Complex Analysis Preliminary Exam Fall 2010
1. Prove that all the roots of z 7 − 5z 3 + 12 = 0 lie between the circles
|z| = 1 and |z| = 2.
First, let us start by seeing how many roots of the polynomial lie
in |z| < 2. Let f (z) = z 7 − 5z 3 + 12 and g(z) = z 7 . On the boundary
of the domain, we have |f (z) − g(z)| = | − 5z 3 + 12| ≤ 5(8) + 12 = 52 ≤
27 = |g(z)|. Since the domain is bounded with a smooth boundary and
f (z) and g(z) are both analytic on D ∪ ∂D, Rouche’s Theorem gives
us that f (z) has the same number of zeros as g(z) in in |z| < 2. Hence
f (z) must have 7 zeros in in |z| < 2 since g(z) has a zero of multiplicity
seven at z = 0.
Now, let’s consider how many roots of f (z) lie in |z| < 1. Let’s
define h(z) = 12. Then on the unit circle |z| = 1 we have |f (z)−h(z)| =
|z 7 − 5z 3 | = 4 < 12 = |h(z)|. Hence, by Rouche’s f (z) and h(z) must
have the same number of roots in in |z| < 1, namely zero. And so, all
the 7 roots of f (z) must lie in the annulus.
2. Evaluate the following integral:
Z
0
∞
cos(3x)
dx
(1 + x2 )2
Carefully justify all your steps.
R ∞ cos(3x)
We know that 0 (1+x
2 )2 dx since this integral is even, it is equal to
R
1 ∞
ei3x
the real part of 2 −∞ (1+x
2 )2 dx. Let D be the half circle in the upper
half-plane with the horizontal edge going from -R to R and labelling
the upper half circle as ΓD . Then, we know that
R
R R ei3x
R
ei3z
ei3z
dz
=
dx
+
dz
2
2
2
2
D (1+z )
−R (1+x )
ΓD (1+z 2 )2
Applying the ML-estimate to the second term on the right hand side
of the equality, we see that as R approaches ∞, this term goes to zero.
Meaning, that as R7→ ∞,
R
R ∞ ei3x
ei3z
dz = −∞ (1+x
2 )2 dx
D (1+z 2 )2
We can calculate the integral on the left-hand side of the equality using
residue calculus:
R
3iz
3iz
ei3z
d
dz = 2πiRes[ (z+i)e2 (z−i)2 , i] = limz→i dz
((z−i)2 (z+i)e2 (z−i)2 )
D (1+z 2 )2
1
=
−π
4
− 8e−3 =
2π
e3
3. Prove that there is no one-to-one conformal map of the punctured disc
{z C : 0 < |z| < 1} onto the annulus {z C : 1 < |z| < 2}
(From Berkeley problems)
Let A = {z C : 1 < |z| < 2} and B = {z C : 0 < |z| < 1}.
Assume that f is a one-to-one conformal map from B to A. Clearly, f
is bounded, so the singularity at the origin must be removable. Let
p = limz→0 f (z). Since f is continuous, p must be in the closure of A.
Suppose that p is on the boundary of A then f(B) = A ∪ {p}. This
gives us that f(B) is not an open set contradicting the Open Mapping
Theorem:
Open Mapping Theorem for Analytic Functions: If f(z) is analytic
on a domain D, and f(z) is not constant, the f(z) maps open sets,
that is, f(U) is open for each open subset U of D
Let p A and a B be such that f (a) = p. Take disjoint open neighborhoods U of 0 and V of a. By the Open Mapping Theorem, f (U ) and
f (V ) will be open sets containing p, which implies that f (U )∩f (V ) 6= ∅.
Let x f (U ) ∩ f (V ) 6= ∅ with x 6= p. Then f (z) = x for some z U
and f (w) = x for some w V. Then z and w are distinct elements of
G with f (z) = f (w), contradicting the injectivity of f.
4. Let f be a complex-valued function in the unit disc D(0, 1) such that
g = f 2 and h = f 3 are both analytic. Prove that f is analytic in D(0, 1).
First notice that f = hg and that h and g must have the same
zeros since they are both just powers of the function f. Let z0 be a
common zero of g and h, then there exists analytic function g1 and h1
which are nonzero at z0 so that h(z) = f (z)3 = (z − z0 )k h1 (z) and
g(z) = f (z)2 = (z − z0 )j g1 (z). Since (f 2 )3 = f 6 = (f 3 )2 we have
(z − z0 )2k h1 (z)2 = (z − z0 )3j g1 (z 3 ). We can rearrange this equality
2
to get hg11(z)
= (z − z0 )3j−2k . Since neither h1 nor g1 is zero at z0 by
(z 3
assumption, the left hand side of this equality must be analytic and
nonzero at z0 , implying that 3j − 2k = 0, and so k > j. This gives us
that z0 has a higher multiplicity as a zero of f 3 than a as a zero of f 2 .
3
This will hold for every zero of f, and so f = ff 2 can be extended to an
analytic function on D.
2
5. Let f be a holomorphic function in the unit disc D(0, 1) and |f (z)| ≤ M
for all z D(0, 1). Prove that |f 0 (z)| ≤ M (1 − |z|)−1 for all z D(0, 1).
(NOT COMPLETELY CORRECT) Cauchy’s Integral Formula gives
us that
R 2π f (eit ) it
R 2π |f (eit )|
1
1
ie
dt|
≤
|dt|
|f 0 (z)| = | 2πi
it
2
2π 0 |(eit −z)2 |
0 (e −z)
R
R
2π
2π
M
1
M
M
1
|dt| = 2π
|dt| = (1−|z|)
≤ 2π
2
0 (||eit |−|z||)2
0 (1−|z|)2
Notice that the exponent of the denominator does not match exactly
what we need to prove in this problem.
6. Suppose that g : C 7→ C is a holomorphic function, k and n are integers,
n
and (2 + |z|k )−1 ddznf is bounded on C.
(a) Prove that f is a polynomial.
(b) Estimate the degree of f in terms of the integers k and n.
(a) Let M be the bound of the expression, then we can rewrite
the inequality in the following way:
dn f
≤ M (1 + |z|k )
dz n
Now divide both sides by |z|k and take the limit as |z| 7→ ∞:
dn f 1
M (1 + |z|k )
≤
lim
=M
|z|→∞ dz n |z|k
|z|→∞
|z|k
lim
n
And so ddznf has a pole at infinity of degree at most k. And so
polynomial of degree at most k.
dn f
dz n
is a
n
(b) Since ddznf can have at most degree k, then f must have degree
at most n + k.
7. Let h : C 7→ R be a harmonic and non-constant function. Show that h
has at least one zero.
Unfinished
8. (a) Prove that the series
function on C.
Pn=∞
1
n=−∞ (z−n)2
3
converges to a meromorphic
(b) Prove that there is an entire function h(z) so that
h(z).
π2
sin2 (πz)
=
Pn=∞
1
n=−∞ (z−n)2 +
Unfinished
(Note: I do not claim to have developed all of the above solution.)
4