# Ahlfors Exercises (1)

```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| &lt; 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 &plusmn;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
&plusmn;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| &lt; 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| &lt; 1. This is
a circle of radius 1 centered at 1. Importantly, Re f (z) &gt; 0, and −π &lt; arg f (z) &lt; 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 = &plusmn;i,
we arrive at
Z
1
dz
=
(2πi − 2πi) = 0.
2+1
z
2i
|z|=2
It is important that &plusmn;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| &lt; ρ and/or ρ2 /|a| &lt; ρ. The latter inequality reduces to ρ &lt; |a|.
So, there are two cases: |a| &lt; ρ and |a| &gt; 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| &lt; ρ
|dz|
=
.
2
2
2
2πρ/(|a| − ρ )
|a| &gt; 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 &gt; 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 &lt; 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 &gt; m, we interpret the above to be zero. Now suppose m &lt; 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 &gt; 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| &lt; ρ, then 1/(z − ρ2 /a)2 is analytic inside |z| = ρ, and if |a| &gt; ρ then
1/(z − a)2 is analytic inside |z| = ρ. We assume |a| &lt; 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| &gt; 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 &gt; 0 such that |f (z)| ≤ |z|n for all |z| &gt; R. Let r &gt; 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| &lt; r. Of course, for any a we can find an r such that |a| &lt; r and R &lt; 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 &lt; 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| &lt; 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)| &gt; 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 | &gt; 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 &gt; 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 &gt; 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)| &gt; 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 &gt; 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 &gt; h, and
b) limz→a |z − a|α |f (z)| = ∞ for all real α &lt; h.
Consider now the function f g. Suppose α &gt; h + k. Then we can write α as α1 + α2 with α1 &gt; h
and α2 &gt; k. Indeed, if = α − h − k &gt; 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 &lt; h + k we can write α = α1 + α2 with α1 &lt; h and α2 &lt; 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| &lt; R, and is therefore bounded by M &lt; ∞ 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 &lt; 0 and goes to zero for α &gt; 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 &gt; 0}
to the unit disc. Consequently, the map Tc (z) = (z − 1 − c)/(z + 1 − c) maps {Re &gt; c} onto the unit
disc. If Im &gt; c, Re &lt; c, Im &lt; c, we can always rotate these domains onto Re &gt; 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 &gt; 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 | &lt; 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 | &lt;
a contradiction. Hence f is injective on |z| &lt; 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
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 | &lt; we can choose a single-valued branch
p if is such that |g̃(z0 ) − g̃(z)| &lt;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 | &lt; R and |w0 | &lt; 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| &lt; 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| &lt; 1 was arbitrary, so long as |f (z0 )| &lt; 1. Consequently,
1
|f 0 (z)|
≤
2
1 − |f (z)|
1 − |z|2
on |z| &lt; 1.
4.3.4.2. If f (z) is analytic and Im f (z) ≥ 0 for Im z &gt; 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 &gt; 0 to |ζ| &lt; 1 and Sw maps Im w ≥ 0
to |Sw| ≤ 1. Since f (z) maps Im z &gt; 0 to Im f (z) ≥ 0, it follows that |F (ζ)| ≤ 1 on |ζ| &lt; 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| &lt; 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| &lt; 1. Hence, the composition F (z) = Sf (z) satisfies |F (z)| &lt; 1 for
|z| &lt; 1. Moreover, Sf (0) = 0 so that F (0) = 0 and F is analytic on |z| &lt; 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| &lt; 1 for |w| &lt; 1 (since also |f (0)| &lt; 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| &lt; 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 &lt; 6 = |f (z)|.
By Rouché’s theorem, f (z) and g(z) have the same number of zeros in |z| &lt; 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| &lt; 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| &lt; 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 &lt; 6 = |f (z)|.
Thus, f (z) and g(z) have the same number of zeros in |z| &lt; 1. Clearly f (z) has one so that g(z)
has one zero. To find the number of zeros in |z| &lt; 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| &lt; 2. We’ve already counted one of these, so that g(z) has three zeros
in 1 ≤ |z| &lt; 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 &gt; 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 = &plusmn;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| &lt; 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 &gt; 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| &lt; 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|
&gt;
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 &lt; α &lt; 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 &lt; |2a| ≤ |2a + 1| + 1 = |b| + 1
so that |b| &gt; 1. Now, suppose b &gt; 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 &lt; −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 = &plusmn;3i, &plusmn;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 = &plusmn;ai are poles of order 3. WLOG assume that a &gt; 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&gt;0
The poles of R(z) occur at z = &plusmn;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&gt;0
y=0
The poles of R(z) are at z = &plusmn;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 &lt; α &lt; 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&gt;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&gt;0
The poles of z 1/3 /(1 + z 2 ) are at z = &plusmn;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 &lt; arg z &lt;
3π/2.
Note that f (z) = log z/(z 2 + 1) has poles at z = &plusmn;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 &lt; α &lt; 1. We wish to find
Z ∞
xβ
dx
1 + x2
0
where 0 &lt; β = 1 − α &lt; 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&gt;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 &gt; α &gt; 1, so we wish to find
Z ∞
1
dx
β
x (1 + x2 )
0
where 0 &lt; β = α − 1 &lt; 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 &lt; α &lt; 2 we have 0 &lt; 2 − α &lt; 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| &lt; ρ, then ρ2 /|a| &gt; ρ, and thus there is only one pole in |z| &lt; 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| &gt; ρ then ρ2 /|a| &lt; ρ. Thus we still have one pole in |z| &lt; 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 &lt; |z| &lt; ρ, show that the origin is a removable singularity
in the sense that u becomes harmonic in |z| &lt; 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 &lt; r &lt; ρ and constants α, β. Since u is bounded, we have that
|α log r + β| ≤ M r
where |u| ≤ M on 0 &lt; |z| &lt; 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 &lt; |z| &lt; 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 &gt; 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| &lt; 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| &lt; 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| &lt; r for all z ∈ Ω and some r &gt; 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 &gt; 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 &gt; 1. Then,
|ζ(z)| ≤
∞
X
|n−z | ≤
n=1
∞
X
1
&lt; ∞.
n
n=1
So, we know that ζ(z) converges on Re z &gt; 1. Moreover, if K is any compact subset of Re z &gt; 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 &gt; 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 &gt; 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
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 &lt; ∞.
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 &gt; 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 &gt; 0, this is finite. Hence, the series in question converges for Re z &gt; 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 &gt; 1. In this case,
#
&quot;
∞
∞
∞
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 &lt; |z − zi | &lt; ρ. By the maximum
modulus principle, |f (z)| &gt; 0 on C. Let δ &gt; 0 be such that |f (z)| &gt; 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)| &lt; δ/2 on C.
Hence,
δ &lt; |f (z)| ≤ |fn (z) − f (z)| + |fn (z)| &lt; δ/2 + |fn (z)|
and |fn (z)| &gt; δ/2 on C. Since g(z) = 1/z is uniformly continuous on |z| &gt; δ/2, |fn (z)| &gt; δ and
|f (z)| &gt; 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 &gt; 0, then there exists a δ &gt; 0 such that |g(z) − g(w)| &lt; whenever |z − w| &lt; δ. With this δ, by uniform convergence of the fn there exists an N such that if n ≥ N
then |fn (z)−fn (w)| &lt; δ. Combining the two, we see that if n ≥ N then |g(fn (z))−g(fn (w))| &lt; for
all z, w ∈ Ω, asserting uniform convergence. We now show that g(z) = 1/z is uniformly continuous
on subsets of |z| &gt; m &gt; 0. Clearly we have
1
1
|z − w|
|z − w|
−
=
&lt;
z
w
|z||w|
m2
whenever |z| &gt; m and |w| &gt; 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| &lt; 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| &lt; 1, we have |z n | &lt; 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| &lt; 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 &times; 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 &plusmn; 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| &lt; 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 &gt; 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| &lt; 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 &gt; 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
≈
&lt; ∞.
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
≤
&lt;∞
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 &lt; 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 | &lt; ∞, 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 ) &gt; 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
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