Math 535: Complex Analysis – Spring 2016 – David Dumas

Final Exam Solutions

Note: The solutions given here are in many places more detailed than the minimum requirements
for full credit.
Z ∞ 1
x5
(1) Calculate: dx
0 x2 + 1
1
1
Solution. Let f (z) = z2z+1
5
where z 5 is the holomorphic branch defined on C \ {z > 0}.
Let γε,R denote the closed “keyhole contour” which is the concatenation of the following
arcs:
• γ1 = the oriented line segment from R1 exp(iε) to R exp(iε),
• γ2 = the counterclockwise arc of |z| = R from R exp(iε) to R exp(i(2π − ε)),
• γ3 = the oriented line segment from R exp(i(2π − ε)) to R1 exp(i(2π − ε)), and
• γ4 = the clockwise arc of |z| = R1 from R1 exp(i(2π − ε)) to R1 exp(iε).
Define Z
I = lim lim f (z)dz.
R→∞ ε→0 γε, R
Z ∞
1
We claim that I = (1 − exp(2πi/5)) f (x)dx. This is because the branch of z 5 we
0
defined above extends continuously when approaching x ∈ R+ from above or from below,
1 1
with respective limits that are x 5 or exp(2πi/5)x 5 . Hence
Z Z R
lim f (z)dz = f (x)dx
ε→0 γ1 1/R
Z Z 1/R
lim f (z)dz = exp(2πi/5) f (x)dx
ε→0 γ3 R

Z and taking R → ∞ gives the claimed value for I, so it suffices to show that
Adding these
lim lim f (z)dz = 0 for i = 2, 4. We use the standard method to get a bound these
R→∞ ε→0 γi
integrals based on the length of the contour and the size of the integrand.
Considering γ2 first, we have
Z
f (z)dz 6 (sup | f (z)|)(length(γ2 )).
γ2 z∈γ2

If z ∈ γ2 then |z| = R and for large R this implies |z 2 + 1| > 1 2

2 RZ. We therefore find
1 4
− 95
| f (z)| 6 2R /R2 = 2R . We also have length(γ2 ) 6 2πR, giving
5 f (z)dz 6 4πR− 5
γi
which goes to zero as R → ∞.
Turning to γ4 , we have
Z
f (z)dz 6 (sup | f (z)|)(length(γ4 ))
γ4 z∈γ4
 1
If z ∈ γ4 then |z| = 1
R and forZlarge R this implies |z 2 + 1| > 1
2 and | f (z)| 6 2R− 5 . Since
6
length(γ4 ) 6 2πR−1 , we find f (z)dz 6 4πR− 5 which also goes to zero as R → ∞.
γi
Now we use the Residue Theorem to compute I. The simple poles of the integrand are
at ±i, points about which the contour has winding number one for all large R and small ε,
and we find
I = 2πi(Res z=i f (z) + Res z=−i f (z)).
We compute
1 1
z5 i5 exp(πi/10)
Res z=i f (z) = lim (z − i) 2 = =
z→i (z + 1)) 2i 2i
1
noting that i has argument π/2 for the purposes of computing our chosen branch of z 5 .
Similarly,
1 1
z5 (−i) 5 exp(3πi/10)
Res z=−i f (z) = lim (z + i) 2 =− =−
z→−i (z + 1)) 2i 2i
Using the above formula for I we have, finally
 exp(πi/10) exp(3πi/10) 
1
Z ∞
x5 2πi 2i − 2i
dx =
0 x +1
2 1 − exp(2πi/5)
Further simplification is not necessary on an exam, but with a bit of algebra this can be
reduced to the tidy expression
Z ∞ 1
x5 sin(π/10)
dx = π .
0 x2 + 1 sin(π/5)

(2) Does there exist an entire function f with the following properties?
• f (1) = 0
• f (2) = 0
• f (z) ∈ R if and only if z ∈ R
Either give an example of such a function, or prove that no such function exists.
Solution. No such function exists. In fact, we claim that any holomorphic function which
has f (1) = 0, f (2) = 0, and which is real for all z ∈ R is also real at some point in C \ R.
If f is identically zero this is immediate, so assume from now on that f is not identically
zero.
Since f is real-valued and differentiable on [1, 2], by Rolle’s theorem there exists c ∈ [1, 2]
such that f 0 (c) = 0. Let g(z) = f (c + z) − f (c), so that g(0) = g0 (0) = 0. Let k > 2 be
the order of this zero of the function g. Then the local standard form for the holomorphic
map g is
g(z) = h(z) k
for some holomorphic function h with h(0) = 0 and h0 (0) , 0. In particular h is conformal
on some small open disk about 0. The condition g(z) ∈ R is equivalent, for z near 0,
to arg h(z) ∈ πk Z. Since the local inverse of h is a conformal map, we find that near
z = 0, the set g −1 (R) consists 2k smooth arcs emanating from 0 whose tangent vectors have
 arguments differing by all integer multiples of π/k. At most two of these arcs are tangent
to R, and k > 2, so one of these arcs contains a non-real point z0 (which by construction
has g(z0 ) ∈ R). Since c and f (c) are both real, we conclude f (c + z0 ) = g(z0 ) + f (c) is
real while c + z0 is not.

Remark. There are lots of different solutions to this problem. Another one is to consider
the image by f of a large circle which encloses 1 and 2. By the argument principle, the
image has winding number 2 about the origin. However, the given conditions on f would
imply that the image crosses R at only two points (the images of the real points on the circle),
which can be used to show that the winding number is zero or one, giving a contradiction.
It is also possible to attack the problem by considering the imaginary part of f , which
is a harmonic function vanishing only on R, and using the Cauchy-Riemann equations to
infer from this that the real part of f cannot have multiple zeros on R.

(3) For n ∈ N let Λn ⊂ C denote the lattice generated by ω1 = 1 and ω2 = ni. Let ℘n denote the
Weierstrass function of Λn . Identify the limit of the meromorphic functions ℘n as n → ∞,
and the region on which the convergence is locally uniform.
Solution. We will show that the limit is π 2 csc2 (πz) − π3 , with locally uniform convergence
2

in C \ Z. First recall that

X 1
π 2 csc2 (πz) =
n∈Z
(z − n) 2
with locally uniform convergence on C \ Z. (It is equivalent to say: “With uniform
convergence on every closed disk, once we omit the terms with poles in that disk”.) Also
recall that
∞
X 1 π2
2
= .
n=1
n 6
Since (−n) 2 = n, we can rewrite this as half of the corresponding sum over Z \ {0}, using
(−n) 2 = n, obtaining
X 1 π2
2
= .
n∈(Z\{0})
n 3
Adding this to the series identity for the cosecant function and renaming n to ω, we find
π2
!
1 X 1 1
π csc (πz) −
2 2
= 2+ − 2 .
3 z ω∈(Z\{0})
(z − ω) 2 ω

Recalling the definition of the function ℘, we see that the sum above consists of the terms
from ℘n that lie on the real axis. To show that this is equal to the limit as n → ∞ of the full
sum, we need to show that the sequence of functions defined by the remaining terms, i.e.
!
X 1 1
f n (z) = − 2
ω∈(Λ \R)
(z − ω) 2 ω
n

converges locally uniformly to zero as n → ∞.

 To show this, recall that the sum defining ℘n itself converges in |z| 6 R because, after
omitting the finitely many terms with |ω| 6 2R, we have
1 1 C|z|
− 2 6
(z − ω) 2 ω |ω| 3
for a fixed constant C, hence the remaining terms of ℘n have as majorant in |z| 6 R the
absolutely convergent series
X 1
CR
ω∈(Λ \{0})
|ω| 3
n

Applying the same reasoning to f n , there is no need to omit any terms if n is large enough,
since ω ∈ (Λn \ R) implies |ω| > n, so we have
X 1
| f n (z)| 6 C R
ω∈(Λ \R)
|ω| 3
n

again on |z| 6 R. Thus our task is reduced to showing that

X 1
ω∈(Λn \R)
|ω| 3

is small when n is large. Define Ξn = {(` + mi) : max(|`|, |m|) > n}, so that Λn ⊂ Ξn and
X 1 X 1
6 .
ω∈(Λ \R)
|ω| 3 ω∈Ξ |ω| 3
n n

We will estimate the sum on the right, which is easily seen to converge, for example since
Ξn ⊂ Λ1 and the sum of |ω| −3 over any lattice converges.
Consider the terms in ω∈Ξn |ω|1 3 with ω = (` + mi) and max(|`|, |m|) = k. The set of
P
such terms is nonempty for each k > n in which case it has 8k elements, each contributing
at most k −3 to the sum. We find,
X 1 ∞ ∞
X 1 X 1
6 8k = 8
ω∈Ξ
|ω| 3 k=n
k3 k=n
k2
n

which is the tail of a convergent series with the first (n − 1) terms omitted. Thus as n → ∞
this remainder goes to zero, as required.

(4) Suppose f is a holomorphic function on |z| < 2 that is even (that is, f (−z) = f (z)). Show
that there exists a holomorphic function F on the annulus 1 < |z| < 2 such that
f (z)
F 0 (z) = .
z2 − 1

Solution. Let A = {z : 1 < |z| < 2} and let g(z) = zf2(z) −1

. Such F exists if and only
if the integral of g over every closed path in A is equal to zero. Any closed path in A is
homologous to an integer multiple of the circle C = {|z| = 32 }, so we need only show that
Z
g(z)dz = 0.
C
 Since g is meromorphic in |z| < 2, we can compute the integral above by residues:
Z
g(z)dz = 2πi(Res z=1 g(z) + Res z=−1 g(z)).
C

Since 1
z 2 −1
has simple poles at z = ±1 with residues ± 21 , the residues of g are
1 1
Res z=1 g(z) = f (1)Res z=1 = f (1)
z2 − 1 2
and
1 1
Res z=−1 g(z) = f (−1)Res z=−1 = − f (−1).
−1 2 z2
Since f is even, these residues sum to zero as required.

(5) Completely describe the convergence of the power series

∞
X z 2n
n=1
2n n3
for z ∈ C. That is, determine the set of all z for which the series converges, and separately,
identify the largest open set in which the convergence is locally uniform.
Solution. This radius of convergence R of a power series satisfies
R−1 = lim sup |a k | 1/k
n→∞

where a k is the coefficient of z k . In this case a k = 0 for odd k and a2n = 2−n n− 3. Thus
1
R−1 = lim sup (2−n n−3 ) 2n .
n→∞
1 1 √
Note that (2−n ) 2n = 2− 2 , so it will follow that R = 2 if we show
3
lim n 2n = 1.
n→∞
Taking the logarithm gives
3 3 log n
log lim n 2n = lim =0
n→∞ 2 n→∞ n
as required.
We therefore
√ conclude that the series in√question converges locally uniformly in the open
disk |z| < 2 and that it diverges if |z| > 2. √
It remains only to consider what happens for |z| = 2. For such z the series becomes
P wn
n 3 where w = 2 z has |w| = 1. Thus at such points the series has the convergent series
1 2
P n1
n n3 as a majorant, and in particular converges.
√
To summarize, the
√ series converges if and only if |z| 6 2, and it converges locally
uniformly in |z| < 2 (and not in any larger open set).

(6) Find all linear fractional transformations T such that T (1) = 1, T (3) = 3, and T (T (z)) = z
for all z.
 Solution. In this solution we use multiplicative notation for composition of linear
fractional transformations, so e.g. ST refers to the composition S ◦ T if S(z) and T (z) are
linear fractional. We also use I to denote the identity map, I (z) = z.
A linear fractional transformation which fixes 0 and ∞ in Ĉ has the form F (z) = λz
for some λ ∈ C∗ . If such a transformation has FF = z then λ 2 = 1 and there are two
possibilities: F = I or F (z) = −z.
Let S(z) = z−3
z−1
. This linear fractional transformation satisfies S(1) = 0 and S(3) = ∞,
so if T is as described in the problem, then ST S −1 is linear fractional, fixes 0 and ∞, and
has ST S −1 ST S −1 = STT S −1 = SI S −1 = I. Thus the possible transformations T are S −1 FS
where F = I or F (z) = −z. The first is simply I, the latter is easily computed to be
2z − 3
T (z) = .
z−2

(7) Find all holomorphic functions on C∗ that satisfy:

| f (z)| < |z| + log |z|

Solution. Let us call this inequality (*).

We will show that the functions satisfying (*) are exactly the linear functions f (z) = az+b
where |a| + |b| < 1.
First, we show that a function satisfying (*) has a removable singularity at the origin,
and hence defines an entire function. Consider the function z f (z). Then for |z| = r < 1
we have |z f (z)| 6 r 2 + r log 1r which goes to zero as r → 0. Thus z f (z) is bounded near
zero, the singularity of z f (z) is removable, and the extended function g vanishes at z = 0.
But then f (z) = g(z)/z has a removable singularity at 0, giving the desired extension of f .
Next, we show f is linear. An entire function with a pole of order k at infinity is a
polynomial of degree k, so it suffices to show that f has at most a simple pole at infinity, or
equivalently that f (1/z) has a simple pole at z = 0. By (*) we have
1 1 1
f (1/z) < + log = + log |z|
|z| |z| |z|
Arguing as above we find |z f (1/z)| is bounded near z = 0, hence extends holomorphically,
and f (1/z) is expressible as 1z g(z) for g holomorphic near 0. That is, f (1/z) has at most a
simple pole at z = 0.
Now we must determine which linear functions az + b satisfy (*). For |z| = 1, inequality
(*) becomes | f (z)| < 1. If either of a or b is zero, this shows the other has absolute value
b|a|
less than one. Otherwise, taking z = a|b| ∈ S 1 we find | f (z)| = |a| + |b| and so again (*)
gives |a| + |b| < 1, and we conclude this condition is necessary.
Finally, we show |a| + |b| < 1 is sufficient for f (z) = az + b to satisfy (*). Note that 1 is
the absolute minimum value of r + | log r | on (0, ∞). If |z| = r 6 1 then
|az + b| 6 |a|r + |b| 6 |a| + |b| < 1 6 r + | log r |
and so (*) is satisfied for such z. On the other hand, if r > 1 then
|az + b| 6 |a|r + |b| 6 (|a| + |b|)r < r < r + | log r |
and (*) is satisfied for these z as well.
(8) Determine whether or not each family of holomorphic functions on the unit disk is normal:
(a) F1 = { f : ∆ → C : f (z) , 0 for all z ∈ ∆}
(b) F2 = { f : ∆ → C : f (z) < [0, 1] for all z ∈ ∆}
(c) F3 = { f : ∆ → C : | f (z)| > 1 for all z ∈ ∆}
Solution.
(a) F1 is not normal.
Consider the sequence of functions f n = (z + 1) n ∈ F1 . Then f n (0) = 1 but for any
x ∈ (0, 1) we have f n (x) → ∞ as n → ∞, so no subsequence of f n can converge to a
function continuous at zero, nor does any subsequence tend to infinity locally uniformly.
(b) F2 is normal, and
(c) F3 is normal.
In fact, since F3 ⊂ F2 , and a subfamily of a normal family is normal, it suffices to show
that F2 is normal.
Recall that h(z) = z + 1z is a conformal map from the complement of the unit disk to the
complement of [−2, 2]. Thus H (z) = 14 h(z) + 1 is a conformal map from the complement
of the unit disk to the complement of [0, 1]. Note that both H and its inverse have a simple
pole at infinity.
If f n is a sequence in F2 , then gn (z) = H −1 ( 1f (z)) is a sequence of holomorphic functions to
n
∆∗ . As these are uniformly bounded, there exists a locally uniformly convergent subsequence
gnk . By Hurwitz’s theorem, the limit function g∞ is either nowhere zero or identically zero.
In the former case we find that
1
f nk (z) = H ( )
gnk (z)
converges locally uniformly to H ( g1∞ ). In the latter case, f nk converges locally uniformly
to infinity, since gn1 → ∞ and H has a pole at infinity. Thus F2 is normal.
k