6 Z ∞ 1/3
2πi
e
πeiπ/6
π
π
x
dx
=
=
= −iπ/6
=√
2
πi/3
iπ/3
iπ/6
1
+
x
2i
1
+
e
1
+
e
e
+
e
3
0
g) Let γ be the contour given by the rightward oriented intervals [−R, −r] and [r, R], as well
as the counterclockwise oriented upper semicircle of radius R and clockwise oriented upper
semicircle of radius −r. This produces a semicircle-like contour avoiding the origin. With
our typical branch cut of log z, this contour meets the cut. Hence we choose a different one,
namely along the negative imaginary axis. That is, we choose arg z so that −π/2 < arg z <
3π/2.
Note that f (z) = log z/(z 2 + 1) has poles at z = ±i, but only one of these is bounded by
the contour for R large enough. Hence,
Z
log z
log z
2πi Resz=i
=
dz
2+1
z2 + 1
z
γ
Z −r
Z R
Z
Z
log x
log x
log z
log z
=
dx +
dx +
dz +
dz
2
2
2
2
−R 1 + x
r 1+x
γr 1 + z
γR 1 + z
where γR and γr are the semicircular arcs of radius R and r respectively. We wish to show
that the second two vanish and the first can be absorbed into the second. For the first
integral,
Z R
Z R
Z R
Z −r
log(−x)
log x
πi
log x
dx
=
dx
=
dx
+
dx.
2
2
2
1
+
x
1
+
x
1
+
x
1
+
x2
r
r
r
−R
In the limit, the second integral becomes π 2 i/2.
Computing the second to last integral, we parametrize it by z(t) = reit with 0 ≤ t ≤ π, of
course with the opposite orientation from usual. Hence,
Z
Z π
log z
(log r + it)eit
dz
=
−ir
dt.
2
1 + r2 e2it
γr 1 + z
0
Ahlfors Exercises
Bounding this gives
Z
Z
Z π
r| log r| π
r
r| log r|π
rπ 2
log z
≤
dz
dt
+
t
dt
=
+
.
2
1 − r2 0
1 − r2 0
1 − r2
2(1 − r2 )
γr 1 + z
Letting r → 0 we see the integral vanishes. Observe that bounding the integral over γR
provides the same estimate with r replaced by R. Then, letting R → ∞, we also see the
integral vanishes.
Finally, we compute the residue of log z/(z 2 + 1) at z = i. This is easily found as
Resz=i
log z
log |z| + iθ
iπ/2
π
= lim
=
= .
2
z→i
z +i
z+i
2i
4
Z
2
log z
π i
dz =
2+1
z
2
γ
and on the other hand,
Z
Z R
Z
Z
log x
log z
π2 i
log z
log z
dz
=
2
dx
+
+
dz
+
dz.
2+1
2
2
z
1
+
x
2
1
+
z
1
+ z2
r
γ
γr
γR
In the limit, the last two integrals vanish. It follows that
Z ∞
log x
dx = 0.
1 + x2
0
h) We begin by applying integration by parts:
Z ∞
Z
∞
log(1 + x2 )
1 ∞ d/dx log(1 + x2 )
log(1 + x2 )
+
dx
=
−
dx
x1+α
αxα
α 0
xα
0
0
Z
x
2 ∞
dx.
=
α
α 0 x (1 + x2 )
There are now three cases. First, if α = 1 then the integral easily evaluates to π. Now
assume 0 < α < 1. We wish to find
Z ∞
xβ
dx
1 + x2
0
where 0 < β = 1 − α < 1. We apply the same method as in part g) and find
β Z ∞
xβ
2πi X
z
dx =
Res
.
2
βπi
1
+
x
1
+
e
1
+
z2
0
y>0
There is only one necessary residue to compute, at z = i. It is found as
β z
eiβπ/2
Resz=i
=
.
1 + z2
2i
Hence, the integral is
Z ∞
xβ
π
πeiβπ/2
π
=
.
dx
=
=
2
βπi
1
+
x
1
+
e
2
cos(βπ/2)
2
sin(απ/2)
0
Now we assume 2 > α > 1, so we wish to find
Z ∞
1
dx
β
x (1 + x2 )
0
where 0 < β = α − 1 < 1. Note the additional pole at z = 0. Let γR be a semicircle in the
upper half plane of radius R. Then,
Z
1
πR
πR2−α
dz
≤
=
.
β
2
Rα−1 |R2 − 1|
|R2 − 1|
γR z (1 + z )
Ahlfors Exercises
Since 1 < α < 2 we have 0 < 2 − α < 1, and thus the integral tends to zero as R → 0 or
R → ∞. Hence we can use the same keyhole contour method as before. The only residue
we capture is at z = i, and is computed as
e−iβπ/2
1
=
.
Resz=i
β
2
z (1 + z )
2i
Hence, the integral is Hence, the integral is
Z
∞
0
xβ
πe−iβπ/2
π
π
dx =
=
=
.
2
−βπi
1+x
1+e
2 cos(βπ/2)
2 sin(απ/2)
In total,
∞
Z
0
π
log(1 + x2 )
dx =
.
1+α
x
α sin(απ/2)
This is consistent when α = 1.
4.5.3.4. Compute
Z
|z|=ρ
|dz|
|z − a|2
where |a| =
6 ρ. Hint: Use zz = ρ2 to convert the integral to a line integral of a rational function.
Solution: As seen in Exercise 4.2.2.3., we have that |dz| = −iρdz/z. Thus,
Z
Z
Z
|dz|
−iρ dz
dz
=
=
iρ
.
2
2
|z|=ρ |z − a|
|z|=ρ z(z − a)(z − a)
|z|=ρ (z − a)(az − ρ )
There are now two cases. First, if |a| < ρ, then ρ2 /|a| > ρ, and thus there is only one pole in |z| < ρ.
The residue at this pole is
Resz=a
1
1
1
=
= 2
.
2
2
(z − a)(az − ρ )
aa − ρ
|a| − ρ2
Hence, by the residue theorem
Z
Z
dz
2πρ
|dz|
= iρ
= 2
.
2
2)
|z
−
a|
(z
−
a)(az
−
ρ
ρ
− |a|2
|z|=ρ
|z|=ρ
On the other hand, if |a| > ρ then ρ2 /|a| < ρ. Thus we still have one pole in |z| < ρ, but at a
different location. The residue here is
Resz=ρ2 /a
1
(z − ρ2 /a)
1
=
lim
= 2
.
(z − a)(az − ρ2 ) z→ρ2 /a a(z − a)(z − ρ2 /a)
ρ − |a|2
So, by the residue theorem
Z
|z|=ρ
|dz|
= iρ
|z − a|2
Z
|z|=ρ
2πρ
dz
= 2
.
(z − a)(az − ρ2 )
|a| − ρ2
Note that in both cases, the integral is positive, as expected.
4.6. Harmonic Functions.
4.6.2. The Maximum Principle.
Ahlfors Exercises
4.6.2.1. If u is harmonic and bounded in 0 < |z| < ρ, show that the origin is a removable singularity
in the sense that u becomes harmonic in |z| < ρ when u(0) is properly defined.
Solution: Recall that the arithmetic mean of u can be written as
Z
1
u dθ = α log r + β
2π |z|=r
for 0 < r < ρ and constants α, β. Since u is bounded, we have that
|α log r + β| ≤ M r
where |u| ≤ M on 0 < |z| < ρ. Taking r → 0+ , we see that α = 0. Now, recall that
Z
Z
∂u
1
1
r
dθ =
?du
0=α=
2π |z|=r ∂r
2π |z|=r
where ?du isRthe conjugate differential. Since |z| = r is a homology basis for the punctured disc, it
follows that γ ?du = 0 for any closed curve γ in the punctured disc. It follows that the conjugate
harmonic function v is single valued.
Define now f (z) = u + iv, which is analytic in the punctured disc. Since Re f (z) is bounded,
by Exercise 4.3.2.5 it follows that any isolated singularity of f (z) is removable. In this way, u has a
well defined harmonic extension.
4.6.2.2. Suppose that f (z) is analytic in the annulus r1 < |z| < r2 and continuous on the closed
annulus. If M (r) denotes the maximum of |f (z)| for |z| = r, show that
M (r) ≤ M (r1 )α M (r2 )1−α
where α = log(r2 /r)/ log(r2 /r1 ) (Hadamard’s three-circle theorem). Discuss cases of equality. Hint:
Apply the maximum principle to a linear combination of log |f (z)| and log |z|.
Solution: By taking logarithms of both sides, we are equivalently asked to prove
log M (r) ≤ α log M (r1 ) + (1 − α) log M (r2 ).
Let a > 0 and consider
| log |f (z)| + a log |z|| ≤ log |f (z)| + a log |z| = log |z a f (z)|.
We can find the maximum of this by looking at the maximum of z a f (z) instead. On the annulus
Ω = {r1 ≤ |z| ≤ r2 },
|z a f (z)| ≤ max{r1a M (r1 ), r2a M (r2 )}.
Evidently, for some a we achieve equality so that
r1a M (r1 ) = r2a M (r2 ).
Taking logarithms,
a log(r1 ) + log(M (r1 )) = a log(r2 ) + log(M (r2 )),
and solving for a yields
a=
log(M (r2 )/M (r1 ))
.
log(r1 /r2 )
Now, we have
|z a f (z)| ≤ ra M (r) ≤ r1a M (r1 )
owing to the maximum modulus principle and equality in the max. Taking logarithms,
a log r + log M (r) ≤ a log r1 + log M (r1 ).
Note that
a log r − a log r1 = a log(r/r1 ) =
log(r/r1 )
log(r/r1 )
log M (r1 ) −
log M (r2 ).
log(r2 /r1 )
log(r2 /r1 )
Ahlfors Exercises
Hence,
log(r/r1 )
log M (r) ≤ 1 −
log(r2 /r1 )
Now, if α = log(r2 /r)/ log(r2 /r1 ), we see that
log M (r1 ) +
log(r/r1 )
log M (r2 )
log(r2 /r1 )
log(r2 /r)
log(r/r1 )
=−
log(r2 /r1 )
log(r2 /r1 )
log(r2 /r)
log(r/r1 )
1−α=1−
=
log(r2 /r1 )
log(r2 /r1 )
α=
so
log M (r) ≤ α log M (r1 ) + (1 − α) log M (r2 )
as desired.
5. Series and Product Developments
5.1. Power Series Expansions.
5.1.1. Weierstrass’ Theorem.
5.1.1.1. Using Taylor’s theorem applied to a branch of log(1 + z/n), prove that
z n
lim 1 +
= ez
n→∞
n
uniformly on all compact sets.
Solution: We equivalently show that
z
=z
lim n log 1 +
n→∞
n
uniformly on all compact sets. The finite Taylor expansion of n log(1 + z/n) is
z
z
z2
z3
n log 1 +
=z−
+ 2 + ... + R
zm
n
2n 3n
n
where R(z/n) is analytic. By the Cauchy integral formula
Z
z
1
R(ζ/n)
R
=
dζ
n
2πi Cr ζ − z
for all |z| < r, where Cr is a circle of radius r centered at the origin. By a change of variables,
Z
z
R(ζ)
1
R
=
dζ
n
2πi Cr/n n(nζ − z)
Since R(ζ) is analytic, and Cr/n is compact, |R(ζ)| ≤ M on Cr/n . Thus,
Z
z
1
MR
M
R
|dζ| = 2
.
≤
n
2nπ Cr/n |n2 r − |z||
n |r − |z||
Choosing, say, |z| < r/2 we find
Z
z
M
1
2M
R
≤
|dζ| = 2 → 0
n
2nπ Cr/n |n2 r − |z||
n
where the convergence is uniform since we bound R(z/n) independently of z.
Now let Ω be compact, so that |z| < r for all z ∈ Ω and some r > 0. It follows that
z
z
r2
r3
n log 1 +
≤r+
+ 2 + ... + R
rm .
n
2n 3n
n
Taking n → ∞, we see that n log(1 + z/n) grows like |z|.
Ahlfors Exercises
5.1.1.2. Show that the series
ζ(z) =
∞
X
n−z
n=1
converges for Re z > 1, and represent its derivative in series form.
Solution: First observe that
n−z = e− Re z log n e−i Im z log n =
1
.
nRe z
Let = Re z > 1. Then,
|ζ(z)| ≤
∞
X
|n−z | ≤
n=1
∞
X
1
< ∞.
n
n=1
So, we know that ζ(z) converges on Re z > 1. Moreover, if K is any compact subset of Re z > 1,
then the series converges. Because each term fn (z) = n−z is analytic, it follows by Weierstrass’
theorem that ζ(z) is analytic on Re z > 1. Not only this, but we can differentiate the series term by
term. Hence,
∞
∞
∞
X
d −z X d −z log n X log n
n =
e
=
− z .
ζ 0 (z) =
dz
dz
n
n=1
n=1
n=1
5.1.1.3. Prove that
(1 − 21−z )ζ(z) = 1−z − 2−z + 3−z − ...
and that the latter series represents an analytic function for Re z > 0.
Solution: There is an interesting trick to this which I’m not sure will be useful anywhere else.
Recalling the alternating series test, we cannot apply it here since our terms do not monotonically
decrease (indeed, there is compatible total ordering of the complex numbers with both multiplication
and addition). Instead, we appeal
P∞ to a generalized alternating series test which states if an and bn
are complex sequences, then n=1 an bn converges when:
(1) There exists an M , independent of N , such that
N
X
bn ≤ M.
n=1
(2) The terms an tend to zero as n → ∞.
P∞
(3) The sequence of an is of bounded variation, that n=1 |an+1 − an | ≤ L < ∞.
P∞
With this in hand, let an = n−z and bn = (−1)n+1 . Clearly the series n=1 an bn represents the
series on the right hand side. The two conditions are clearly satisfied, since the partial sums of bn
alternate between zero and one, and since |n−z | = n− Re z , with Re z > 0. The third criteria is the
most interesting. Indeed, we have
Z ∞
Z ∞
Z ∞
∞
X
1
|z|
|z|
d 1
−z
1
−
dt =
≤
dt
=
dt
=
z
z
z
z+1
1+Re
z
(n + 1)
n
dt t
t
t
Re z
1
1
1
n=1
where we recognize the first series as an approximation of the arc length of the curve 1/tz . Since
Re z > 0, this is finite. Hence, the series in question converges for Re z > 0, and also on compact
subsets. Since each term is analytic, it represents an analytic function.
We are thus left to prove equality, which only makes sense on the domain Re z > 1. In this case,
#
"
∞
∞
∞
X
X
X
X
X
−z
−z
1−z
−z
(1 − 2 )
n =
n +
n
−2
(2n)−z =
(−1)n+1 n−z
n=1
as desired.
n odd
n even
n=1
n=1
Ahlfors Exercises
5.1.1.4. As a generalization of Theorem 2, prove that if the fn (z) have at most m zeros in Ω, then
f (z) is either identically zero or has at most m zeros.
Solution: Suppose f is not identically zero. Let {z1 , z2 , ...} be the distinct zeros of f . Fix an i ∈ N.
We form a circle C of radius ρ around zi such that f (z) 6= 0 on 0 < |z − zi | < ρ. By the maximum
modulus principle, |f (z)| > 0 on C. Let δ > 0 be such that |f (z)| > δ on C. Since fn (z) → f (z)
uniformly on compact sets, there exists an N such that if n ≥ N then |fn (z) − f (z)| < δ/2 on C.
Hence,
δ < |f (z)| ≤ |fn (z) − f (z)| + |fn (z)| < δ/2 + |fn (z)|
and |fn (z)| > δ/2 on C. Since g(z) = 1/z is uniformly continuous on |z| > δ/2, |fn (z)| > δ and
|f (z)| > δ/2, and the fn converge uniformly to f on C, we have that 1/fn (z) converges uniformly to
1/f (z) on C. For full details, see the end of this solution. By Weierstrass’ theorem, we additionally
have fn0 → f 0 uniformly on C. Finally, because 1/f and f 0 are bounded on C, the product fn0 /fn
also uniformly converges. Consequently,
Z
Z
f 0 (z)
1
fn0 (z)
1
dz =
lim
dz
2πi C f (z)
2πi n→∞ C fn (z)
n
} be the distinct zeros of fn , where mn ≤ m for all n.
Let {z1n , ..., zm
n
We now show that if g is a uniformly continuous function on E and fn → f uniformly on Ω, then
g ◦ fn → g ◦ f uniformly on Ω. Let > 0, then there exists a δ > 0 such that |g(z) − g(w)| < whenever |z − w| < δ. With this δ, by uniform convergence of the fn there exists an N such that if n ≥ N
then |fn (z)−fn (w)| < δ. Combining the two, we see that if n ≥ N then |g(fn (z))−g(fn (w))| < for
all z, w ∈ Ω, asserting uniform convergence. We now show that g(z) = 1/z is uniformly continuous
on subsets of |z| > m > 0. Clearly we have
1
1
|z − w|
|z − w|
−
=
<
z
w
|z||w|
m2
whenever |z| > m and |w| > m. So, if δ = /m2 , we are done.
5.1.1.5. Prove that
∞
∞
X
X
nz n
zn
=
1 − zn
(1 − z n )2
n=1
n=1
for |z| < 1. (Develop in a double series and reverse the order of summation.)
Solution: We recognize in the first sum a term that looks like 1/(1 − z), which can be written
as a geometric series. Since |z| < 1, we have |z n | < 1 for any fixed n. Exploiting this allows us to
rewrite the series on the left as a doubly indexed sum:
∞
∞
∞
∞ X
∞
∞ X
∞
X
X
X
X
X
nz n
n
nk
n(k+1)
=
nz
z =
nz
=
nz nk .
n
1
−
z
n=1
n=1
n=1
n=1
k=0
k=0
k=1
It is not immediately obvious to me that we can interchange the order of summation – indeed, the
series is not absolutely convergent. However, proceeding forth we get
∞ X
∞
X
n=1 k=1
nz nk =
∞ X
∞
X
k=1 n=1
nz nk =
∞ X
∞
X
nz nk .
k=1 n=0
Now, notice that
∞
∞
X
X
z d X kn
z =z
nz kn−1 =
nz kn
k dz n=0
n=0
n=0
Ahlfors Exercises
for any k ≥ 1. We can also evaluate the left hand side via a geometric series,
z
zk
z d X kn
z d
1
kz k−1
=
=
.
z =
k
k
2
k dz n=0
k dz 1 − z
k (1 − z )
(1 − z k )2
In total,
∞
∞ X
∞
X
X
zk
nz n
z d X kn
z d
1
nk
=
=
.
nz
=
z
=
1 − zn
k dz n=0
k dz 1 − z k
(1 − z k )2
n=1
n=0
k=1
5.1.2. The Taylor Series.
5.1.2.1. Develop 1/(1 + z 2 ) in powers of z − a, a being a real number. Find the general coefficient
and for a = 1 reduce to simplest form.
Solution: Recall by the geometric series that
n X
∞ ∞
1
1
1
1X
z−a
(−1)n (z − a)n
=
=
−
=
b + (z − a)
b 1 + (z − a)/b
b n=0
b
bn+1
n=0
which converges when |z − a| < b. Now, notice that
1
1
1
1
1
1
1
−
=
−
.
=
1 + z2
2i z − i z + i
2i (a − i) + (z − a) (a + i) + (z − a)
So, we apply the above twice with b = a − i then b = a + i. This gives
1
1
=
2
1+z
2i
=
1
2i
=
1
2i
!
∞
∞
X
(−1)n (z − a)n X (−1)n (z − a)n
−
(a − i)n+1
(a + i)n+1
n=0
n=0
!
∞ X
(a + i)n+1 − (a − i)n+1
(−1)n (z − a)n
(a − i)n+1 (a + i)n+1
n=0
!
∞ X
(a + i)n+1 − (a − i)n+1
n
n
(−1) (z − a)
(a2 + 1)n+1
n=0
By the binomial theorem we have
(a + i)n+1 − (a − i)n+1 =
n+1
X
k=0
=2
n+1 n + 1 k n+1−k X n + 1
(−1)k ik an+1−k
i a
−
k
k
k=0
X
n + 1 k n+1−k
i a
k
1≤k≤n+1, odd
=2
N X
n+1
k=0
2k + 1
2k+1 n−2k
i
a
= 2i
N X
n+1
k=0
2k + 1
(−1)k an−2k
where N is the largest integer such that 2N +1 ≤ n+1. In the case a = 1, this simplifies dramatically
to 2i[2(n+1)/2 sin(π(n + 1)/4)]. One way of showing this is to probably write out the difference in
terms of exponentials instead. I found this by trial and error. Thus, for a = 1 we have
∞
X
(−1)n sin(π(n + 1)/4)
1
=
(z − 1)n
(n+1)/2
1 + z2
2
n=0
5.1.2.2. The Legendre polynomials are defined as the coefficients Pn (α) in the development
(1 − 2αz + z 2 )−1/2 = 1 + P1 (α)z + P2 (α)z 2 + ...
Find P1 , P2 , P3 , and P4 .
Ahlfors Exercises
Solution: The simplest way to do this is likely by differentiation. Letting f (z) = (1 − 2αz + z 2 )−1/2 ,
we have
1
α−z
d
=
f 0 (z) =
2
1/2
dz (1 − 2αz + z )
(1 − 2αz + z 2 )3/2
−(1 − 2αz + z 2 )3/2 − 3(α − z)(1 − 2αz + z 2 )1/2 (z − α)
(1 − 2αz + z 2 )3
1
3(α − z)2
=−
+
2
3/2
(1 − 2αz + z )
(1 − 2αz + z 2 )5/2
f 00 (z) =
3(z − α)
6(z − α)(1 − 2αz + z 2 )5/2 − 15(z − α)3 (1 − 2αz + z 2 )3/2
+
(1 − 2αz + z 2 )5
(1 − 2αz + z 2 )5/2
3
9(z − α)
15(z − α)
=
−
2
5/2
(1 − 2αz + z )
(1 − 2αz + z 2 )7/2
f 000 (z) =
9(1 − 2αz + z 2 )5/2 − 45(z − a)2 (1 − 2αz + z 2 )3/2
(1 − 2αz + z 2 )5
45(z − α)2 (1 − 2αz + z 2 )7/2 − 105(z − α)4 (1 − 2αz + z 2 )5/2
−
(1 − 2αz + z 2 )7
90(z − α)2
105(z − α)4
9
−
+
=
2
5/2
2
7/2
(1 − 2αz + z )
(1 − 2αz + z )
(1 − 2αz + z 2 )9/2
f 0000 (z) =
Consequently,
f 0 (0) = α,
f 00 (0) = 3α2 − 1,
f 000 (0) = 15α3 − 9α,
f 0000 (0) = 105α4 − 90α2 + 9.
The associated Legendre polynomials are thus
5α3 − 3α
3α2 − 1
,
P3 (α) =
,
2
2
5.1.2.3. Develop log(sin z/z) in powers of z up to the term z 6 .
P1 (α) = α,
P2 (α) =
P4 (α) =
35α4 − 30α2 + 3
.
8
Solution: The Taylor expansion of sin z is
z3
z5
z7
+
−
+ ...
3!
5!
7!
Next, we write log(sin z/z) as log(1 + [sin z/z − 1]), where the Taylor expansion for f (z) = sin z/z − 1
is easily derived from the above as
sin z = z −
z4
z6
z2
+
−
+ ...
3!
5!
7!
Set P (z) = −z 2 /3! + z 4 /5! − z 6 /7!. Now, the Taylor expansion of log(1 + z) (choosing the branch
which is zero at the origin) is
sin z/z − 1 = −
z2
z3
z4
+
−
+ ...
2
3
4
Since we want up to the z 6 term, we can neglect anything beyond it. In particular, since the
expansion of f (z) starts with a z 2 term, we can ignore everything beyond the z 3 term in log(1 + z).
Thus we let Q(z) = z − z 2 /2 + z 3 /3. Furthermore, we included the z 6 term in P (z) since log(1 + z)
starts with a linear term. Now, expanding Q(P (z)) gives
2
2
2
2
3
z
z4
z6
1
z
z4
z6
1
z
z4
z6
Q(P (z)) = − +
−
−
− +
−
+
− +
−
3!
5!
7!
2
3!
5!
7!
3
3!
5!
7!
2
4
6
4
6
6
z
z
z
1 z
2z
1
z
= − +
−
−
−
+ [z 7 ] +
− 3 + [z 7 ]
3!
5!
7!
2 3!2
3!5!
3
3!
z2
z4
z6
=− −
−
+ [z 7 ]
6
180 2835
log(1 + z) = z −
Ahlfors Exercises
It follows that the first three terms form the Taylor expansion of log(sin z/z) in powers of z up to
the z 6 term.
5.1.2.4. What is the coefficient of z 7 in the Taylor development of tan z?
Solution: We continue the development presented in the text. Namely, since
z5
z7
z3
+
−
+ [z 9 ],
3
5
7
we can find the Taylor expansion of z = tan w by repeated substitution of
arctan z = z −
z3
z5
z7
−
+
− [z 9 ]
3
5
7
into itself. Consider the relevant integer partitions of 7:
z=w+
7 = 7, 1 + 1 + 5, 1 + 2 + 4, 2 + 2 + 3, 1 + 3 + 3, 1 + 1 + 1 + 1 + 3, 1 + 1 + 1 + 2 + 2
Only four of these are relevant: 7, 1 + 1 + 5, 1 + 3 + 3, 1 + 1 + 1 + 1 + 3, since these contain only odd
terms; so too does our expansion for z. The w7 terms corresponding to these are
w7
w7 w7
w7
, − ,
, −
7
15
27
15
taking into account permutations of these sums. XXX
5.1.2.5. The Fibonacci numbers are defined by c0 = 0, c1 = 1,
cn = cn−1 + cn−2 .
Show that the cn are Taylor coefficients of a rational function, and determine a closed expression
for cn .
Solution: Consider the function P (z) =
P (z) =
∞
X
P∞
n=0 cn z
cn z n = c0 z + c1 z +
n=0
=z+
n
. We then have
∞
X
cn z n
n=2
∞
X
(cn−1 + cn−2 )z n = z + z
∞
X
cn z n + z 2
n=0
n=2
X
cn z n
n=0
2
= z + zP (z) + z P (z)
Rearranging this gives
z
.
1 − z − z2
Next, note the sequence obeys the matrix product
1 1
cn−1
cn
=
1 0
cn−2
cn−1
P (z) =
Repeated application of this then yields
n−1 1 1
c1
cn
=
1 0
c0
cn−1
for n ≥ 1. We now diagonalize the 2 × 2 matrix. The eigenvalues are roots of
1−λ 1
det
= (1 − λ)(−λ) − 1 = λ2 − λ − 1 = 0,
1
−λ
√
√
√
i.e. λ = 1/2(1 ± 5). Note that if ϕ = 1/2(1 + 5) then 1/2(1 − 5) = −ϕ−1 . Manual verification
shows the eigenvectors respectively are
−1 ϕ
−ϕ
e1 =
,
e2 =
1
1
Ahlfors Exercises
and since
1 −ϕ−1
1 ϕ
c1
−√
=√
1
c0
5 1
5
it follows that
cn
1
=
cn−1
1
n−1 n−1 1 1
1 1 1
ϕ
1
c1
−√
=√
1
0
c0
5 1 0
5 1
ϕn−1 ϕ
(−ϕ−1 )n−1 −ϕ−1
√
= √
−
1
1
5
5
n−1 −1 1
−ϕ
0
1
owing to the fact the two vectors are eigenvectors. Hence,
cn =
ϕn − (−ϕ)−n
√
5
5.2. Partial Fractions and Factorization.
5.2.1. Partial Fractions.
5.2.1.1.
5.2.1.2. Express
∞
X
n=−∞
z3
1
− n3
in closed form.
Solution: We can factor z 3 − n3 as
(z 3 − n3 ) = (z − n)(z − ne2πi/3 )(z − ne4πi/3 ).
Let α = e2πi/3 . Notice that
(z − nα)(z − n/α) + (z − n)(z − n/α) + (z − n)(z − nα) = 3z 2 − 2n(1/α + 1 + α)z + n2 (1/α + 1 + α)
but since α is a root of unity, we have 1 + α + α2 = 0. Thus,
(z − nα)(z − n/α) + (z − n)(z − n/α) + (z − n)(z − nα) = 3z 2 .
We therefore obtain the partial fraction decomposition
3z 2
1
1
1
1
1/α
α
=
+
+
=
+
+
.
3
3
z −n
z − n z − nα z − n/α
z − n z/α − n αz − n
Recalling that
π cot(πz) = lim
m→∞
m
X
1
,
z
−
n
n=−m
by summing the above identity from n = −m to m and taking limits, we obtain
m
m
m
X
X
X
3z 2
1
1
1
1
=
lim
+
lim
+
α
lim
3
3
m→∞
m→∞
m→∞
z −n
z − n α m→∞ n=−m z/α − n
αz − n
n=−m
n=−m
n=−m
π
= π cot(πz) + cot(πz/α) + πα cot(απz)
α
Consequently,
∞
X
1
π cot(πz) + π/α cot(πz/α) + πα cot(απz)
=
.
3 − n3
z
3z 2
n=−∞
lim
m
X
Note: We technically must, in each of the three sums prior, add a constant term like 1/n, 1/(αn),
and α/n respectively to ensure the sums converge absolutely. Then, we can rearrange them into the
valid cotangent terms.
Ahlfors Exercises
5.2.1.3. Use (13) to find the partial fraction development of 1/ cos(πz), and show that it leads to
π/4 = 1 − 1/3 + 1/5 − 1/7 + ...
Solution: Recall (13) states
m
X
π
(−1)n
= lim
.
sin(πz) m→∞ n=−m z − n
Since cos(πz) = sin(π(z + 1/2)), we obtain
m
m
X
X
π
(−1)n
(−1)n
= lim
= 2 lim
.
m→∞
cos(πz) m→∞ n=−m z + 1/2 − n
2z + 1 − 2n
n=−m
Thus, the partial fraction decomposition of 1/ cos(πz) is
m
m
X
X
1
(−1)n
2
(−1)n
= lim
=
lim
.
cos(πz) m→∞ n=−m z + 1/2 − n
π m→∞ n=−m 2z + 1 − 2n
Letting z = 0, we obtain
π
= lim
m→∞
2
= lim
m→∞
= lim
m→∞
!
m
0
X
X
(−1)n
(−1)n
= lim
+
m→∞
1 − 2n n=−m 1 − 2n
n=1
!
m−1
m
X −(−1)n
X
(−1)n
+
1 − 2(n + 1) n=0 2n + 1
n=0
!
!
m−1
m
m−1
X (−1)n
X
X (−1)n
(−1)n
(−1)m
+
= lim 2
+
m→∞
2n + 1 n=0 2n + 1
2n + 1 2m + 1
n=0
n=0
m
X
(−1)n
1 − 2n
n=−m
!
Since |(−1)m /(2(2m + 1))| → 0 as m → ∞, we finally get
!
m−1
m−1
X (−1)n
X (−1)n
π
(−1)m
= lim
+
= lim
m→∞
m→∞
4
2n + 1 2(2m + 1)
2n + 1
n=0
n=0
as desired.
5.2.1.4. What is the value of
∞
X
1
?
2 + a2
(z
+
n)
n=−∞
Solution: Note the factorization
(z + n)2 + a2 = (z + n + ia)(z + n − ia)
which leads to the partial fraction decomposition
1
1
1
1
=
−
.
(z + n)2 + a2
2ia z − ia + n z + ia + n
Summing from n = −m to m, applying an index change, and taking limits gives
m
m
X
X
1
1
1
1
1
=
lim
−
lim
2 + a2
m→∞
m→∞
m→∞
(z
+
n)
2ia
(z
−
ia)
−
n
2ia
(z
+
ia)
−n
n=−m
n=−m
n=−m
π
=
(cot(π(z − ia)) − cot(π(z + ia))) .
2ia
lim
m
X
5.2.2. Infinite Products.
Ahlfors Exercises
5.2.2.1. Show that
∞ Y
1
n2
1−
n=2
1
.
2
=
2
Solution: Let an = −1/n for n ≥ 1. We therefore have an infinite product of the following form
Y
∞ ∞
Y
1
1− 2 =
(1 + an ) .
n
n=2
n=2
Hence, the series converges simultaneously with the series
X
∞
∞
∞
X
X
(n − 1)(n + 1)
1
.
log
log(1 + an ) =
log 1 − 2 =
n
n2
n=2
n=2
n=2
At this point, we evaluate the partial sums directly:
X
N
N
N
N
X
X
X
(n − 1)(n + 1)
log
=
log(n
−
1)
+
log(n
+
1)
−
2
log(n)
n2
n=2
n=2
n=2
n=2
=
N
−1
X
N
+1
X
log(n) +
n=1
log(n) − 2
n=3
N
X
log(n)
n=2
= log(1) − log(2) + log(N + 1) + 2
N
−1
X
log(n) − 2
N
X
log(n)
n=3
n=3
1
− log(2)
= log(N + 1) − log(N ) − log(2) = log 1 −
N
Evidently, this tends to − log(2) as N → ∞. We therefore establish the infinite product converges,
and moreover that
! X
∞
∞ Y
1
1
log 1 − 2 = − log(2)
=
log
1− 2
n
n
n=2
n=2
as desired.
5.2.2.2. Prove that for |z| < 1
(1 + z)(1 + z 2 )(1 + z 4 )(1 + z 8 )... =
1
.
1−z
Solution: Consider the infinite product
P =
∞
Y
n
(1 + z 2 ).
n=0
Let N > 0 and consider the partial product PN . We claim that
PN =
N
Y
2n
(1 + z ) =
+1
2NX
−1
n=0
zn.
n=0
Clearly this holds when N = 1. Suppose it holds for N . Then by definition of PN +1 and the
inductive hypothesis,
PN +1 = 1 + z
2N +1
PN = 1 + z
2N +1
+1
−1
2NX
n
z =
n=0
=
+1
2NX
−1
n=0
zn +
N +1
2N +1 +2
X −1
n=2N +1
zn =
+1
2NX
−1
n=0
+1
2NX
−1
n
z +
n=0
zn +
+2
2NX
−1
n=2N +1
+1
2NX
−1
N +1
z n+2
n=0
zn =
+2
2NX
−1
zn
n=0
as desired. Hence the partial products form a geometric series, and it follows easily from this that
P = 1/(1 − z) when |z| < 1.
Ahlfors Exercises
5.2.2.3. Prove that
∞ Y
n=1
1+
z −z/n
e
n
converges absolutely and uniformly on every compact set.
Solution: Let an be defined by
an = e−z/n +
ze−z/n
−1
n
so that the above product can be written as
∞ ∞
Y
z −z/n Y
1+
e
=
(1 + an ) .
n
n=1
n=1
P∞
It follows that this converges absolutely so long as n=1 |an | does. Let r = Re z; if r = 0 then we
easily have convergence, since the product is just of complex numbers on the unit circle. As n → ∞,
the polar angle tends to zero, and thus for large n there is little rotation. If r 6= 0 we have
e−z/n = e−r/n → 1
as n → ∞. It follows that e−z/n − 1 → 0 as n → ∞. Furthermore,
|z|e−r/n
C
ze−z/n
≤
≤
→0
n
n
n
where C is some constant making the inequality valid for large n. XXX
5.2.2.4. Prove that the value of an absolutely convergent product does not change if the factors are
reordered.
Q∞
Solution:
P∞Recall a product n=1 (1 + an ) converges absolutely if and only if the corresponding
series n=1 |an | converges (absolutely).
If σ(n) is a permutation of N, we therefore
have by the
P∞
Q∞
Riemann Rearrangement theorem that n=1 |aσ(n) | converges. It follows that n=1 (1 + aσ(n) ) converges. Denote the original product’s value by L and this new product’s value by L0 . We wish to
show L = L0 . XXX
5.2.3. Canonical Products.
5.2.3.1. Suppose that an → ∞ and that the An are arbitrary complex numbers. Show that there
exists an entire function f (z) which satisfies f (an ) = An . Hint: Let g(z) be a function with simple
zeros at the an . Show that
∞
X
eγn (z−an ) An
.
g(z)
z − an g 0 (an )
n=1
converges for some choice of the numbers γn .
Solution: First let us see how the hint is useful. Indeed, if g(z) has simple zeros at the an , then each
g(z)/(z − an ) is an entire function. Consequently, each term in the series is an entire function, and
by Weierstrass’ theorem the series itself is entire. Next, by L’hopitals rule we have
g(z)
g 0 (z)
= lim
= g 0 (an )
z→an z − an
z→an
1
so if f (z) denotes the series, then f (an ) = An . Thus we need only show convergence of the above
series for a choice of γn . To achieve this, we show convergence on every disc of radius R. Let hn (z)
denote the function
g(z)An
hn (z) = 0
g (z)(z − an )
so that the series is
∞
X
eγn (z−an ) hn (z).
lim
n=1
Ahlfors Exercises
If we choose γn appropriately so the series is bounded by a geometric series, then we show uniform
convergence. Let DR denote the disc of radius R; we want the γn to be such that
eγn (z−an ) ≤
2−n
2−n
=
supz∈DR |hn (z)|
supz∈CR |hn (z)|
where CR is the circle |z| = R. What this tells is is the real part of γn (z − an ) must be negative,
since |ez | = eRe z , which must be small. Since the an → ∞, we can conceivably hope for this. Let
N be such that for all n ≥ N , |an | ≥ 2R. Set θn = arg(an ) and let γn = ηn e−θn for ηn > 0. Write
z ∈ CR as z = Reiϕ . We find then that
γn (z − an ) = Rηei(ϕ−θn ) − |an |ηn .
Notice first that Rηei(ϕ−θn ) is a point on |w| = Rη, and as ϕ varies we traverse this circle. By
subtracting |an |ηn , we translate this circle to the left. Moreover, |an |ηn ≥ 2Rηn so that
Re(γn (z − an )) ≤ Rηn − |an |ηn ≤ −Rηn .
Hence, if ηn is large enough so that
e−Rηn ≤
2−n
supz∈CR |hn (z)|
then we are done. Choosing ηn for n ≥ N obeying this, we split the series into a finite sum and an
infinite series dominated by a convergent geometric series.
5.2.3.2. Prove that
sin(π(z + α)) = e
πz cot(πα)
∞ Y
1+
n=−∞
z
n+α
e−z/(n+α)
whenever α is not an integer. Hint: Denote the factor in front of the canonical product by g(z) and
determine g 0 (z)/g(z).
Solution: The zeros of sin(π(z + α)) are given by z = n − α for n ∈ Z. Since α is not an integer, the sin(π(z + α)) is nonzero at z = 0. Letting an = 1/(n − α), we see that
∞
∞
X
X
1
1
≈
=∞
|a
|
n
n
n=−∞
n=−∞
whereas
∞
X
∞
X
1
1
≈
< ∞.
2
|an |
n2
n=−∞
n=−∞
Hence, we write
sin(π(z + α)) = e
h(z)
∞ Y
n=−∞
z
1−
n−α
ez/(n−α) .
By an index change n 7→ −n we get
sin(π(z + α)) = eh(z)
∞ Y
1+
n=−∞
z
n+α
e−z/(n+α) ,
which is almost what we want. We show now that h(z) = eπz cot(πα) . For fixed n, we have that
d
z
1
e−z/(n+α)
z
e−z/(n+α) 1 +
= e−z/(n+α)
−
1+
dz
n+α
n+α
n+α
n+α
−z/(n+α)
−z/(n+α)
e
z
ze
=
1− 1+
=−
n+α
n+α
(n + α)2
So, taking the derivative on both sides in the Weierstrass expansion of sin(π(z + α)), we get
∞
X
ze−z/(k+α) Y
z
0
h(z)
π cos(π(z + α)) = h (z) sin(π(z + α)) − e
1+
e−z/(n+α) .
(k + α)2
n+α
k=−∞
n6=k
Ahlfors Exercises
The logarithmic derivative is
π cos(π(z + α))
π cot(π(z + α)) =
sin(π(z + α))
Q
z
∞
1 + n+α
e−z/(n+α)
−z/(k+α)
X
n6
=
k
h0 (z) sin(π(z + α))
ze
=
− e−h(z) eh(z)
sin(π(z + α))
(k + α)2 Q∞
1+ z
e−z/(n+α)
k=−∞
= h0 (z) −
0
= h (z) −
∞
X
−z/(k+α)
ze
(k + α)2 1 +
k=−∞
∞
X
k=−∞
z
k+α
1
n=−∞
n+α
e−z/(k+α)
∞ X
z
1
1
1
0
= h (z) +
−
(k + α) (k + α) + z
z + (k + α) k + α
k=−∞
Recall the series expansion for π cot(πz) is
1 X
1
1
1 X
1
1
π cot(πz) = +
+
= +
−
.
z
z−k k
z
z+k k
k6=0
k6=0
Hence, the series for π cot(π(z + α)) is
X
1
π cot(π(z + α)) =
+
z+α
k6=0
1
1
−
z + (α + k) k
.
We therefore see that
∞ X
X
1
1
1
1
1
0
+
−
= h (z) +
−
z+α
z + (α + k) k
z + (k + α) k + α
k=−∞
k6=0
1
1 X
1
1
0
= h (z) +
− +
−
z+α α
z + (k + α) k + α
k6=0
and, consequently
1 X
h (z) = +
α
0
k6=0
1
1
−
k+α k
.
Recalling our formula for π cot(πz), we find that h0 (z) = π cot(πα). Thus, h(z) = π cot(πα) + c for
some complex number c, and
∞ Y
z
sin(π(z + α)) = Aeπz cot(πα)
1+
e−z/(n+α)
n
+
α
n=−∞
with A = ec . This is actually a mistake in the book – we must further determine this constant A.
When z = 0, we easily find A = sin(πα). Thus,
∞ Y
z
πz cot(πα)
sin(π(z + α)) = sin(πα)e
1+
e−z/(n+α) .
n
+
α
n=−∞
√
5.2.3.3. What is the genus of cos z?
√
Solution: We first find the zeros of cos z, and to do so it suffices to find the zeros of cos z. Recall
that
| cos z|2 = cosh2 (y) − sin2 (x)
where z = x + iy as usual. Note that 0 ≤ sin2 (x) ≤ 1 and cosh2 ≥ 1 so that | cos z|2 = 0 if and only
if cosh2 (y) = 1 and sin2 (x) = 1. The former only happens
√ when y = 0,√while the latter occurs when
x = π/2 + nπ = π/2(2n + 1) for integers k. So, if cos z = 0 then z = π/2(2n + 1) and hence
z = π 2 /4(2n + 1)2 . However, these roots are double counted for n ≤ −1, so we require n ≥ 0. Then,
∞
X
∞
X
4
4
≤
<∞
2 (2n + 1)2
2 n2
π
π
n=0
n=1
Ahlfors Exercises
so the canonical product for cos
√
z has genus zero. We can therefore write cos
∞ Y
√
4z
.
1− 2
cos z = eg(z)
π (2n + 1)2
n=0
√
z as
To find the genus, we must find g(z). By a change of variables we get
∞ Y
p
4(z − π/2)2
h(z)
2
sin z = cos(z − π/2) = cos (z − π/2) = e
1− 2
π (2n + 1)2
n=0
∞
∞ Y
Y
2z − π
2π(n + 1) − 2z
2z − π
2nπ + 2z
h(z)
h(z)
=e
1−
1+
=e
π(2n + 1)
π(2n + 1)
π(2n + 1)
π(2n + 1)
n=0
n=0
∞
∞
∞
∞ Y 2π(n + 1) − 2z Y 2nπ + 2z
zeh(z) Y 2π(n + 1) − 2z Y 2nπ + 2z
= eh(z)
=
π(2n + 1)
π(2n + 1)
π n=0
π(2n + 1)
π(2n + 1)
n=0
n=0
n=1
P
where h(z) = g((z −π/2)2 ). All the above product manipulations are valid since the sum n |z|/|an |
is convergent.
5.2.3.4. If f (z) is of genus h, how large and how small can the genus of f (z 2 ) be?
Solution: If the zeros of f (z) are denoted zk = Rk eiθk , then the zeros of f (z) are ak =
So, if f (z) is of genus h then
∞
∞
X
X
1
1
=
η+1
η+1
|zk |
Rk
k=1
√
Rk eiθk /2 .
k=1
converges for η = h but diverges for η < h. The corresponding series for f (z 2 ) is
∞
X
k=1
∞
X
1
1
.
=
η+1
η/2+1/2
|ak |
R
k=1
k
We want to find the smallest integer η such that η/2+1/2 ≥ h+1. Evidently, this is η = 2h+1. Thus
the genus of the canonical product is 2h − 1. Now, if g(z) in the canonical product representation
of f (z) is a degree h polynomial, then g(z 2 ) is a degree 2h polynomial. It follows that the genus of
f (z 2 ) is either 2h + 1 or 2h.
5.2.3.5. Show that if f (z) is of genus 0 or 1 with real zeros, and if f (z) is real for real z, then all
zeros of f 0 (z) are real. Hint: Consider Im f 0 (z)/f (z).
Solution: Suppose first f (z) is of genus zero. Then we can write f (z) as
∞ Y
z
m
f (z) = Cz
1−
an
n=1
P
with
1/|an | < ∞, m ≥ 0, and all the an are real. The logarithmic derivative is
P∞
Q∞
Q∞
Cmz m−1 n=1 (1 − z/an ) Cz m k=1 1/ak n=1,6=k (1 − z/an )
f 0 (z)
Q
Q
=
−
∞
∞
f (z)
Cz m n=1 (1 − z/an )
Cz m n=1 (1 − z/an )
∞
∞
∞
m X 1
1
m X 1
mz X z − ak
=
−
=
+
= 2+
z
ak 1 − z/ak
z
z − ak
|z|
|z − ak |2
k=1
k=1
k=1
Since the ak are real, Im(z − ak ) = Im(z) = − Im(z). Thus,
!
0 ∞
X
f (z)
m
1
Im
=−
+
Im(z).
f (z)
|z|2
|z − ak |2
k=1
Notice the term in the parenthesis is strictly positive. Hence, if Im(f 0 (z)/f (z)) = 0 then Im(z) = 0.
Suppose z is a common zero of f (z) and f 0 (z) – then z is real. On the other hand, if f 0 (z) = 0 but
f (z) 6= 0 then Im(f 0 (z)) = Im(f 0 (z)/f (z)) = 0. Hence, by the above, Im(z) = 0 and z is real.
Ahlfors Exercises
Suppose now f (z) is of genus one. Then we can write f (z) as
∞ Y
z
m αβz
f (z) = Cz e
1−
eγz/an
a
n
n=1
where all the an are real, α, C ∈ C, and β, γ ∈ {0, 1}. The logarithmic derivative is
Q∞
Q∞
αβCz m eαβz n=1 (1 − z/an )eγz/an
Cmz m−1 eαβz n=1 (1 − z/an )eγz/an
f 0 (z)
Q
Q
+
=
∞
∞
f (z)
Cz m eαβz n=1 (1 − z/an )eγz/an
Cz m eαβz n=1 (1 − z/an )eγz/an
P
Q
∞
∞
Cz m eαβz k=1 (γ − 1 − γz/ak )eγz/ak /ak n=1,6=k (1 − z/an )eγz/an
Q
+
∞
Cz m eαβz n=1 (1 − z/an )eγz/an
∞
∞
X γ − 1 − γz/ak
X
m
m
γz/ak + 1 − γ
=
+ αβ +
=
+ αβ +
z
ak (1 − z/ak )
z
z − ak
k=1
k=1
We must now make the distinction between the γ = 0 and γ = 1 cases:
(m
P∞
1
γ=0
f 0 (z)
k=1 z−a
z + αβ +
P∞ 1 k
= m
1
f (z)
γ=1
k=1 ak + z−ak
z + αβ +
P
where the convergence of k 1/ak is assumed in the γ = 1 case.
5.5. Normal Families.
5.5.5. The Classical Definition.
5.5.5.1. Prove that in any region Ω the family of analytic functions with positive real part is normal.
Under what added condition is it locally bounded? Hint: Consider the functions e−f .
Solution: Let F be the family defined by
F = {f : Ω → C | Re(f ) > 0, f is analytic}.
We wish to show F is normal in the classical sense. By Theorem 17, this amounts to showing
ρ(f ) = 2|f 0 (z)|/(1 + |f (z)|2 ) is locally bounded. Consider the map