Math 185 Homework 4. Due Saturday 10/3 1. (a) Assume that f(z) is an everywhere holomorphic function which is periodic with period π, so that f(z + π) = f(z). Show that if f is bounded on the strip...

Math 185 Homework 4. Due Saturday 10/3
1. (a) Assume that f(z) is an everywhere holomorphic function which is periodic
with period π, so that f(z + π) = f(z). Show that if f is bounded on the strip
{a+bi | −π/2 ≤ a ≤ π/2} then f is bounded on all of C. By Liouville’s theorem
(which we will prove later in the week), this implies f is constant!
(b) Show that the sum f(z) :=
∑∞
n=−∞
1
(z+π·n)2 converges for all z othe
than integer multiples of π (for which one of the 1z+π·n will blow up), and
show that it is periodic with period π. Hint: it is enough to prove absolute
convergence, i.e. convergence of the series of real numbers 1|z+π·n| . These can
e summed in any order, as this is a sum of positive numbers. Reduce this
statement to the same statement for z in a strip −π ≤ Re(z) ≤ π, and use
dominated convergence on this real series with one term removed.
(c) You may assume f(z) defined above is holomorphic (on the domain
z 6= πn). Show that f is bounded for z satisfying |Im(z)| ≥ 1.
Remark. In a later set we will see by studying singularities of the function
1
sin(iz)2 and comparing it to the function f(z) above that f(z) =
1
sin(iz)2 . This
will let us find the value of the zet function ζ(2).
2. The solution to this problem is mostly pictures, and can be done
y hand. Please either include the picture in your LaTeX document
using the includegraphics command, or convert it to pdf and combine
it with your LaTeX document into a single file before submitting to
gradescope. Clearly label the problem number of any supplementary
pages attached this way.
Let f be a function which is defined and holomorphic on
Ω := C \ 2Z,
the complement to the set of even integers. Let γ(t) = 3 exp(it) be the circle
of radius 3 and let γ−1, γ0, γ1 be small circles of radius 1/2 centered around
−1, 0, 1, respectively: so γ−1(t) = 12 exp(it)− 1, etc.
(a) Draw a picture of the four loops γ, γ−1, γ0, γ1.
(b) Prove, using pictures, that∫
γ
f(z)dz −
∫
γ−1
f(z)dz −
∫
γ0
f(z)dz −
∫
γ1
f(z)dz = 0,
1
for f defined on Ω as above. Do this in analogy with 3.7 in the notes. Namely,
draw a simple closed curve depending on a parameter � whose integral in the
� → 0 limit approaches the difference of integrals above. Rigorous proof not
necessary but include enough arguments to show you know what you are doing,
including the relationship between the contour and the region Ω.
(c) Give an example of a function f whose
3. Do the following problem (a-g) from Gamelin.
4. (a) Show that
z3 + i = (z + i)(z +
√
3− i
2
)(z +
−
√
3− i
2
).
(b) Let f(z) = 1z3+i . Show that the real integral
∫∞
−∞ f(x)dx is equal to
−2π. Imitate Example 4.13 in the notes (integrating 1(x2+1)2)dx using a complex
semicircle contour), for example using the following steps.
1. Write a formula for g(z) = f(z)z−i . Show that it has no singularities in the
upper half-plane.
2. For R > 0 a real number relate the sum
∫ R
−R f(z)dz +
∫
CR
f(z)dz for R
the semicircle of radius R (as in Example 4.13) with g(i) using Cauchy’s
integral formula.
3. Use |
∫
γ
f(z)dz|
∫
|f(z)||dz| to show that
∫
CR
f(z)→ 0 as R→∞. Here∫
γ
|f(z)||dz| is defined as
∫ T
t=0
|f(γ(t))||dγdt |dt.
5. Extra credit This problem can be done instead of one of the above problems
for full credit. Indicate which one you are replacing if you do so. Alternatively,
this problem can be done as extra credit for 1/2 points.
Do 2.7.1 in Stein-Shakarchi, i.e. “Problem 1” (distinct from “Exercise
1”!) of Chapter 2 on page 67
2

Math 185 Problem Set 7. Due Sunday 10/25 at 3pm
1 Inverting power series
Recall that if f(z) =
∑
an(z − z0)n is a Taylor series expansion for a nonzero holomorphic function f , then the
degree of zero of f at z0 is the minimum n such that an 6= 0. For each of the following functions f(z) in parts
a-d below
• (Optional but helpful): write down the first two or three nonzero terms of the Taylor series expansion of f
at z0 = 0 (i.e., in terms of powers of z) .
• Find the degree n of f , and write f(z) = czn(1 − �(z)), where c ∈ C∗ is a nonzero constant and �(z) is a
holomorphic function with no constant term (i.e., �(0) = 0).
• Use the expansion 11−�(z) =
∑∞
k=0 �(z)
k to get a formula for 1f(z) in terms of z and a power series in �(z).
• Compute the first few terms of the Laurent series akz
k, namely find all ak with k ≤ 0.
• Find the residue of 1f(z) at z0 = 0 (this is defined to be the term a−1.)
Example: for f(z) = 2z + z2, the degree is 1, the factorization is f(z) = 2z(1− (−z/2)), the expansion is
1
f(z)
=
1
2z
∑(−1
2
)k
zk =
1
2
z−1 +
−1
4
z0 + . . . ,
the residue is the z−1 term, 12 .
(a) f(z) = ez − 1
(b) g(z) = sin(z)
(c) h(z) = z2 + z3.
(d) j(z) = sin(z)2. Hint: the answer you should get is 1j(z) = z
−2 + 13 + . . . .
2 Logarithmic derivatives
Define the logarithmic derivative of a holomorphic function f(z) to be the function
Lf :=
f ′(z)
f(z)
.
(It is called the logarithmic derivative because if the image of f is contained in a
anch of the logarithm,
Lf (z) = log(f(z))
′.) Note that it is only defined at points such that f(z) 6= 0.
(a) Compute Lf for f(z) = exp(z), g(z) = cos(z), and h(z) =
1
cos(z) .
(b) If f, g are holomorphic on a domain Ω, prove (using the chain rule, the inverse rule, etc.) that Lfg = Lf +Lg
and L1/f = −Lf .
(c) Recall (see also problem set 5) that any complex polynomial f(z) = anz
n + · · · + a0zn of degree d can be
(uniquely) decomposed as a product
f(z) = ad · (z − z1)n1 · (z − z2)n2 · · · · · (z − zk)nk ,
1
where zj are the roots of f, nk are their multiplicities.
Show that Lf =
∑k
j=1
nj
z−zj , a sum over the roots. (Note that in particular it does not depend on the leading
term ad.)
(d) Assume that γ is a simple closed curve that does not go through any root zj of f . Deduce from (c) that∮
γ Lf (z)dz is the sum of multiplicities of roots of f which are in the interior of γ, so∮
γ
Lfdz = 2πi
∑
j|zj∈Intγ
nk.
(e) Show that the functions h(z) = 12z and j(z) =
1
z2
cannot be the logarithmic derivative of a polynomial f(z).
(Hint: There are three ways of doing this, one using d involving loops integrals, and two ways using c with
keywords “Laurent series” or “partial fraction decomposition”. Use whichever of these methods you’re most
comfortable with.)
(f) Recall that another definition of degree of zero is this. A function f(z) has a zero of degree d at z0 if (at
least outside of z0,) we can write f(z) = g(z)(z− z0)d, for g(z) a function with g(z0) 6= 0 (i.e., “g(z) is invertible
at z0”). Recall that a function f has a pole of order d at z0 if, outside of z0, f(z) =
g(z)
(z−z0)d
, with g a function
which is holomorphic at z0 and g(z0) 6= 0. Let
D :=
{
d z0 is a zero
−d z0 is a pole.
(from the above, D is the unique number such that f(z) = g(z)(z − z0)D with g(z) holomorphic at z0 and
g(z0) 6= 0). Show that the logarithmic derivative Lg has a Laurent series expansion at z0
Lg(z) =
D
z − z0
+ l0 + l1(z − z0) + l2(z − z0)2 + . . . .
Hint: equivalently, this is asking to show that Lg(z) is given by
D
z−z0 plus a function which is holomorhpic at z0).
3 Sum of inverse squares and Laurent series
Here we will finish a computation involving periodic functions from Problem Set 4 (don’t wo
y, I will remind
you of everything you need from there). From 1 (d), we know that the function
f(z) :=
1
sin(z)2
has Laurent series expansion at 0 given by f(z) = 1
z2
+ 13z
XXXXXXXXXXLet
g(z) =
∑
n∈Z
1
(z − nπ)2
.
(a) Show that f(z) − g(z) is holomorphic at 0. (Hint: you may use that g(z) − 1
z2
=
∑
n∈Z,n 6=0
1
(z−nπ)2 is
holomorphic at z = 0: this follows from the fact that this sum is an absolutely convergent sum of holomorphic
functions in some neighborhood of 0).
(b) Show that the function h(z) = f(z)− g(z) is periodic, i.e., satisfies h(z + π) = h(z), and is holomorphic
at z if z is not a multiple of π. Since we have just shown that h(z) is holomorphic at 0, periodicity implies that
it is also holomorphic when z is a multiple of π, i.e., h(z) is holomorphic. From Problem Set 4 we can see h(z)
is bounded (no need to prove this), and then Liouville’s theorem implies that h(z) is a constant function.
(c) Prove that limy→∞ f(iy) = 0 and limy→∞ g(iy) = 0 (for the latter, you may use the following trick.
For N ≥ 1, let cN =
∑
|n|≥N
1
π2n2
= 2
π2
∑
n≥N
1
n2
. Then absolute convergence of
∑ 1
n2
implies that cN → 0 as
2
N → ∞. Now show by decomposing the sum in the definition of g(z) and using the triangle inequality that fo
|y| ≥ N, we have |g(iy)| ≤ 2 1N + cN , which approaches 0).
Deduce from (b) that f(z)− g(z) = 0, i.e.∑
n∈Z
1
(z − nπ)2
=
1
sin(z)2
,
when both sides are defined.
(d) By comparing the values at zero of f(z)− 1z =
1
XXXXXXXXXXand
g(z)− 1
z
=
∑
n∈Z,n 6=0
1