Download Some homework solutions

HOMEWORK #1, #2,# 3, # 4 SELECTED SOLUTIONS
I’ve written up the questions that I got asked about most during office hours.
HW #1 : Ahlfors p. 33 # 4 What is the general form of a rational function
which has absolute value 1 on the circle |z| = 1? In particular, how are the zeros
and poles related to each other?
Solution:
Let R be our rational function. Lets do an easy case, where R is in fact a
fractional linear function, i.e.
R(z) =
z − z0
.
cz + d
Now, |R(z)| = 1 on the unit circle, by assumption. And |R
1 = 1
|S(z)| := R̄ z̄1 1
z̄
| = 1. Thus
and in fact, S(z) = R(z). But, z0 is a root for R(z), and thus 1/z¯0 is a pole for
S(z). But, since R(z) = S(z), we then have that 1/z¯0 is a pole for R(z), i.e. the
denominator of R(z) is a constant times z̄0 z − 1.
The case where R(z) = P (z)/Q(z) proceeds in exactly the same way.
HW #2 Problem 5 Find all functions f analytic in C\0 such that |f | has constant
value on all circles x2 + y 2 − ax = 0.
Solution:
First, note that if |f | has constant value, then f has the form ceiθ . So we are
looking for all analytic functions that take each circle in this family of circles to a
circle centered at the origin.
We change the problem a little. Note that x2 + y 2 − ax = 0 is an “ear-ring” of
circles intersecting at 0, and extending into the left half of the plane. We want to
turn this into something more tractable.
Note that since 0 is missing from our domain, we can apply 1/z with impunity.
This transforms the cirlces into lines parallel to the y axis.
Thus, we are reduced to the problem of finding all functions, now analytic on
all of C, that take lines of the form x0 + iy into a circle centered at the origin.
Functions of the form recz where c, r are constants, are the only functions that do
this.
HW # 3 Ahlfors #1 p. 123 Compute
Z
z n (1 − z)n dz
|z|=2
Solution: I was asked about this problem a number of times. I think the
problem is that it is in the Cauchy integral formula chapter, and doesn’t have the
form of something you compute with the Cauchy integral formual. But note that
z n (1 − z)n is just a polynomial and sol it’s holomorphic! Thus, it’s integral around
any closed curve, and in particular |z| = 2 is 0.
1
2
HOMEWORK #1, #2,# 3, # 4 SELECTED SOLUTIONS
HW # 4 Problem 2 let f (z) be continuous in {|z| ≥ 1} and limz→∞ f (z) = 0.
Show that for any m > 0 and any a > 0,
Z
lim
eimz f (z)dz = 0
R→∞
ΓR
where R = {|z| = R, Im z ≥ a} is an arc of the circle radius R.
Solution: This problem was hard because (I believe) there is a missing hypothesis. One should want f (z) ∼ R12 , that is, at some point f (z) is shrinking as fast
as R12 is. Once this is in place, you use the usual estimates.
That is,
Z
Z
imz
≤
e
f
(z)dz
|eimz ||f (z)|dz
Note that |e
imz
|=e
ΓR
−mimz
ΓR
, which is bounded. So the above is
1
C|f (z)|dz ≤ CπR max |f (z)| ≤
≤
Γ
R
R
ΓR
Z
so as R → ∞ the integral goes to 0.
HW # 4 Problem 6 Let p(z) be a polynomial and γ = {|z − z0 | = R} be a circle.
Show, without using the Cauchy integral formula, that
Z
1
p(z)
= p(z0 )
2πi γ z − z0
Solution: I had a lot of questions on this problem, but it is arguably the most
elementary on the set (which is probably what tripped people up). The point is
that since p(z) is just a polynomial, we can divide, so that
p(z)
= q(z) + Remainder
z − z0
and the remainder is just p(z0 ), from high school calculus.