Math 5120: Complex analysis. Homework 1 Solutions
1.1.1.2 Omitting details of the computation, with z = x + iy the answers are:
• <z4 = x4 − 6x2 y2 + y4 , =z4 = 4xy(x2 − y2 ).
• <z−1 = x/(x2 + y2 ), =z−1 = −y/(x2 + y2 ).
• <(z − 1)/(z + 1) = (x2 + y2 − 1)/((x + 1)2 + y2 ), =(z − 1)/(z + 1) = 2y/((x + 1)2 + y2 ).
• <z−2 = (x2 − y2 )/(x2 + y2 )2 , =z−2 = −2xy/(x2 + y2 )2 .
1.1.1.3 This is an easy direct computation, or can be done by noting that all of these numbers have modulus 1, the first
pair have argument π ± π3 (so multiplication by 3 gives multiples of 2π), while the other pair in the second set
have arguments ± π3 (so multiplication by 6 gives multiples of 2π).
1.1.2.4 One may apply the quadratic formula.
1.1.4.3 If |a| = 1 then
a − b = |a| a − b = 1
2
1 − āb
a − |a| b
and an analogous argument using b̄ holds if |b| = 1. The only exception is when the denominator vanishes,
which occurs iff āb = 1. In the case that either |a| = 1 or |b| = 1 we conclude immediately that this occurs iff
a = b.
1.1.4.4 An equation in complex variables is two equations in real variables, in this case two simultaneous linear equations. A fast way to extract them is to take the complex conjugate, so we have
az + bz̄ + c = 0
āz̄ + b̄z + c̄ = 0.
Multiplying by ā and b respectively,
|a|2 z + ābz̄ + āc = 0
ābz̄ + |b|2 z + bc̄ = 0
the difference eliminates z̄ and we obtain
|a|2 − |b|2 z + āc − bc̄ = 0
so that there is a unique solution iff |a| , |b|, in which case the solution is
z=
bc̄ − āc
|a|2 − |b|2
1.1.5.1 Each of the following statements is equivalent.
a − b 2
< 1
1 − āb
a − b ā − b̄
<1
1 − āb 1 − ab̄
|a|2 + |b|2 − bā − ab̄ < 1 − āb − ab̄ + |a|2 |b|2
|a|2 1 − |b|2 < 1 − |b|2
|a| < 1 and |b| < 1 or |a| > 1 and |b| > 1 .
1.1.5.3 By induction and the triangle inequality
n
n n X
X
X
λ a <
λ = 1.
λ j a j ≤
j
j
j
j=1
j=1
j=1
1.1.5.4 It is easiest to solve this problem if you recognize it as the equation of an ellipse with foci at ±a, as then you
know what to expect. In any case it is easy to see
2|a| = |a + a| = |a − z + a + z| ≤ |z − a| + |z + a| = 2|c|
so that |a| ≤ |c| is necessary for the existence of z satisfying the equation. Then for |c| ≥ |a| we may set z = |c|a/|a|
and find that |z − a| = |c| − |a|, |z + a| = |c| + |a|, so |z − a| + |z + a| = 2|c|, whence it is also sufficient.
To find the maximum of |z| on the curve is easy, because repetition of our first argument shows |z| ≤ |c| on the
curve, but the point z = |c|a/|a| has |z| = |c|, so this maximum is achieved. The minimum is a little trickier. One
way is as follows:
4|z|2 = 4zz̄ = (z − a) + (z + a) (z̄ − ā) + (z̄ + ā) = |z − a|2 + |z + a|2 + 2|z|2 − 2|a|2
so 2 |z|2 − |a|2 = |z − a|2 + |z + a|2 . Then use the general inequality 2xy ≤ x2 + y2 for x, y ∈ R to see
2|z − a||z + a| ≤ |z − a|2 + |z + a|2 with equality iff |z − a| = |z + a|, and substitute into the square of the equation
, so |z|2 ≥ |c|2 − |a|2 ,
for 2|c| to obtain 2|c|2 ≤ |z − a|2 + |z + a|2 . Combining these we have 2|c|2 ≤ 2 |z|2 − |a|2 p
with equality iff |z − a| = |z + a|. The latter fact allows us to guess that a point at distance |c|2 − |a|2 from 0 in
a direction perpendicular to a will lie on the curve, a guess that is rapidly verified using Pythagoras’ theorem.
1.2.1.1 The easy way to do this is to note that symmetry is preserved under rotation by π/4, such that the lines become
the axes, then use that the symmetric points with respect to the axes are obtained by ± and conjugation, and
rotate back. The points are then
e−iπ/4 (±eiπ/4 a) = ±a,
e−iπ/4 (±eiπ/4 a) = ±e−iπ/4 e−iπ/4 ā = ∓iā.
1.2.1.2 The vertices of an equilateral triangle are mapped to the vertices of an equilateral triangle by any map of the
form z 7→ α(z + β), and for any given equilateral triangle there is a map of this form that takes it to the triangle
with vertices 1, e2πi/3 , e−2πi/3 . The correct choice is 3β = a1 + a2 + a3 and α = 3/(2a1 − a2 − a3 ). It is obvious
that the equality
a21 + a22 + a23 = a1 a2 + a2 a3 + a3 a1
is preserved by the map z 7→ αz, and readily verified that it is preserved by z 7→ z + β. Thus it suffices to verify
the equality for the triangle with vertices 1, e2πi/3 , e−2πi/3 , in which case both sides are easily checked to sum to
2.
1.2.2.2 If z = eiφ then
1 + cos φ + . . . + cos nφ = <(1 + z + . . . + zn )
1 − zn+1
1−z
1 − z̄ − zn+1 + zn
=<
1 + |z|2 − z − z̄
1 − cos φ − cos(n + 1)φ + cos nφ
=
2 − 2 cos φ
1 − z̄ − zn+1 + zn
sin φ + . . . + sin nφ = =
1 + |z|2 − z − z̄
sin φ − sin(n + 1)φ + sin nφ
=
2 − 2 cos φ
=<
Page 2
1.2.2.4 Let w = cos(2πi/n) + i sin(2πi/n) = e2πi/n . Provided h is not a multiple of n we have wnh , 1 and thus
1 + wh + . . . + w(n−1)h =
1 − wnh 1 − e2πhi
=
= 0.
1 − wh
1 − wh
1.2.3.1 To see when az + bz̄ + c = 0 represents a line, recall exercise 1.1.4.4. We must have infinitely many solutions
for the simultaneous equations, so they must be the same, so |a| = |b| and āc = bc̄. An equivalent form of the
latter is that c is a real multiple of (a + b); we may find this by summing it with its complex conjugate to find
(a + b)c̄ = (ā + b̄)c, from which the ratio (a + b)/c is real. To be certain that the solution set is one dimensional
not two dimensional we must insist that |a| , 0.
1.2.3.5 Rotation around the origin preserves the circle |z| = 1, preserves angles, and may be used to move a to lie on
the positive real axis. WLOG a > 1. Then the circle through a and 1a has center 12 a + 1a and radius 21 a − 1a .
The circles intersect at right angles iff the radii to the intersection point are at right angles iff (by Pythagoras’
theorem) the sum of the squares of the radii of the circles is the square of the distance to the center of the the
circle through a and 1a . Written algebraically, this is equivalent to
1+
1
1
1
1
1
1
1
1
a − 2 = 1 + a2 + 2 − 2 = a2 + 2 + 2 = a + 2
4
a
4
4
4
a
a
a
1.2.4.5 I think it likely that this problem will have caused difficulties. The reason is that the spherical distance in
equation (28) on page 20 of the book is not a geodesic metric! It is the metric corresponding to chords of the
sphere, so the triangle inequality is always strict for three distinct points because they cannot all lie on a chord
(the chord intersects the sphere at two points only). This and the fact that the center of the disc in the plane is
not the same as the center of the circle on the Riemann sphere mean that you cannot do either of the following:
(1) take the distance d(a, a + Ra/|a|), or d(a, a − Ra/|a|) (because a is not the center in the spherical metric), (2)
take half of d(a − Ra/|a|, a + Ra/|a|) (because the metric is not geodesic, so half of the diameter is not the radius).
You can do one thing to simplify your calculations, which is to rotate the plane (an isometry of the spherical
metric) so a ∈ R. At this point you have the option of finding the images on the sphere (using equations (25)
and (26) on page 18), then find the center, then find the length of the chord corresponding to the radius. This
latter can also be done by Pythagoras’ theorem, which makes the problem much simpler.
Suppose that the length of the chord joining the images√of a ± R on the sphere is 2s. Then the distance from
the center of the sphere to the midpoint
of the chord is 1 − s2 (by Pythagoras), so the distance radially from
√
this midpoint to the sphere is 1 − 1 − s2 . The landing point of the ray is the spherical center of the disc, and
the spherical radius we seek is the hypotenuse of the right triangle with vertices the landing point, the midpoint
of the
√ chord, and the image of a + R. Since √we have just seen the sides of this triangle have lengths s and
1 − 1 − s2 the hypotenuse has length 2(1 − 1 − s2 ) (Pythagoras again). So it suffices to find s. By formula
28 on page 20,
2|(a + R) − (a − R)|
4R
2s =
=
.
2
2
2
1 + (a + R) 1 + (a − R)
1 + (a + R) 1 + (a − R)2
We can then compute
1 + (a + R)2 1 + (a − R)2 − 4R2
1 + (a + R)2 1 + (a − R)2
1 + (a + R)2 + (a − R)2 + (a2 − R2 )2 − 4R2
=
1 + (a + R)2 1 + (a − R)2
1 + 2(a2 − R2 ) + (a2 − R2 )2
=
1 + (a + R)2 1 + (a − R)2
1 + a2 − R2 2
=
1 + (a + R)2 1 + (a − R)2
1 − s2 =
Page 3
so that
Spherical radius of disc = 2(1 −
=2 1−
√
1 − s2 )
1 + a2 − R2 1 + (a + R)2 1/2 1 + (a − R)2 1/2
p(1 + a2 − R2 )2 + 4R2 − |1 + a2 − R2 | =2
.
p
(1 + a2 − R2 )2 + 4R2
Page 4
Math 5120: Complex analysis. Homework 2 Solutions
2.1.2.4 Suppose f = u + iv is analytic and | f | is constant, so | f |2 = u2 + v2 = c. Analyticity
gives existence of the partial derivatives u x , uy , v x , vy and validity of the CauchyRiemann equations. Differentiating | f |2 = c gives uu x + vv x = uuy + vvy = 0;
Cauchy-Riemann lets us rewrite these as
! !
!
u v ux
0
=
.
v −u v x
0
Since the determinant is | f |2 = c, we discover that either c = 0, whence f = 0
everywhere, or the matrix is invertible and u x = v x = 0, from which also uy = vy =
0 by Cauchy-Riemann and thus u and v, and hence f , are constant.
2.1.2.5 Each of the following statements is equivalent.
f (z) is analytic at z = z0 .
f (z + z0 ) − f (z0 )
exists.
lim
z→0
z
f (z + z0 ) − f (z0 )
lim
exists.
z→0
z̄
f (w̄ + w̄0 ) − f (w̄0 )
lim
exists. (Set w = z̄, w0 = z̄0 )
w→0
w
f (w̄) is analytic at w = w0 .
2.1.4.1(a) Let R(z) = z4 (z3 − 1)−1 . It has poles at ∞, 1, e2πi/3 , e2πi/3 . Set τ = 2πi/3. We need
to expand at each pole, for which purpose (using the notation in the book) perform
each of the following expansions. We include only the relevant terms, noting the
presence of those with lower order using an ellipsis.
z
R(z) = z + 3
so G(z) = z
z −1
R(1 + z−1 ) =
R(eτ + z−1 ) =
R(e2τ + z−1 ) =
(1 + z−1 )4
(z + 1)4
z4 + · · ·
z
=
=
= + ···
(1 + z−1 )3 − 1 z(z + 1)3 − z4 3z3 + · · · 3
1 1
G1
=
z−1
3(z − 1)
(eτ + z−1 )4
(eτ z + 1)4
e4τ z4 + · · ·
e2τ z
=
=
=
+ ···
3
(eτ + z−1 )3 − 1 z(eτ z + 1)3 − z4 3e2τ z3 + · · ·
1 e2τ
Geτ
=
z − eτ
3(z − eτ )
(e2τ + z−1 )4
(e2τ z + 1)4
e8τ z4 + · · ·
e4τ z
eτ z
=
=
=
+
·
·
·
=
+ ···
3
3
(e2τ + z−1 )3 − 1 z(e2τ z + 1)3 − z4 3e4τ z3 + · · ·
1 eτ
Ge2τ
=
2τ
z−e
3(z − e2τ )
Thus
R(z) = z +
1
e4πi/3
e2πi/3
+
+
.
3(z − 1) 3(z − e2πi/3 ) 3(z − e4πi/3 )
1
2
2.1.4.1(b) In a similar but more tedious calculation, R(z) = z−1 (z + 1)−2 (z + 2)−3 has poles at
0, −1 and −2. We perform the following expansions, including only the relevant
terms.
R(z−1 ) =
R(−1 + z−1 ) =
1
z
z6
z6
= + ···
=
=
z−1 (z−1 + 1)2 (z−1 + 2)3 (1 + z)2 (1 + 2z)3 8z5 + · · · 8
1
1
G0
=
z
8z
1
(−1 +
z6
z6
= 4
= −z2 + 2z + · · ·
3
+
(z − 1)(1 + z)
−z − 2z3 + · · ·
1 2
−1
+
G−1
=
2
(z + 1)
(z + 1)
(z + 1)
z−1 )z−2 (z−1
1)3
=
z6
(−2 +
+
(−2z + 1)(−z + 1)2
z6
−z3 5z2 17z
=
=
−
−
+ ···
3
2
2
4
8
−2z + 5z − 4z + · · ·
R(−2 + z−1 ) =
G−2
1
z−1 )(−1
z−1 )2 z−3
=
5
17
1 −1
−
−
=
(z + 2)
2(z + 2)3 4(z + 2)2 8(z + 2)
Thus
R(z) =
−1
2
1
5
17
1
+
+
−
−
−
.
8z (z + 1)2 (z + 1) 2(z + 2)3 4(z + 2)2 8(z + 2)
2.1.4.4 Suppose R(z) is a rational function with |R(z)| = 1 on |z| = 1. Then
1
S (z) =
R
1
z̄
is also a rational function, and since w = w̄1 whenever |W| = 1 we see that R(z) =
S (z) on |z| = 1. But then R(z) = S (z), because their difference is a rational function
with zeros at every point of the unit circle. Now observe that if R(z) has a root at
a of order α then S (z) has a pole at z = (ā)−1 with the same order, so R(z) = S (z)
has a pole of this order at this location. This argument is reversible; if R(z) has
a pole at b of order β then S (z) has a zero at z = (b̄)−1 with the same order, so
R(z) = S (z) has a zero of this order at this location. Thus the poles and zeros of
R are bijectively paired by the map z 7→ (z̄)−1 . (Note that this implies there are
neither poles nor zeros at points of |z| = 1, because for these points z = (z̄)−1 and
there cannot simultaneously be a zero and a pole at a single point.) An equivalent
z−a
formulation is that R(z) is necessarily a product of factors of the form 1−āz
with
|a| , 1. Since we proved in Exercise 1.1.4.3 that such factors have modulus 1
on the unit circle, this condition is also sufficient. Grouping the factors and using
negative powers α j when a zero is outside the unit disc we see that the general
form of the rational function we seek is
m Y
z − a j α j
R(z) = czk
1 − ā j z
j=1
with k and α j in Z and |a j | < 1 for all j.
3
2.1.4.6 Suppose R(z) = cP(z)/Q(z) is a rational function, where P and Q are monic polynomials without common zeros. Let m = deg(P) and n = deg(Q). It will be
convenient at first to assume neither P nor Q is constant. We have
P0 (z)Q(z) − P(z)Q0 (z)
R0 (z) = c
.
Q2 (z)
It is an easy observation that deg(P0 ) = m−1 and deg(Q0 ) = n−1, thus deg(P0 Q) =
deg(PQ0 ) ≤ m + n − 1. Observe that the lead term in P0 Q has coefficient n while
that in PQ0 has coefficient m. If we assume m , n then these terms cannot cancel,
so deg(P0 Q − PQ0 ) ≤ m + n − 1 with equality at least when m , n. To find the
degree of R0 (z) we must then cancel common factors. Any root of Q with order α
is a root of P0 Q with the same order, so to be a root of P0 Q − PQ0 it must divide
P or Q0 . By assumption P and Q have no common roots, so it must be a root of
Q0 . An easy computation shows that Q and Q0 have common roots if and only if
Q has multiple roots, and that the order of the root in Q0 is α − 1. We conclude that
Q0 /Q is of the form S /T , where T is the (monic) product of the distinct factors of
Q and has order k ≤ n, and deg S = k − 1. It is then apparent that
P0 (z)T (z) − P(z)S (z)
R0 (z) = c
Q(z)T (z)
0
with deg(P T − PS ) ≤ m + k − 1 (with equality if m , n), deg(QT ) = n + k, and
the numerator and denominator having no common roots. It follows that
deg(R0 ) ≤ max{m+k−1, n+k} = max{m−1, n}+k ≤ max{m−1, n}+n ≤ deg R+n ≤ 2 deg(R)
and that equality holds at the first inequality if m , n, at the second if k = n (i.e.
all poles of R are distinct), at the third and fourth if deg R = n ≥ m. Thus we have
deg(R0 ) ≤ 2 deg(R) and conditions under which equality holds.
We may also get lower bounds. If m , n then we have equality in the first
inequality above, so deg(R0 ) = max{m + k − 1, n + k} ≥ max m, n + 1 ≥ deg(R)
because we assumed Q non-constant and thus k ≥ 1. If m = n then deg(P0 T −
PS ) ≤ m + k − 1 < n + k = deg(QT ), so deg(R0 ) = n + k ≥ deg(R) + 1. (Note this
also implies that deg(R0 ) = 2 deg(R) when m = n = k
What remains are the special cases where either P or Q is constant. If Q is
constant and P is not then R is a polynomial, so deg(R0 ) = deg(R) − 1. If P is
constant and Q is not then the above reasoning implies R0 = S /QT , so deg(R0 ) =
deg(QT ) = k + n > n = deg(R). If both are constant then deg(R0 ) = −∞.
We may summarize our results as follows. If R(z) is a non-constant rational
function, then deg(R) − 1 ≤ deg(R) ≤ 2 deg(R), with equality on the left iff R is a
polynomial and equality on the right at least when the poles of R are distinct and
there is no pole at infinity.
P
2.2.3.3 Suppose |a j | converges. A re-ordering of the sum is given by a bijection η : N →
P
P
N, where the new sum is j aη( j) . Both sums converge absolutely; set S = j a j
P
and T = j aη( j) . Given > 0 let N
be so large that for any K ≥ N, each of
P
S − P a ≤ and T − P a ≤ . Let M be so large that
|a
|
≤
,
j≤K j
j≤K η( j)
j>K j
η( j) : j ≤ M ⊃ { j ≤ N} (Note that then M ≥ N.)
X
X
X
|S − T | ≤ aj −
aη( j) + 2 ≤
|a j | + 2 ≤ 3
j≤M
j≤M
j>N
where middle inequality used that the sum is over those j in { j ≤ M} \ {η( j) : j ≤
M} ∪ {η( j) : j ≤ M} \ { j ≤ M} , which does not include any j ≤ N.
4
2.2.3.4 For fixed z with |z| ≥ 1 we have |nzn | = n|z|n → ∞, while if |z| < 1 we have
n|z|n = exp(log n + n log |z|) → 0 because log |z| < 0 and (n log |z|)/ log n → −∞ as
n → ∞ (using, for example, L’Hopital’s rule). Hence {nzn } converges pointwise
on |z| < 1. A slight improvement is that on |z| ≤ r < 1 we have n|z|n ≤ nrn → 0, so
that the convergence is uniform on this disc. Any compact set that lies in |z| < 1 is
in such a disc, so we have {nzn } converges uniformly to the zero function on any
compact subset of the open unit disc. Moreover, this is the largest collection of
sets on which the convergence is uniform. To see this, suppose S is a set on which
{nzn } converges uniformly, from which we see S is in the unit disc and n|z|n → 0.
If S is not a compact subset of the disc then it contains points zk such that |zk | → 1.
If n is so large that n|z|n < 1/2 on S then the fact that n|zk |n → n as k → ∞ provides
a contradiction.
2.2.3.6 There are a number of ways of doing this, the most usual being to use summation
by parts. I will show a method that is messier, but perhaps easier to visualize. Let
P
P
P
U = j u j , V = j v j ; suppose WLOG that j u j is the absolutely convergent
P
series and set W = j |u j |. It is helpful to arrange the terms in a doubly-infinite
list.


u1 v1 u1 v2 u1 v3 . . .
u2 v1 u2 v2 u2 v3 . . .
(u j vk ) j,k = u3 v1 u3 v2 u3 v3 . . .


..
..
 ..

.
.
.
P
th
We see immediately that n−1
j=1 u j vn− j is the sum along the n upward diagonal,
Pm Pn−1
so n=1 j=1 u j vn− j is the sum over the upper left triangle j + k ≤ m in the list.
We can also see that UV may be approximated by the sum over
j ≤ p,
a square
P
U − j≤p u j ≤ ,
k ≤ p, because
for
fixed
>
0
we
can
take
N
so
p
≥
N
implies
V − P v ≤ , and so
k≤p k
X X X
X
X X
X X u ≤ |V|+ |u | ≤ |V|+W .
UV− u v = UV−V
≤
|V|+
u
v
u
−
u
+V
j
j
j
k j
j
j
k j≤p
k≤p
j≤p
j≤p
j≤p
k≤p
j≤p
j≤p
Thus the difference between UV and the sum we are considering is controlled by
a small term plus the sum over a region inside an upper left triangle and outside a
square, i.e. j + k ≤ m but also j ≥ p, k ≥ p.
X
m X
n−1
m X
n−1
X
X X UV−
≤ |V|+W +
u
v
−
u
v
≤
|V|+W
+
u
v
j
n−
j
j
n−
j
j
k
n=1 j=1
n=1 j=1
j≤p
k≤p
X
{ j≥p,k≥p, j+k≤m}
We split this sum into two pieces, one with k ≥ p and one with k ≤ p. We want a
bound independent of m, and you should think that m p. Since it is a finite sum
it can be rearranged any way we like; we will sum first along the rows and then
down the columns. For the piece with k ≥ p the sum along the jth row is bounded
as follows:
X
m− j
u
v
k ≤ |u j |
j
k=p+1
P
provided p is so large that qp+1 vk ≤ for all q ≥ p; this can be achieved by (if
P
necessary) increasing N, because vk is convergent. Summing over all relevant
u j vk .
5
rows we have a contribution bounded as follows:
m−p
m− j
m−p
∞
X
X
X
X
≤ u
|u j | = W.
v
|u
|
≤
j
k
j
j=1
j=1
k=p+1
j=1
P
P
Now for the piece with k < p we reason as follows. Since k vk converges, k≤r vk
is a convergent sequence in r, so has absolute value bounded by a constant X. Now
the sum across each row in the second piece is of size
min{p−1,m−
X j} u
vk ≤ X|u j |
j
1
and summing over the rows we have
X
min{p−1,m−
m
m
∞
X j} X
X
≤ X
u
v
|u
|
≤
X
|u j | ≤ X
j
k
j
j=p+1
1
j=p+1
j=p+1
provided p is large enough that j≥p+1 |u j | ≤ ; again this can be achieved by
increasing N. Combining our estimates we now have
P
m X
n−1
X
UV −
u j vn− j ≤ |V| + 2W + X
n=1 j=1
and the result follows.
By the way, if you are wondering how this is connected to the material in this
P
chapter, the following may be of interest. Consider the functions U(z) = j u j z j
P
and V(z) = j v j z j . These converge at z = 1, so have radius of convergence at
least 1; moreover Abel’s limit theorem says that U(r) → U(1) and V(r) → V(1)
as r → 1, where r ∈ (0, 1). On the disc |z| < 1 they are absolutely convergent, and
P
their product U(z)V(z) has a power series expansion j cn zn . We can compute by
the Leibniz rule
cn =
n
n−1
1 dn
1 X
n!
d j U(z) dn− j V(z) X
=
U(z)V(z)
=
u j vn− j .
z=0
n! dzn
n! j=0 j!(n − j)! dz j z=0 dzn− j
j=1
P
P P
If we now knew that n cn = n n−1
j=1 u j vn− j was convergent, then Abel’s limit
P
theorem would imply n cn = limr→1 U(r)V(r) = limr→1 U(r) limr→1 V(r) =
U(1)V(1), which is the theorem we seek. This suggests that there is a proof essentially by the proof of Abel’s limit theorem, which is the standard summation by
parts proof.
2.2.4.2
2z + 3 2(z − 1) + 5
1
1
=
=2+
z+1
(z − 1) + 2
2 1 + (z − 1)/2
∞
1 X (−1) j
=2+
(z − 1) j if |z − 1| < 2.
2 j=0 2 j
The radius of convergence is 2.
6
2.2.4.4 The following statements are equivalent.
X
an zn has radius of convergence R
n
lim sup |an |1/n = R−1
lim sup |an |1/2n = R−1/2 and lim sup |an |2/n = R−2
X
√
an z2n has radius of convergence R, and
n
X
a2n zn has radius of convergence R2
n
P n
−n
2.2.4.8 The series ∞
converges if and only if |z(1 + z)−1 | < 1, which is if and
0 z (1 + z)
only if |z| < |1 + z|, if and only if <(z) > 12 .
2.3.2.2
e−z + ez
= cosh z
2
e−z − ez
sin iz =
= i sinh z
2i
cos iz =
We can get addition formulae for cosh and sinh by computing
ea + e−a eb + e−b ea+b + e−(a+b) + ea−b + eb−a
=
2
4
4
ea − e−a eb − e−b ea+b + e−(a+b) − ea−b − eb−a
sinh a sinh b =
=
2
4
4
ea + e−a eb − e−b ea+b − e−(a+b) − ea−b + eb−a
cosh a sinh b =
=
2
4
4
cosh a cosh b =
elementary algebra implies
cosh(a + b) = cosh a cosh b + sinh a sinh b
sinh(a + b) = sinh a cosh b + cosh a sinh b
cosh 2a = cosh2 a − sinh2 a
sinh 2a = 2 sinh a cosh a
and substitution of the above expressions for cos and sin retrieves the usual trigonometric addition formulae, which could alternatively be used to obtain the above.
2.3.2.3
cos(x + iy) = cos x cos iy − sin x sin iy = cos x cosh y − i sin x sinh y
sin(x + iy) = sin x cos iy + cos x sin iy = sin x cosh y + i cos x sinh y
7
2.3.4.4
ez = 2 when z = log 2 + 2kπi,
k∈Z
e = −1 when z = (1 + 2k)πi, k ∈ Z
1
ez = i when z =
+ 2k πi, k ∈ Z
2
3
−i
z
when z = − log 2 +
+ 2k πi, k ∈ Z
e =
2
2
5
1
z
e = −1 − i when z = log 2 +
+ 2k πi, k ∈ Z
2
4
1
z
e = 1 + 2i when z = log 5 + arctan 2 + 2k πi, k ∈ Z and arctan in (0, π/2).
2
z
2.3.4.5 Let z = x + iy, so ez = e x cos y + ie x sin y. Then the real and imaginary parts of
exp(ez ) are obtained from
exp(ez ) = exp(e x cos y + ie x sin y) = exp(e x cos y) cos(e x sin y) + i exp(e x cos y) sin(e x sin y)
2.3.4.6
2i = exp(i log 2) = cos(log 2) + i sin(log 2) single valued because 2 ∈ R
−π
− 2kπ), k ∈ Z
ii = exp(i log i) = exp
2
(−1)2i = exp(2i log(−1)) = exp(−2π − 4πk), k ∈ Z.
Note that the last example shows something we have lost in making the convention
that the logarithm is single valued on the positive reals and multivalued elsewhere,
because (−1)2i , ((−1)2 )i = 1, but simply contains 1 as one of its values.
Math 5120: Complex analysis. Homework 3 Solutions
3.1.2.6 R and C are separable metric spaces, so see Exercise 3.1.3.7 below.
3.1.3.4 Let A = {(0, y) : |y| ≤ 1} and B = {(x, sin(1/x) : x > 0} in R2 . Note that both A and B are continuous images of
connected subsets of R, so are connected. If there is a disconnection of A ∪ B, then the set containing a point
of A must contain all of A because A is connected, and similarly for B. So the disconnection must consist of
open sets U ⊃ A and V ⊃ B. But then U contains a disc around 0, and any such disc contains points from B, for
example those of the form (2πn)−1 , sin(2πn) for all sufficiently large n ∈ N, so U ∩ V , ∅ and there is no such
disconnection. We conclude that A ∪ B is connected. (Note: A ∪ B is not path connected).
3.1.3.5 Let En = {(x, 1/n) : 0 ≤ x ≤ 1} and E∞ = {(x, 0) : 0 ≤ x ≤ 1} in R2 . Let E = E∞ ∪ ∪n En . For any n let Un
−2
be the open set consisting of all points within distance (n + 1) of En , and Vn be the interior of the complement
of Un . Then Un ∩ Vn = ∅, Un ⊃ En and Vn ⊃ E \ En . We conclude that the maximal connected set containing a
point of En is contained in En . Since En is connected (because it is a curve), we conclude that En is a component
of E for each n. For a point in E∞ we see that the component containing this point must contain E∞ , because
E∞ is connected; from the above it does not contain any point of En for any n, so E∞ is a component. We have
therefore found the decomposition of E into its components. We have also seen that the components of the form
En are closed in R2 and relatively open in E; the component E∞ is closed in R2 , but not relatively open in E
because any open neighborhood of E∞ in R2 intersects some En .
Finally, a set is locally connected if every neighborhood of a point in the set contains a connected neighborhood
of the point. Consider (0, 0) ∈ E∞ . Any neighborhood of this point is the intersection of E with an open set in
R2 , and therefore cannot be connected because it contains points from another component En . Since this point
has no connected neighborhoods it is not locally connected. (Remark: local connectivity of a set is equivalent
to all components being open.)
3.1.3.7 S is discrete in a metric space (M, d), so for all x ∈ S there is r x such that the ball around x of radius r x , denoted
B(x, r x ), has B(x, r x ) ∩ S = {x}. Observe that then B(x, r x /2) ∩ B(y, ry /2) = ∅ if x , y, because if there is z in the
intersection we would have
|x − y| ≤ |x − z| + |y − z| < (r x + ry )/2 ≤ max{r x , ry }
contradicting either x 3 B(y, ry ) or y 3 B(x, r x ).
Now if A ⊂ M is a dense set then for each x there is a x ∈ B(x, r x /2) ∩ A, and a x , ay if x , y because
B(x, r x /2) ∩ B(y, ry /2) = ∅. The map x 7→ a x is then an injection from S to A, so the cardinality of S cannot
exceed that of A. In particular if M is separable we may take A to be a countable dense subset and conclude S
is countable.
√
√
√
3.2.2.1 We want to give a domain and a definition of · so that 1 + z + 1√− z is a single-valued analytic function.
In the book there is a definition of a single valued analytic branch of w for w ∈ C \ (−∞, 0]. It suffices then
that we ensure 1 + z √
and 1 − z √
are in this set, which is true if z ∈ C \ (−∞, −1] ∪ [1, ∞) . Nothing more need be
done, as on this set 1 + z + 1 − z is a sum of single-valued analytic functions.
Math 5120: Complex analysis. Homework 4 Solutions
3.3.1.1 The map z 7→ z̄ fixes 0,1 and ∞. Consider a fractional linear transformation z 7→
az+b
cz+d fixing these points. Fixing 0 implies b = 0, fixing ∞ implies c = 0, fixing 1
implies a/d = 1, so the map is the identity. Thus z 7→ z̄ is not a fractional linear
transformation.
3.3.3.1 A reflection z 7→ z∗ = g(z) is defined via (z∗ , z1 , z2 , z3 ) = (z, z1 , z2 , z3 ). Let φ be
the fractional linear transformation so φ(z1 ) = 1, φ(z2 ) = 0 and φ(z3 ) = ∞, and let
f (z) = z̄. Then φ(z∗ ) = φ(z), so g = φ−1 ◦ f ◦ φ. Now we know from the book
that φ maps circles to circles (on the Riemann sphere), so it suffices to see that the
same is true of f . A circle not through ∞ has the form |z − a| = r, so is mapped to
r = |z̄ − a| = |z − ā|, which is also a circle. A circle through ∞ is z = x + iy with
ax + by + c = 0, a, b, c ∈ R, and is mapped to z = x + iy with ax − by + c = 0, which
is also a circle through ∞.
3.3.3.5 It is easy to think of some fractional linear transformations that fix |z| = R as a set.
For example, any rotation around the origin (i.e. z 7→ az with |a| = 1), the map
z−a
z 7→ R2 /z. In the case R = 1 we also know from prior homeworks that z 7→ 1−āz
2
(z/R)−a
for |a| , 1 preserves |z| = 1, so z 7→ R 1−ā(z/R)
= Rz−aR
R−āz preserves |z| = R. The
more difficult question is whether we have found all of the maps or not. One way
to approach the problem is to take an arbitrary fractional linear transformation that
fixes |z| = R, compose it with some of the maps we know, and try to get the result
to fix 3 points, so that the composition is the identity.
Suppose that φ is a fractional linear transformation that fixes |z| = R as a set.
Let eiθ = φ(R)/R, so z 7→ e−iθ φ(z) fixes the point z = R. Let a = e−iθ φ(0)/R and
2
−iθ
ψ(z) = Rz−aR
R−āz . Then z 7→ ψ(e φ(z)) fixes 0 and R, and also fixes |z| = R as a
set. Since the mapping of reflection in a circle depends only on the circle, this
map must preserve pairs of points that are symmetric under reflection in |z| = R.
Now {0, ∞} has such reflective symmetry and ψ(e−iθ φ(0)) = 0, so we conclude
ψ(e−iθ φ(∞)) = ∞. Thus we have a fractional linear transformation fixing three
points, which we know is the identity, and so ψ(e−iθ φ(z)) = z, from which φ(z) =
2
eiθ ψ−1 (z). It is easy to compute that ψ−1 (z) = Rz+aR
R+āz . We have thus shown that
φ(z) = eiθ
Rz + aR2
.
R + āz
3.3.3.7 If we had a fractional linear transformation taking |z| = 1 and |z − 14 | = 14 to
concentric circles, then by composing with a translation to move their common
center to 0 and a dilation we could assume that the fractional linear transformation
preserves the unit circle; note that neither operation affects the ratio of radii of
the image circles. By exercise 3.3.3.5 with R = 1, such a map is of the form
z−a
= f (z), |a| , 1. The rotation preserves both the concentric property
z 7→ eiθ 1−āz
and the ratio of radii, so we are free to choose it, for example by setting θ = 0.
Now observe that the initial and final configurations are mapped to themselves by
z−ā
z 7→ z̄, so g(z) = f (z̄) = e−iθ 1−az
also preserves |z| = 1 and takes |z − 14 | = 14 to a
circle with center at 0. If the image of |z − 14 | = 41 under f is |z| = r, it follows from
exercise 3.3.3.5 that
f ◦ g−1 (z) =
z+ā
1+az
−a
z+ā
1 − ā 1+az
=
ā−a
(1 − a2 )z + (ā − a)
1 − a2 z + 1−a2
=
a−ā
(1 − |a|2 ) + (a − ā)z 1 − |a|2 1 + 1−|a|
2z
1
2
2
b
is a fractional linear transformation fixing |z| = r, so is of the form z 7→ eiφ rz+r
.
r+b̄z
We conclude that a ∈ R. Knowing this, we see that f preserves R, and therefore
that f (0) and f ( 12 ) are the points ±r. This gives us that simultaneously −a = ±r
and ( 21 − a)/(1 − a2 ) = ∓r, which we reduce to r2 ± 4r + 1 = 0, or (finding only
√
solutions with r > 0), a = r = 2 ± 3. We have found necessary conditions for
f to map |z − 14 | = 14 to |z| = r, and it remains to see they are sufficient. But f
is a fractional linear transformation so maps circles to circles; by construction the
image of |z − 41 | = 14 passes through ±r, and since a ∈ R the image is mapped
to itself by z 7→ z̄, so it is |z| = √
r. We have therefore determined that the map is
z−a
one of z 7→ eiθ 1−az
for a = 2 ± 3 and the ratio of the smaller to larger radius is
√
√
2 − 3 : 1 = 1 : 2 + 3.
3.3.3.8 We are now asked to repeat this problem for |z| = 1 and x = 2. All of the same
z−a
arguments apply to say that the map is z 7→ eiθ 1−az
for a ∈ R. The points z = 2
and z = ∞ are invariant under z 7→
z̄,
so
go
to
±r.
Computing
as in the previous
√
√
1
problem we find again r = 2 ± 3, but with a = r = 2 ∓ 3. Alternatively
we could notice that the map z 7→ 1z takes the circle in 3.3.3.7 to the line in this
problem (to see this, note what happens to 0 and ∞ and that the symmetry under
z 7→ z̄ is preserved), so the whole problem is obtained from 3.3.3.7 by composition
with z 7→ 1z .
3.4.2.2 The circles intersect at z = 1 and are parallel there, so any fractional linear transformation mapping 1 to ∞ will give us the region between two circles that intersect
at ∞ with zero angle, meaning a strip between two parallel lines. There are many
fractional linear transformations taking 1 to ∞, but it is convenient to take one that
maps one of the circles to the real axis; for example z 7→ i z+1
z−1 takes |z| = 1 to R
because it takes −1 to 0 and i to 1. The image of the other line must have the form
z = x + ic for a constant c; by computing 0 7→ −i we find c = −i. Now we can
multiply by −π and exponentiate to get the upper half plane. Our final map is
z + 1
z 7→ exp −πi
.
z−1
3.4.2.3 The arc |z| = 1, y ≥ 0 can be mapped to a segment on the real line by a fractional
linear transformation, and we know how to map the complement of a segment on
the complement of the unit disc (or rather we know how to map the complement of
the unit disc on the complement of the segment [−2, 2] using the map z 7→ z + z−1 ).
For the first step we should take a point on |z| = 1 to ∞, and since it is convenient
to make the image segment symmetric around 0 we may as well take i to 0 and −i
to ∞. After multiplying by 2 so our arc goes to [−2, 2], the map is z 7→ 2i z−i
z+i .
−1
Now the trickier issue is how to invert
√ z 7→ z+z . Formally using the quadratic
equation the inverse is z 7→ 12 (z ± z2 − 4), but we have to make sense of the
square root. If z2 − 4 mapped the complement of [−2, 2] to the complement of a
ray connecting 0 and ∞ we could use a version of the usual square root, but it does
not – the image in this case is the complement of the interval [−4, 0]. However,
we can do whatever algebraic manipulation we desire to the expression for the
formal inverse and still have a formal inverse. Knowing that we want the bit inside
the square root to omit a ray from 0 to ∞ (preferably the negative real axis) and
that right now what is in there omits [−4, 0], tends to suggest we should divide by
either z or z − 2 to move an endpoint to ∞. A little playing with the formula yields
3
a different formal expression for the inverse branches
p
z
f± (z) = 1 ± 1 − 4/z2
2
in which 1 − 4/z2 is easily seen to omit the negative real axis (and the ray (2, ∞)
√
on the positive real axis). We can therefore use the usual definition of · taking
C \ (−∞, 0] to the right half plane. The result is that the branches f± (z) are each
well defined on the complement of [−2, 2].
Now note that the two branches of the inverse map either to the interior or the
exterior of the unit disc, and that we want
√ the one which maps to the exterior. We
z
can verify that the desired one is 2 (z + 1 − 4z−2 ) by observing that it has image
converging to ∞ as z → ∞.
Our final result is therefore that
4(z + i)2 1/2 1 + iz (z + i)2 1/2 1 z − i 1
+
1+ 1−
=
1
+
z 7→ 2i
2 z+i
z+i
−4(z − i)2
(z − i)2
2
(z+i)
where we readily determine that for z not on our arc, 1+ (z−i)
2 is not on the negative
real axis.
3.4.2.7 We wish to map the exterior of an ellipse on the interior of the unit disc with
preservation of symmetries, those being z 7→ z̄ and z 7→ −z. Recall that in studying
the map z 7→ z + z−1 = f (z) we saw that the circle z = reiθ , is mapped to z = x + iy
with x = (r + r−1 ) cos θ and y = (r − r−1 ) sin θ, which is an ellipse centered at
0 with semi-major axis along the real line and of length a = r + r−1 and semiminor axis along the imaginary axis and of length b = |r − r−1 |; this is the ellipse
(x/a)2 + (y/b)2 = 1, a > b. The map also preserves the symmetries of the ellipse,
in that f (z̄) = f (z) and f (−z) = − f (z).
The map z 7→ f (z) is a double covering; if r > 1 then both |z| > r and |z| < 1/r
are mapped onto the the exterior of the ellipse. It is easy to see that when r > 1
we should have r = (a√− b)/2 and thus 1/r = 2/(a − b). The inverse map has two
branches z 7→ 2z (1 ± 1 − 4z−2 ) which were discussed in the previous problem.
√
Here we need the branch z 7→ 2z (1 − 1 − 4z−2 ) which maps the exterior of the
ellipse to the interior of the disc of radius 2/(a − b). There are only a few more
difficulties. The first is that we need the ellipse to have a > b, so that a preliminary
rotation is needed if b > a. The second is that we need r = (a − b)/2 > 1 in
the above analysis. To ensure that this is the case it is convenient to perform a
preliminary dilation, for example one ensuring that r = 2. Then the inverse branch
of f will map to the disc |z| < 1/2, and an additional dilation will bring us to
|z| < 1. At this point it is simplest to split into cases.
Case 1: a > b. Then (x/a)2 + (y/b)2 = 1 has its longer axis along the real line.
We perform the dilation z 7→ 4z/(a − b), so that the length of the semi-major axis
becomes 4a/(a − b) and of the semi-minor axis becomes 4b/(a − b);
√ the exterior of
this is the image of |z| < 1/2 under f (z), so applying z 7→ 2z (1 − 1 − 4z−2 ) takes
it to |z| < 1/2 and dilating by a factor of 2 gives |z| < 1. The result is
4z 4(a − b)2 1/2 4z (a − b)2 1/2 =
z 7→ 2
1− 1−
1− 1−
2
2(a − b)
a−b
16z
4z2
Case 2: b > a. We can use the same argument as above, but must first rotate by
π/2 to get the semi-major axis along the real line, and must undo this at the end.
The maps are z 7→ iz 7→ 4iz/(b − a), then the inverse branch, then dilation by 2 and
4
rotation by −i, resulting in
4(b − a)2 1/2 (a − b)2 1/2 4iz 4z 1− 1−
1
−
1
+
z 7→ −2i
=
2(b − a)
b−a
−16z2
4z2
Case3: b = a. Here we have just the circle |z| = a = b, so we can scale
by z 7→ z/a to the unit circle and invert to map the exterior to the interior. The
resulting map is
a
z 7→ .
z
Math 5120: Complex analysis. Homework 5 Solutions
4.1.3.2 Using the parametrization, z = reiθ , so z = x + iy with x = r cos θ, y = r sin θ we
have
Z
Z 2π
x dz =
r cos θireiθ dθ
|z|=r
0
Z
2π
cos2 θ + i cos θ sin θ dθ
Z 2π
Z 2π
1
1
= ir2
(cos 2θ + 1) dθ − r2
sin 2θ dθ
2
2
0
0
= πir2 .
R
Using x = 12 (z + r2 z−1 ) as suggested in the text we use that |z|=r z dz = 0 by
R
equation (11) on pg 107 and |z|=r z−1 dz = 2πi by the following discussion to
R
conclude |z|=r x dz = πir2 .
4.1.3.3 The function (z2 − 1)−1 does not have an antiderivative on the disc |z| ≤ 2, so we
cannot directly apply the results from section 3.3. Since we do not yet have the
Cauchy theorem, we must integrate directly using a parameter. This is messy, but
the point is to get you to appreciate the Cauchy theorem and residue calculus once
we have them. Take z = 2eiθ , so the integral is
Z π
Z π
Z π
Z π
2ieiθ
6i cos θ + 10 sin θ
2ieiθ (4e−2iθ − 1)
8ie−iθ − 2ieiθ
dθ
=
dθ.
dθ
=
dθ
=
2iθ − 1
2iθ − 1)(4e−2iθ − 1)
17
−
8
cos
2θ
17 − 8 cos 2θ
4e
(4e
−π
−π
−π
−π
The real part of the integral has an antisymmetric integrand on a symmetric interval
so is zero. For the imaginary part we note from symmetry and periodicity of cosine
that
Z π
Z π
cos θ
cos θ
dθ = 2
dθ
17
−
8
cos
2θ
17
−
8 cos 2θ
−π
0
Z π/2
Z π
cos θ
cos θ
=2
dθ + 2
dθ
17
−
8
cos
2θ
17
−
8 cos 2θ
0
pi/2
Z 0
Z π/2
cos(π − θ)
cos θ
dθ + 2
(−dθ)
=2
17 − 8 cos 2θ
pi/2 17 − 8 cos(2π − 2θ)
0
Z π/2
Z π/2
cos θ
− cos(θ)
=2
dθ + 2
dθ
17
−
8
cos
2θ
17
− 8 cos(2θ)
0
0
=0
= ir2
0
so that the whole integral is zero.
4.1.3.6 We know that f 0 / f should have antiderivative log f , provided that log f is well
defined. Suppose then that | f (z) − 1| < 1 for z ∈ Ω. We know that log is a singlevalued analytic function on C \ (−∞, 0], with derivative z−1 . Since the values of
f avoid (−∞, 0] it follows that log f is well defined and analytic, with derivative
(by the chain rule) f 0 (z)/ f (z). Hence the integrand is the derivative of an analytic
function on Ω and by the result on page 107 the integral around any closed curve
in Ω is zero. R
4.1.3.8 In order that γ log z = 0 is meaningful and true, we must first require that γ
lies in a region where log z is a single-valued function (otherwise the integral is
not well-defined). We know Ω = C \ (−∞, 0] is such a region (in fact, so is the
1
2
complement
of any connected set containing 0 and ∞). Now we might hope to find
R
that γ log z = 0 for any closed curve in Ω by proving that log z is the derivative of
an analytic function. Since we know that (log z)0 = z−1 on Ω, it is easy to check
that (z log z − z)0 = log z + z/z − 1 = log z there. Hence a suitable condition is that
γ is a closed curve in Ω.
4.2.2.1 Using Cauchy’s integral formula we have
Z
ez
dz = 2πie0 = 2πi.
|z|=1 z
1
1
1
4.2.2.2 Writing z21+1 = (z+i)(z−i)
= 2i1 z−i
− z+i
we can apply the Cauchy integral formula
to each of the below integrals
Z
Z
Z
1
1
1
1
1
dz =
dz −
dz = π − π = 0.
2+1
2i
z
−
i
2i
z
+
i
z
|z|=2
|z|=2
|z|=2
Compare this to exercise 4.1.3.3 above!
Math 5120: Complex analysis. Homework 6 Solutions
z−n f (z) dz = 2πi f (n−1) (z)/(n − 1)! proR
vided f is analytic in the unit disc. Applying this to f (z) = ez gives |z|=1 z−n ez dz =
2πi/(n − 1)!.
R
4.2.3.1(b) If n and m are both non-negative then the integrand in |z|=2 zn (1 − z)m dz is a
polynomial, hence analytic, and so the integral is zero by the Cauchy formula.
If n is negative and m non-negative then our previous reasoning says the result
m
is 2πi f (|n|−1)(0)/(|n|
− 1)! for f (z) = (1 − z) , which is 0 if m < |n| − 1 and
|n|−1 m
2πi(−1)
|n|−1 otherwise. Similarly, if n is non-negative and m negative then
n
we get 0 if n < |m| − 1 and 2πig(|m|−1) (1)/(|m| − 1)! = 2πi |m|−1
otherwise (with
n
g(z) = z ). If both m and n are negative it is a little more work, basically because we do not have the Cauchy formula for general curves, so cannot reduce
to a circle around 0 and another around 1. However there is a still a relatively
quick argument of the same type as above. We may use partial fractions to write
zn (1 − z)m = P(z)zn + Q(z)(1 − z)m , where P has degree at most |n| − 1 and Q
has degree at most |m| − 1; note that P(z)(1 − z)−m + Q(z)z−n = 1. Integrating the
P(z)zn term gives 2πiP(|n|−1) (0)/(|n| − 1)!, which is just 2πi times the leading coefficient p of P(z). Similarly integrating the Q(z)(1 − z)m term gives (−1)m 2πi times
the leading coefficient q of Q. These leading terms give the power of z−m−n−1 in
P(z)(1 − z)−m + Q(z)z−n to be (−1)m (p + q), and this must be zero because m, n < 0
implies −m − n ≥ 2, so −m − n − 1 ≥ 1. We may summarize our results as
m 


if n < 0 and m ≥ −n − 1
2πi(−1)−n−1 −n−1
Z



 n n
m
z (1 − z) dz = 
2πi −m−1
if m < 0 and n ≥ −m − 1



|z|=2

0
otherwise
2
4.2.3.1(c) We write |z − a| = (z − a)(z̄ − ā) = (z − a) (ρ2 /z) − ā on |z| = ρ and use the fact
that on z = ρeiθ we have |dz| = ρdθ and dz = ireiθ dθ = izdθ, so |dz| = ρdz/iz. Thus
the integral is
Z
Z
Z
−2 ρ
1 ρ2
z
ρ
−4
|z − a| |dz| =
dz.
− ā
dz =
2
2
z
iz
i |z|=ρ (z − a) (ρ2 − āz)2
|z|=ρ
|z|=ρ (z − a)
4.2.3.1(a) By the Cauchy formula for derivatives,
R
|z|=1
Now by assumption, |a| , ρ. The integrand has singularities at z = a and z = ρ2 /ā,
which are reflection-symmetric in the circle |z| = ρ, so only one is inside the circle.
The Cauchy formula then says that the integral is 2πρ f 0 (a) with f (z) = z/(ρ2 − āz)2
if |a| < ρ, which evaluates to 2πρ(ρ2 + |a|2 )/(ρ2 − |a|2 )3 . Similarly if |a| > ρ
it becomes 2πρg0 (ρ2 /ā) with g(z) = z/(ā)2 (z − a)2 , which is the negative of the
previous answer. Summarizing
 2πρ(ρ2 +|a|2 )
Z


if |a| < ρ
 (ρ2 −|a|2 )3
−4
|z − a| |dz| = 
2
2
2πρ(ρ
+|a|
)

− 2 2 3
if |a| > ρ.
|z|=ρ
(ρ −|a| )
4.2.3.2 We assume f is globally analytic and there is C so | f (z)| ≤ C|z|n for all sufficiently
large |z|. Let P(z) be the (n−1)th order Taylor polynomial of f (z) at 0, so f (z)−P(z)
has a zero of order n at 0 and therefore f (z) − P(z) = zn g(z) for some globally
analytic g(z). Since P is an order n − 1 polynomial we have |P(z)|/|z|n ≤ 1 for all
sufficiently large |z|; using the hypothesis we get
| f (z)| |P(z)|
|g(z)| ≤
+
≤C+1
|z|n
|z|n
1
2
for all sufficiently large |z|, whence continuity of g implies it is bounded and Liouville’s theorem implies it is a constant a. Thus f (z) = azn + P(z) is a polynomial
of degree at most n.
4.2.3.5 Suppose f is analytic at z0 . Let g(z) = f (z) − f (z0 ). If n ≥ 1 is an integer and r > 0
is sufficiently small (so f is analytic on |z| ≤ r), then by the Cauchy formula
(n)
(n)
n! Z
n!
n!
g(z)
f (z0 ) = g (z0 ) = dz ≤ n max |g(z)| = n max | f (z)− f (z0 )|.
n+1
r |z−z0 |=r
2πi |z−z0 |=r (z − z0 )
r |z−z0 |=r
Putting r = 1/n we see that n!nn f (n) (z0 ) ≤ max|z−z0 |=1/n | f (z) − f (z0 )| → 0 as
n → ∞. This is strictly stronger than the statement that f (n) (z ) > n!nn cannot
0
occur for an unbounded sequence of n ∈ N.
4.3.2.1 The algebraic order of f at a is defined to be that h such that



α<h
∞
α
lim |z − a| | f (z)| = 

0
z→a
α > h.
If f has order h and g has order k at z = a, then it is immediate that |z−a|α | f (z)g(z)| =
∞ for α < h + k and = 0 for α > h + k, so the order of f (z)g(z) is h + k. It is also
easy to see that the order of 1/g is −k; combining these results we see that the
order of f (z)/g(z) is h − k.
To see that the order of f + g cannot exceed max{h, k} it suffices to note that
|z − a|α | f (z) + g(z)| ≤ |z − a|α | f (z)| + |z − a|α |g(z)| → 0 as z → a, by the triangle
inequality.
4.3.2.3 The map ez takes any infinite horizontal strip of width 2π to C \ {0}, so it takes
any punctured neighborhood |z| > R of ∞ to C \ {0}. It follows that ez does not
have a limit in C̄ as z → ∞, and therefore that the isolated singularity of ez at ∞ is
essential. Similar arguments apply to cos z and sin z. All we need verify is that the
image of |z| > R cannot converge in C̄ as R → ∞, which is true for cos because it
maps all lines z = 2πk + iy onto [0, ∞) ⊂ R and for sin because it maps the same
lines onto the imaginary axis.
4.3.2.5 Suppose f has an isolated singularity at a, so f is analytic on 0 < |z − a| <
δ. If either < f or = f is bounded on this punctured disc, then the image of the
punctured disc cannot be dense in C, so the Casorati-Weierstrass theorem implies
the singularity cannot be essential. The singularity is therefore removable or a
pole. However, if it is a pole then f (z) = (z − a)−k g(z) for some k ∈ N and g
analytic on |z − a| < δ with g(a) , 0. Then we can easily verify that both < f
and = f are unbounded on 0 < |z − a| < δ as follows. Take z = a + reiθ , so
f (z) = r−k e−ikθ g(z). For r > 0 so small that |g(z) − g(a)| < |g(a)|/2 and θ chosen
so e−ikθ g(a) is real we have < f (z) ≥ r−k |g(a)|/2. A similar argument works for
the imaginary part. It follows that the singularity cannot be a pole, so it must be
removable.
4.3.2.6 Let f have an isolated singularity at a, so f is analytic on 0 < |z − a| < δ. Consider
g(z) = e f (z) , which also has an isolated singularity at a. Suppose the singularity of
g(z) is a pole. Then limz→a |g(z)| = ∞, and |g(z)| = e< f (z) , so limz→a < f (z) = ∞.
In particular, f (z) is unbounded on 0 < |z − a| < δ, so the singularity of f is not
removable. Also, f (z) omits a left half-plane, so the Casorati-Weierstrass theorem
implies the singularity of f cannot be essential. Thus f has a pole at a. Reasoning
as in the previous question we see that for any sufficiently small r > 0, the image
of any 0 < |z − a| < r under f covers a set of the form |z| > R. However the
3
exponential map takes any horizontal infinite strip of height 2π to C \ {0}. Since a
region of the form |z| > R contains such a horizontal strip we conclude that g maps
any 0 < |z − a| < r to all of C \ {0}, contradicting the assumption that g has a pole
at a. We conclude that if f (z) has an isolated singularity at a then e f (z) cannot have
a pole at a.
Math 5120: Complex analysis. Homework 7 Solutions
4.3.3.1 Observe that f (0) = 0 and f 0 (0) , 0, so that f is injective on some neighborhood
of 0. Also f (z) is not injective on |z| < 2, because f (0) = f (−1) = 0. It follows
that R = sup{r : f is injective on |z| < r} is positive and finite.
One easy
√ way to find R is by direct computation. The roots of f (z) − ζ are
1+4ζ 2
z = − 21 ±
, so they are at opposite points of a diameter of a circle radius
2
√1+4ζ 2 s = 2 around − 21 . One of these points has real part less than − 12 , so is
distance at least 21 from 0. We conclude that no open disc |z| < r with r ≤ 12
contains two roots of f (z) − ζ. Moreover if r > 21 we may choose ζ to make s so
small that |z − 21 | < s is inside |z| < r, at which point there will be two roots of
f (z) − ζ in |z| < r. This proves that R = 12 .
A more general approach is to first recognize that an analytic f cannot be injective on an open set containing a critical point. The reason is that if f 0 (z0 ) = 0,
then f (z) − f (z0 ) has a zero of some order n ≥ 2 at z0 . The argument principle
then says that if w is sufficiently close to f (z0 ), the function f (z) − w must have n
simple roots in a neighborhood of z0 , so f cannot be injective. (See book, top of
page 132.) For our problem, the critical point is at 2z + 1 = f 0 (z) = 0, so z = − 12
and we discover R ≤ 21 .
Now if f is a polynomial, then we know that all critical points lie in the convex
hull of the roots (see Theorem 1 on page 29 of the book). So if we have a polynomial f of degree n and a w so f (z) − w has n roots (counting multiplicity) in |z| < r,
then f also has a critical point in |z| < r. In the situation we face, with n = 2, we
discover that if there are 2 roots of f (z) − ζ in |z| < r then r > 12 . Thus R ≥ 12 , and
we conclude R = 12 .
It is interesting to think how, if at all, this idea could be generalized.
4.3.3.4 We are given f analytic at 0 and f 0 (0) , 0. Then f (z) − f (0) = zh(z) with h
analytic at 0 and h(0) = f 0 (0) , 0. We may then take an open disc U around 0
which is so small that h is analytic on this disc and |h(z) − h(0)| < |h(0)| for z ∈ U.
This condition on h implies log h(z) and hence k(z) = h(z)1/n are well-defined
(single-valued) analytic functions on U, from which f (zn ) = f (0) + zn h(zn ) =
f (0) + zk(zn ) 1/n on U, so taking g(z) = zk(zn ) completes the proof.
1
Math 5120: Complex analysis. Homework 8 Solutions
4.3.4.1 Suppose | f (z)| ≤ 1 on |z| ≤ 1 and f analytic. Fix z0 in the open unit disc and
let w0 = f (z0 ). Using a composition of fractional linear transformations and the
Schwarz lemma, it is proved in the book that
f (z) − w0 ≤ z − z0 .
(1)
1 − w̄0 f (z) 1 − z¯0 z With a little rewriting we can extract f 0 (z0 ):
f (z) − w0
| f 0 (z0 )|
1
≤ lim 1 =
(2)
=
lim
2
z→z
z→z
1 − | f (z)|
1 − |z0 |2
0 1 − z¯0 z
0 (z − z0 )(1 − w̄0 f (z))
and z0 was an arbitrary point in the disc, so the estimate is valid for all |z0 | < 1.
4.3.4.2 Suppose f is non-constant analytic and = f (z) ≥ 0 if =z ≥ 0. Composition with
a fractional linear map gives a map from the unit disc to itself to which we can
apply the Schwarz lemma. Specifically, for any z0 with =z0 > 0 we set w0 = f (z0 )
and have =w0 > 0 by the maximum principle. For α in the upper half-plane let
−1
gα (z) = z−α
z−ᾱ so gz0 maps the unit disc to the upper half-plane with 0 7→ z0 and gw0
maps the upper half-plane to the unit disc with w0 7→ 0. Thus gw0 ◦ f ◦ g−1
z0 takes
the unit disc to itself and fixes 0, whence |gw0 ◦ f ◦ g−1
(z)|
≤
|z|
by
the
Schwarz
z0
lemma, and so |gw0 ◦ f (z)| ≤ |gz0 (z)|. Substituting gives
f (z) − f (z0 ) z − z0 .
(3)
≤ z + z̄0 f (z) + f (z0 )
As in the previous question, we can obtain a bound for | f 0 (z0 )| by dividing both
sides of the previous equation by |z − z0 | and sending z → z0 . Since z + z̄ = 2=z,
we obtain
| f 0 (z0 )|
1
(4)
≤
= f (z0 ) =z0
valid at all points in the upper half-plane.
4.3.4.3 Let
( f (z) − w0 )(1 − z̄0 z)
.
F(z) =
(z − z0 )(1 − w̄0 f (z))
Then F is analytic on the open unit disc except for a removable singularity at z0 ,
and (1) says |F(z)| ≤ 1. If equality holds in (2) then |F(z0 )| = 1, so the maximum
principle implies F is constant of modulus 1, which we may write as eiθ , θ ∈
z−α
iθ
[0, 2π). Multiplying out gives that f (z) = h−1
w0 (e hz0 (w)) where hα = 1−ᾱz .
Similarly, if we let
( f (z) − w0 )(z + z̄0 )
G(z) =
(z − z0 )( f (z) + w̄0 )
then we obtain a function analytic on =z > 0 except for the removable singularity
at z0 , and |G(z)| ≤ 1 by (3). Equality in (4) implies G ≡ eiθ for some θ ∈ [0, 2π) by
the same maximum principle argument as before, and inverting the maps we have
iθ
f (w) = g−1
w0 (e gz0 (w)) with g as in the previous exercise.
In either of these two cases we see that f is a composition of fractional linear
maps, so is fractional linear.
4.4.7.3 Let γ be a closed curve. Its complement is open, so the connected components
of the complement are open, and there are thus at most countably many of them.
Label them {U j }∞j=0 , and let U0 be the unique unbounded one; it is unique because
1
2
compactness of γ implies the complement of some large disc is in C \ γ, and this
is a connected set so is contained in a single component.
For each j, let V j = C̄ \ U j and observe V j = ∪k, j U j ∪ γ ∪ {∞}. As these sets are
connected, each must lie in exactly one component of V j . However connectivity
of Uk implies connectivity of its closure, which intersects γ, so all Uk and γ must
lie in a single component of V j . Moreover if j , 0 then U0 ⊂ V j and ∞ is its
closure, so is in this same component of V j . We conclude that if j , 0 then V j has
only one component, so U j is simply connected by definition. Also V0 has at most
two components, one being {∞} and the other ∪ j ≥ 1U j ∪ γ. Taking r so |z| = r
does not intersect γ, we see that the disjoint open sets |z| > r and |z| < r separate
V0 into these two components, so V0 is doubly connected.
4.4.7.5 Let Ω be a region not intersecting a connected set E with ±1 ∈ E. If φ(z) is a
1−z2
globally analytic function such that f (z) = (φ(z))
2 maps E to a connected region
containing 0, then f has a well-defined logarithm and therefore well-defined roots
on on C \ f (E). Provided
√ f (Ω) does not intersect f (E) we may define a singlevalued analytic function 1 − z2 on Ω by
s
p
1 − z2
1 − z2 = φ(z)
(φ(z))2
and it is a legitimate branch of the square root because squaring both sides leads
to an equality.
It remains to find such a φ(z), but writing 1 − z2 = (1 − z)(1 + z) immediately
suggests φ(z) = (1 + z), as then f (z) = 1−z
1+z , which is a fractional linear transformation with 1 7→ ∞ and 1 7→ 0, from which f (E) is a connected set containing 0 and
∞. Our definition is then
r
p
1−z
2
1 − z = (1 + z)
.
1+z
Now consider the integral
Z
Z
1
dz
=
dz
√
q
2
1−z
γ (1 + z)
γ
1−z
1+z
where γ is a closed curve in a component of C \ E. The integrand has no singularities in this component, so if γ does not wind around any point of E then the
integral is zero. In particular if ∞ ∈ E then γ cannot wind around E, so the integral
is zero in this case.
We therefore suppose that E is bounded and take r > 0 so large that E ⊂ {z :
|z| < r}. In this case the winding number n(γ, z) is a constant 2πiN on E and γ is
homologous in the unbounded component of C \ E to Nγr , where γr is the circle
of radius r around 1. The Cauchy theorem then implies
Z
Z
Z 2π
Z 2π
1
ireiθ dθ
dθ
dz
=N
dz. = N
= iN
q
√
q
q
1−z
2
2
γ
γr (1 + z)
0
0
1 − z2
reiθ reiθ − 1
−1
1+z
reiθ
It is tempting at this point to attempt to use the residue theorem, but that theorem
is valid only for isolated singularities, and our square root has singularities along
a connected set joining ±1. Instead we observe that we may take r → ∞ without
√
changing the value of the integral. We find that the integrand converges to 1/ −1,
which is one of ±i (we cannot determine which without knowing more about the
3
set E; draw some pictures to see why). Therefore the possible values of the integral
are ±2πN, or any element of 2πZ.
4.5.2.1 Let f (z) = z7 − 2z6 − z + 1. Then | f (z)| ≤ 5 on |z| = 1, so Rouché’s theorem implies
6z3 + f (z) has the same number of roots as 6z3 in the unit disc, namely three.
4.5.2.2 We use Rouché’s theorem twice. For |z| = 2, |z4 | = 16 > | − 6z + 3|, so z4 − 6z + 3
has 4 roots in |z| ≤ 2. For |z| = 1, |6z| = 6 > |z4 + 3|, so there is one root in |z| ≤ 1.
We conclude that there are 3 roots in 1 < |z| < 2.
Math 5120: Complex analysis. Homework 9 Solutions
4.5.3.1.a
f (z) =
1
1
=
z2 + 5z + 6 (z + 2)(z + 3)
which has
• a pole of order 1 at z = −2 with residue limz→−2 (z + 2) f (z) = 1
• a pole of order 1 at z = −3 with residue limz→−3 (z + 3) f (z) = −1
4.5.3.1.b
f (z) =
1
1
=
(z2 − 1)2 (z − 1)2 (z + 1)2
which has
d
• a pole of order 2 at z = 1 with residue limz→1 dz
(z − 1)2 f (z) = − 14
d
• a pole of order 2 at z = −1 with residue limz→−1 dz
(z + 1)2 f (z) = 14
1
4.5.3.1.c f (z) = sin z has poles at the zeros of sin z, so at the points πk, k ∈ Z. These zeros
are simple, because f 0 (z) = cos z = ±1 at these points, and consequently the poles
are simple. The residue at πk may be computed by L’Hopital’s rule
z − πk
1
= lim
= (−1)k
z→πk sin z
z→πk cos z
lim
so that f (z) has simple poles with residue (−1)k at each πk, k ∈ Z.
z
4.5.3.1.d As cos z is entire, f (z) = cot z = cos
sin z can only have poles at the zeros of sin z,
meaning the points z = πk, k ∈ Z. Since cos z , 0 at these points, there is a pole at
each such point, and since the zeros of sin z are simple the poles are also simple.
The residue at πk is
(z − πk) cos z
1
lim
= cos(πk) lim
=1
z→πk
z→πk cos z
sin z
so cot z has simple poles with residue 1 at each πk, k ∈ Z.
4.5.3.1.e f (z) = sin12 z has poles at each of the zeros z = πk, k ∈ Z of sin z. These zeros are
order 1, so the zeros of sin2 z are order 2. The residues may be computed using
L’Hopital
d (z − πk)2 2(z − πk) sin z − 2(z − πk)2 cos z
lim
=
lim
z→πk dz
z→πk
sin2 z
sin3 z
sin z − (z − πk) cos z z − πk = 2 lim
lim
z→πk sin z z→πk
sin2 z
cos z − cos z + (z − πk) sin z = 2 lim
z→πk
2 sin z
= lim (z − πk) = 0.
z→πk
We determine that f (z) has poles of order 2 at each of the points πk, k ∈ Z and has
residue zero at each pole.
4.5.3.1.f f (z) = z−m (1 − z)−n , m, n ∈ N has a pole of order m at 0 and a pole of order n at 1.
We may compute the residue at 0 using
lim
z→0
dm−1
(n + m − 2)!
1
1
(1 − z)−n =
(1 − 0)−n−(m−1)
m−1
(m − 1)! dz
(m − 1)! (n − 1)!
!
!
(n − 1) + (m − 1)
(n − 1) + (m − 1)
=
=
m−1
n−1
1
2
The residue at 1 may be computed the same way
1
dn−1
1
dn−1
lim n−1 (−1)n z−m
(z − 1)n f (z) =
n−1
z→1 (n − 1)! dz
(n − 1)! z→1 dz
(m + n − 2)!
= (−1)n (−1)n−1
(1)−m−(n−1)
(n − 1)!(m − 1)!
!
!
(n − 1) + (m − 1)
(n − 1) + (m − 1)
=−
=−
m−1
n−1
so we conclude that f has a pole of order m at zero with residue (n−1)+(m−1)
and
n−1
(n−1)+(m−1)
a pole of order n at 1 with residue −
. It is worth noting that this is
n−1
consistent with exercise 4.2.3.1(b).
4.5.3.3 There are a few things we will use repeatedly in computing these integrals. The
curve γR will be the semicircle |z| = R in the upper half-plane, with the usual
(increasing angle) orientation. For R, S , T ∈ (0, ∞) we also let Γ1 = {S + iy : 0 ≤
y ≤ T }, Γ2 = {x + iT : −R ≤ x ≤ S }, Γ3 = {−R + iy : 0 ≤ y ≤ T }, oriented such
that Γ1 + Γ2 + Γ3 and the interval [−R, S ] ⊂ R form a positively oriented closed
curve. We will frequently use that (A) if the integrand f (x) is bounded by |x|−2 as
R∞
RR
|x| → ∞ then −∞ = limR→∞ −R f (x), and (B) if the integrand f (z) is bounded by
R
|z|−2 as |z| → ∞ then limR→∞ γ f (z)dz = 0
lim
R
4.5.3.3.a Note that sin2 x = sin2 (−x) = sin2 (π − x) = sin2 (π + x) implies
Z π/2
Z
dx
1 2π
dx
=
2
4
a
+
sin
x
a
+
sin2 x
0
0
and this may be seen as an integral on the unit circle with respect to the angle
dx = dz/iz where sin z = (z − z−1 )/2i. Thus
Z π/2
Z
dx
dz
1
1
=
2 iz .
2
−1
4
a
+
sin
x
0
|z|=1 a + (z − z )/2i
2
We simplify (z − z−1 )/2i = −(2z)−2 (z4 − 2z2 + 1) and find the integrand becomes
4iz
. At this point we can make our lives a little easier by making the
z4 −(2+4a)z2 +1
substitution w = z2 . Notice that when z winds once around the unit circle, w winds
around twice. We therefore find
Z
Z
Z π/2
1
2i
dw
dx
=
dw
=
i
.
2
2
2
a + sin x 2 |w|=1 w − (2 + 4a)w + 1
|w|=1 w − (2 + 4a)w + 1
0
Now we √
would like to say that the poles of w2 − (2 + 4a)w + 1 are at w± =
1+2a±2 a2 + a by the quadratic formula, but this requires that we make
√ sense of
the square root. Fortunately, |a| > 1 by hypothesis, so |1/a| <√1 and 1 √
+ (1/a) is
2 + a by
well-defined.
We
may
therefore
define
an
analytic
branch
of
a
a2 + a =
√
√
a 1 + (1/a), obtaining w± = 1 + 2a ± 2a 1 + (1/a). By construction, each of
w± is a branch of the inverse of the map w 7→ (w + w−1 )2 /4, evaluated at a. Since
w 7→ (w + w−1 )2 /4 takes the unit circle to the interval [−1, 1], and |a| > 1, we see
that |w± | , 1 (this is important for applying the residue theorem). It also follows
that only one of w± can lie inside the unit disc, and that which one does so is independent of a. Taking a ∈ (1, ∞) we readily see |w− | < 1 and |w+ | > 1, so this
must be true for all a. Thus the result of the integration can be computed from the
3
√
residue at the simple pole w− , which has value 1/(w− − w+ ) = −1/(4a 1 + (1/a)).
Finally
π/2
Z
0
dx
π
2πi2
=
.
=
√
√
2
a + sin x −(4a 1 + (1/a)) 2a 1 + (1/a)
4.5.3.3.b Using that the integrand is even and (A), then the residue theorem, and then (B)
∞
Z
0
Z
1 ∞
x2 dx
z2 dz
=
4
2
4
x + 5x + 6 2 −∞ z + 5z2 + 6
Z R
1
z2 dz
= lim
4
2 R→∞ −R z + 5z2 + 6
Z
X
1
z2 dz
Resz j − lim
= πi
4
2 R→∞ γR z + 5z2 + 6
j
= πi
X
Resz j
j
z2
z4 + 5z2 + 6
where the sum is over residues in the upper half-plane. √Now z4 + 5z2√+ 6 =
(z2 + 2)(z2 +√3), so the integrand
has simple
√
√ poles at z = ±i
√ 2 and z = ±i 3. The
residue at i 2 is −2/(2 2i)(1), and at i 3 is −3/(−1)(2 3i), so the result is
Z
0
∞
√
√
− 2
√ π
√
3
x2 dx
= πi
+
= ( 3 − 2) .
4
2
2i
2i
2
x + 5x + 6
4.5.3.3.c By (A), the residue theorem, and (B)
Z
∞
−∞
x2 − x + 2
dx
x4 + 10x2 + 9
z2 − z + 2
dz
R→∞ −R z4 + 10z2 + 9
Z
X
z2 − z + 2
= 2πi
Resz j − lim
dz
R→∞ γ z4 + 10z2 + 9
R
j
= lim
= 2πi
Z
X
j
R
Resz j
z2 − z + 2
z4 + 10z2 + 9
where the sum is over residues in the upper half-plane. The zeros of z4 +10z2 +9 =
(z2 + 1)(z2 + 9) are at ±i and ±3i; each produces a simple pole in the integrand,
and there are no others. The residue at i is (−1 − i + 2)/(2i)(−1 + 9) and at 3i is
(−9 − 3i + 2)/(−9 + 1)(6i), so the result is
Z
∞
−∞
1 − i 7 + 3i x2 − x + 2
π
5π
dx
=
2πi
+
= (3 − 3i + 7 + 3i)
=
.
4
2
16i
48i
24
12
x + 10x + 9
4
4.5.3.3.d Using that the integrand is even and (A), then the residue theorem, and then (B)
∞
Z
0
Z
x2 dx
1 ∞ z2 dz
=
(x2 + a2 )3 2 −∞ (z2 + a2 )3
Z R
1
z2 dz
= lim
2
2 R→∞ −R (z + a2 )3
Z
X
1
z2 dz
= πi
Resz j − lim
2
2 R→∞ γR (z + a2 )3
j
= πi
X
j
Resz j
(z2
z2
+ a2 )3
where the sum is over residues in the upper half-plane. Factoring (z2 + a2 )3 =
(z + ai)3 (z − ai)3 , a ∈ R, we see that there is a single pole of order 3 in the upper
half-plane, at i|a|. The residue there is
d2
z2
1
=
2
3
z→|a|i dz (z + |a|i)
8|a|3 i
lim
so that the result is
Z
0
∞
x2 dx
π
=
(x2 + a2 )3 8|a|3
4.5.3.3.e Using that the integrand is even and (A), that cos z = <eiz , then the residue theorem for R = S and T sufficiently large,
Z ∞
Z
cos x dx 1 ∞ cos z dz
=
2 −∞ z2 + a2
x2 + a2
0
Z R iz
1
e dz
= lim <
2 + a2
2 R→∞
z
−R
!
Z
Z
Z
X
eiz
1
eiz dz
eiz dz
eiz dz
Resz j 2
= πi
−
lim
<
+
2
2
2
2
2
2
z + a2 2 R,T →∞
Γ1 z + a
Γ2 z + a
Γ3 z + a
j
where the sum is over residues in the upper half-plane. However the integrand is
bounded by a constant multiple of e−y /|z|2 for z = x + iy and |z| sufficiently large.
Writing
f (z)
integrand, and taking R = S and T large enough we find that
R
for the
RT
−2
Γ f (z) dz ≤ R 0 e−y dy ≤ R−2 , and similarly for Γ3 . Now on Γ2 we have that
1
(z2 + a2 )−1 is integrable
(with
integral bounded by constant C) if T is large enough,
R
and therefore Γ f (z) dz ≤ Ce−T . Sending R and T to ∞ we find
2
Z
0
∞
cos x dx
π
= <πi Res|a|i f (z) =
2|a|
x 2 + a2
where at the last step we computed that z2 + a2 = (z + ai)(z − ai), has one simple
cos z
a
pole in the upper half-plane, at i|a|, with residue limz→|a|i z+|a|i
= cos
2|a|i .
2
2
iz
2
4.5.3.3.f We use that the integrand is even and x sin x/(x + a ) = =ze /(z + a2 ). Taking
R, S , T large enough that the curve (−R, S ) ∪ Γ1 ∪ Γ2 ∪ Γ3 encloses the simple pole
5
iz
Z
0
∞
−|a|
−|a|
ze
e
at |a|i, where the residue is limz→|a|i z+|a|i
= |a|ie
2|a|i = 2 we obtain
Z S
x sin x
1
zeiz
dx
=
dz
lim
=
2
2
2 R,S →∞
x2 + a2
−R z + a
!
Z
Z
Z
e−|a|
zeiz
zeiz
zeiz
= =πi
− lim =
dz +
dz +
dz
2
2
2
2
2
2
R,S →∞
2
Γ1 z + a
Γ2 z + a
Γ3 z + a
valid for all sufficiently large T . However, the integrand f (z) satisfies | f (z)| ≤
RT
|z|e−y
R
R
−y
for z = x+iy. The integral for Γ1 can be bounded by R2 −|a|
dy = R2 −|a|
2 0 e
2
|z|2 −|a|2
S
and similarly that for Γ3 can be bounded by S 2 −|a|
2 . The integral for Γ2 can be
−T
e
bounded by SS2 −|a|
2 (R + S ). If we first send T → ∞ so the Γ2 integral goes to 0,
and then send R, S → ∞ we find that they make no contribution to the result, and
therefore
Z ∞
x sin x
π
e−|a|
= |a|
dx
=
=πi
2 + a2
2
x
2e
0
4.5.3.3.g We will do this for general β ∈ (−1, 1), as it will be useful later. Take δ > 0, > 0,
and R > 2. Let Γ± = {rei±δ , r ∈ (, R)} be rays at angle ±δ, and also take arcs
ΓR = {Reiθ : θ ∈ (δ, 2π − δ)} and Γ = {eiθ : θ ∈ (δ, 2π − δ)}. Let zβ be a branch on
C \ [0, ∞), so it is well-defined and analytic in a simply connected neighborhood
of the closed curve Γ+ + ΓR − Γ− − Γ . Provided δ and are sufficiently small, this
curve winds once around the simple poles of f (z) = zβ (1 + z2 )−1 , which are at ±i,
and where there are residues iβ /2i and (−i)β / − 2i respectively.
Z
zβ
dz = 2πi(iβ − (−i)β )/2i = π(eiπβ/2 − ei3πβ/2 ).
2
Γ+ +ΓR −Γ− −Γ z + 1
Now on Γ+ we have zβ = rβ eiδβ , while on Γ− , zβ = rβ ei(2π−δ)β . It follows that
Z
Z R
zβ
xβ
i2πβ
lim δ → 0
dz
=
(1
−
e
)
dx.
2
2
Γ+ −Γ− z + 1
x +1
At the same time, we see that on Γ the integrand has the bound | f (z)| ≤ 2 β , and
the length of the curve is less than 2π, so the integral is bounded by 4π 1+β → 0
as → 0, provided β > −1. On ΓR we have | f (z)| ≤ Rβ /(R2 − 1), and the curve
has length less than 2πR, so the integral is bounded by 2πR1+β /(R2 − 1) → 0 as
R → ∞ provided β < 1. We conclude that if β ∈ (−1, 1) then
Z ∞
Z
zβ
xβ
1
dx
=
lim
dz
2
i2πβ
2
→0,R→∞,δ→0 1 − e
x +1
0
Γ+ +ΓR −Γ− −Γ z + 1
eiπβ/2 − ei3πβ/2
1 − ei2πβ
eiπβ (e−iπβ/2 − eiπβ/2 )
= π iπβ −iπβ
e (e
− eiπβ )
πβ sin(πβ/2)
sin(πβ/2)
π
=π
=π
= sec
sin πβ
2 sin(πβ/2) cos(πβ/2) 2
2
√
In the special case β = 13 , we have sin(π/6)/ sin(π/3) = 1/ 3, so that
Z ∞ 1/3
x
π
dx = √ .
2+1
x
0
3
=π
6
4.5.3.3.h For this problem, let Γ+ = (, R) ⊂ R and Γ− = (−R, −) ⊂ R, Γ and ΓR be the
semicircles of radius and R (respectively) in the upper half-plane. Define log z to
be the branch of the logarithm on the complement of the negative imaginary axis.
Taking Γ+ + ΓR − Γ− − Γ arranged to form a curve winding once around the simple
z
pole of (zlog
2 +1) at z = i, we find from the residue theorem that
Z
iπ2
log z
dz
=
2πi(log
i)/2i
=
.
2
2
Γ+ +ΓR +Γ− −Γ z + 1
Z
Γ+ +Γ−
The computations showing that the contributions from Γ and ΓR vanish in the
limit are essentially the same as in exercise 4.5.3.3.g. All that is different is we use
the bound | log z| ≤ (log |z| + 2π). Since log z is log |z| on Γ+ and log |z| + πi on Γ−
we find
Z R
Z −
Z R
Z R
log x
log x + πi
log x
1
log z
dz
=
dx
+
dx
=
2
dx
+
dx.
2+1
2+1
2+1
2+1
z2 + 1
x
x
x
x
−R
Combining these facts we see
iπ2
= lim
→0,R→∞
2
=
Z
log z
dz
z2 + 1
Γ+ +ΓR +Γ− −Γ
Z
0
log z
dz
z2 + 1
Z ∞
iπ
log x
dx
+
dx
x2 + 1
x2 + 1
0
log x
iπ2
dx +
2
2
x +1
Z
∞
lim
Z ∞
→0,R→∞
=2
0
=2
∞
Z
so that
0
Γ+ +Γ−
log x
dx = 0.
x2 + 1
4.5.3.3.i Let us first observe that f (x) = x(−1−α) log(1 + x2 ) is integrable on [0, ∞) because
it is bounded by Cα x−1−(α/2) as x → ∞ and α > 0, while as x ↓ 0 one has | log(1 +
x2 )| ≤ 2x2 so | f (x)| ≤ x1−α and α < 2. It follows that we can write the integral as a
limit and can integrate by parts
" −α
#R
Z ∞
Z R
Z R 1−α
log(1 + x2 )
log(1 + x2 )
x
1
2x
2
log(1
+
x
lim
dx
=
lim
dx
=
lim
)
+
dx.
1+α
1+α
R→∞ 1
R→∞ −α
α R→∞ R1 1 + x2
x
x
1
0
R
R
We observe that as R → ∞, R log(1 + R ) → 0, and also |Rα log(1 + R−2 )| ≤
2R2−α → 0, so the boundary term from the integration makes no contribution in
the limit. The remaining term may be dealt with by the computation in 4.5.3.3.g.
Indeed, from that problem with β = (1 − α) ∈ (−1, 1), we have
Z ∞
Z
(1 − α)π π
απ log(1 + x2 )
1 ∞ 2x1−α
π
dx
=
dx = sec
= csc
1+α
2
α 0 1+x
α
2
α
2
x
0
−α
2
4.5.3.4 Parameterizing |z| = ρ by z = ρeiθ we have dz = izdθ and |dz| = ρdθ, so |dz| =
2
ρdz/iz. Also |z − a|2 = (z − a)(z̄ − ā) = (z − a)( ρz − ā). Hence we find
Z
Z
Z
ρ
|dz|
ρ
=
dz =
dz
2
2 − āz)
ρ2
|z
−
a|
i(z
−
a)(ρ
|z|=ρ
|z|=ρ iz(z − a)(
|z|=ρ
z − ā)
7
which can be computed by the residue theorem. There are simple poles at a and
ρ2 /ā. By hypothesis, |a| , ρ; if |a| < ρ then a is inside |z| = ρ and ρ2 /ā is not, and
the reverse is true if |a| > ρ.
−ρ
ρ
2
The residue at z = a is i(ρ2 −|a|
2 ) and that at ρ /ā is i(ρ2 −|a|2 ) . We conclude from
the residue theorem that
 2πρ
Z

 ρ2 −|a|2 if |a| < ρ
|dz|
2πρ

.
=
= 
−2πρ

2
2

|z − a|
if |a| > ρ
ρ − |a|2 |z|=ρ
ρ2 −|a|2
Math 5120: Complex analysis. Homework 10 Solutions
4.6.2.2 If M(r) = 0 for some r > 0 then f vanishes on a set containing a limit point, so
f ≡ 0 and the result is trivial. Hence there is no loss of generality in assuming
M(r) > 0 for r > 0, in which case the statement
M(r) ≤ M(r1 )α M(r2 )(1−α)
for α =
log(r2 /r)
log(r2 /r1 )
is equivalent to
log M(r) ≤ α log M(r1 )+(1−α) log M(r2 ) =
log r − log r1
log r2 − log r
log M(r1 )+
log M(r2 )
log r2 − log r1
log r2 − log r1
which is the same as
log r2 −log r1 log M(r) ≤ log r2 log M(r1 )−log r1 log M(r2 )+ log M(r2 )−log M(r1 ) log r
or
log M(r1 )−log M(r2 ) log r + log r2 −log r1 log M(r) ≤ log r2 log M(r1 )−log r1 log M(r2 )
and it is this that we will prove.
It is suggested in the book that we apply the maximum principle (for harmonic
functions) to a linear combination of log |z| + log | f (z)|. Of course we cannot do
this directly if f has zeros, because log | f (z)| is not harmonic in any neighborhood
of a zero of f (in fact it is subharmonic, and there is still a maximum principle for
subharmonic functions, but we have not proved that). We will therefore need to do
something about points where f is zero, but let us begin by assuming that no such
points exist.
If f is analytic on the annulus 0 < r1 < |z| < r2 then A log |z| + B log | f (z)| is
harmonic there, and the maximum principle for harmonic functions implies that
the maximum occurs on the boundary. We obtain
(1)
A log r+B log M(r) = max A log |z|+B log | f (z)| ≤ max{A log r1 +B log M(r1 ), A log r2 +B log M(r2 )}
|z|=r
Taking A = log M(r1 ) − log M(r2 ) and B = log r2 − log r1 we find that the terms on
the right are both equal to log r2 log M(r1 ) − log r1 log M(r2 ). Thus
log r2 log M(r1 )−log r1 log M(r2 ) ≥ log M(r1 )−log M(r2 ) log r + log r2 −log r1 log M(r)
which is what we needed to prove.
Now we deal with the points {z j } where f (z) = 0. Such points are can accumulate only at the boundary. Suppose that around each we place a small disc of radius
δ j (small enough that it is inside the annulus), and delete these discs from our domain. Then (1) must be modified so that for each j there is a term on the right
side corresponding to the maximum of A log |z| + B log | f (z)| on the new boundary
circle |z − z j | = δ j . However B log | f (z)| → −∞ as z → z j , so we may choose δ j so
small that this new term is less than the right side of (1), and therefore need not be
included. It follows that (1) is still valid when f has zeros, and therefore the result
holds for general f .
Note that there is a degenerate case we did not consider, namely r1 = 0. In
this situation one should interpret the formula for α as corresponding to α = 0,
whereupon the result follows directly from the usual maximum principle for the
harmonic function | f (z)|.
1
2
5.1.1.1 Let
K ⊂ C be compact and M = maxz∈K |z|. Observe that for n > M we have
z < 1 when z ∈ K. The principal branch of the logarithm is well-defined on
n
{w : |1 + w| < 1}, so we conclude log 1 + nz is well-defined on K for all n > M
and has Taylor expansion
∞
z X (−1) j+1 z j
log 1 +
=
.
n
j
n
j=1
The series is readily seen to be convergent on 1+ nz < 1, thus uniformly convergent
on compact subsets of this region, and in particular on K for n > M. Uniformity
of the convergence implies we can exchange the limits in
∞
(−1) j+1 z j
z X
lim
=z
=
lim n log 1 +
n→∞
n→∞
n
j n j−1
j=1
Exponentiating both sides and using continuity of the exponential we get that
z
z n
ez = lim exp n log 1 +
= lim 1 +
n→∞
n→∞
n
n
uniformly on K, and since K was arbitrary the convergence is uniform on all compact sets in C.
5.1.2.3 We wish to develop log sinz z around 0 up to terms of order z6 . Since sin z has a
simple zero at 0 the function sinz z has a removable singularity at 0 and its extension
(which is equal to 1 at 0) is entire. It is helpful to recall the series for sin z and
divide by z to obtain a series convergent uniformly on all compact sets to sinz z
∞
∞
∞
X (−1) j z2 j
sin z 1 X (−1) j z2 j+1 X (−1) j z2 j
=
=
=1+
.
z
z j=0 (2 j + 1)!
(2 j + 1)!
(2 j + 1)!
j=0
j=1
(−1)k+1 k
k w
for log(1 + w), which is conj 2j
P
z
vergent for |w| < 1. This amounts to setting w = ∞j=1 (−1)
(2 j+1)! , which we note
satisfies |w| < 1 on a neighborhood of 0. Observe that since the lead z-term is z2 it
suffices to consider the 3rd -order polynomial in w. We have
sin z sin z
1 sin z
2 1 sin z
3
log
=
−1 −
−1 +
− 1 + [z8 ]
z
z
2 z
3 z
 2
2
 1
3
3
X (−1) j z2 j
1 X (−1) j z2 j 
1 X (−1) j z2 j 
 + 
 + [z8 ]
=
− 
(2 j + 1)! 2 
(2 j + 1)! 
3
(2 j + 1)! 
Next we may compose with the series
j=1
P∞
k=1
j=1
j=1
!3
z
z
z
1 z
z
1 z2
=− +
−
−
− +
+
−
+ [z8 ]
3! 5! 7! 2 3! 5!
3 3!
1
1
1 4 1
1
1 6
= − z2 +
−
z
+
−
+
−
z + [z8 ]
3!
5! 2(3!)2
7! (3!)(5!) 3(3!)3
z3 (3 − 5)z4 (−9 + 63 − 70)z6
+
+ [z8 ]
=− + 3 2
3!
23 5
24 34 5 · 7
z3
z4
z6
=−
− 2 2 − 4
+ [z8 ]
2·3 2 3 5 3 5·7
2
4
6
2
!
4 2