Math 814 Final Exam Solutions
1. Let f be a nonconstant entire function. Show that f (C) is dense in C.
First suppose f is a polynomial. If a ∈ C, the nonconstant polynomial f (z) − a
has a root, by the Fundamental Theorem of Algebra. Hence if f is a polynomial, we have the stronger result that f (C) = C. Assume from now on that f is
entire, but not a polynomial. Then from class notes, we know that the function
g(z) = f (1/z) has an essential singularity at z = 0. By the Casorati-Weierstrass
Theorem, for any δ > 0 the image of {z : 0 < |z| < δ} under g(z) is dense in
C. So we have the stronger result that for any R > 0, the image under f (z) of
{z : |z| > R} is dense in C.
Here is another proof, inspired by one of yours. If f (C) is not dense in C,
then there is w ∈ C and > 0 such that |f (z) − w| > for all z ∈ C. Then
(f (z) − w)−1 is bounded entire, hence constant, which implies that f is constant.
2. Suppose f is analytic and one-to-one on a region U . Show that f 0 (z) 6= 0 for
all z ∈ U . Discuss the converse.
Let z0 ∈ U and suppose f 0 (z0 ) = 0. Then f vanishes to some order m ≥ 2 at z0 .
In class, we have seen that this means f is m-to-1 near z0 . This contradicts the
assumption that f is 1 − 1 on U .
The converse is false: The function f (z) = ez has f 0 (z) never zero on U = C,
but f (0) = f (2πi), so f is not 1 − 1. However, the converse is locally true, in the
sense that if f 0 (z) is never zero on U , then around each z0 ∈ U there is an open
disk B(z0 , δ) on which f is 1 − 1. This follows from the same result used in the
previous paragraph. For example, with f (z) = ez , one can take δ = 2π.
3. Suppose f and g are analytic on a region U , with the same zeros and multiplicities of zeros.
a) Show that f /g is analytic and nowhere zero on U .
The only possible points of nonanalyticity or vanishing are at one of the common
zeros of f and g. Suppose f and g vanish to order m > 0 at some point a ∈ U .
Then f (z) = (z−a)m f1 (z) and g(z) = (z−a)m g1 (z), where f1 and g1 are analytic
on U and f1 (a) 6= 0 6= g1 (a). Hence f /g = f1 /g1 is analytic nonvanishing at a.
b) Suppose a ∈ U , is a zero of f and g of multiplicity m. Show that
f (m) (a)
f (z)
= (m) .
z→a g(z)
g (a)
lim
1
Use only facts which we have proved in class. L’Hôpital’s rule is not one of them.
The power series expansion of f (z) is
f (z) =
f (m) (a)
(z − a)m + [ higher powers of (z − a) ],
m!
so f1 (a) = f (m) (a)/m!. Likewise, g1 (a) = g (m) (a)/m!. Hence
f (z)
f1 (a)
f (m) (a)
=
= (m) .
z→a g(z)
g1 (a)
g (a)
lim
4. Let U = {z : 0 < |z| < 1}, and let γ be a closed path in U .
a) Let f (z) be analytic
R on U and assume limz→0 zf (z) = 1. Show, by calculating it, that the integral γ f depends only on γ, not on f .
Let
f (z) =
∞
X
an z n
n=−∞
be the Laurent expansion of f on U . Then zf (z) has the Laurent expansion
zf (z) =
∞
X
an−1 z n .
n=−∞
Since zf (z) → 1, there are no negative powers of z in the series for zf (z), and its
constant term is a−1 = 1. Hence Res[f (z), 0] = a−1 = 1, so
Z
f = 2πi · n(γ, 0) · 1
γ
depends only on γ.
b) Find two functions g and h, both analytic on U , for which
lim z 2 g(z) = 1 = lim z 2 h(z),
z→0
yet
R
γ
g 6=
R
γ
z→0
h.
P
The functions
and h(z) must have Laurent expansions g(z) =
bn z n ,
P g(z)
n
h(z) =
cn z such that b−2 = c−2 = 1, and no lower powers of z appear.
We have to find g, h such that b−1 6= c−1 . For example, if we take
g(z) =
ez
,
z2
h(z) =
2
sin z
,
z3
then b−2 = 1 = c−2 , yet b−1 = 0 and c−1 = 1.
5. In class we proved for λ ∈ C − Z, that
∞
X
π2
1
=
.
2
sin πλ k=−∞ (λ + k)2
R
To do this, we integrated γn π cot πz/(z + λ)2 over certain paths γn . Apply the
R
same method and paths to γn π cot πz/(z 2 − λ2 ) to show that
∞
π cot πλ =
1 X 2λ
+
.
λ k=1 λ2 − k 2
Choose n large enough so that ±λ are both inside γn . Since z 2 − λ2 is again
quadratic in z, the same estimates that we used for the first integral show that
Z
π cot πz
dz = 0.
lim
n→∞ γ (z 2 − λ2 )
n
The poles λ, −λ, 1/k of the integrand inside γn are all simple, with residues
π cot πz
π cot πz
π cot πλ
Res
,
λ
=
Res
,
−λ
=
,
(z 2 − λ2 )
(z 2 − λ2 )
2λ
1
π cot πz
,k = 2
.
Res
2
2
(z − λ )
k − λ2
Hence
n
X
k=−n
or
k2
1
π cot πλ
+2·
→ 0,
2
−λ
2λ
n
X
1
2
π cot πλ
− 2+
+
→ 0,
2
2
λ
k
−
λ
λ
k=1
as we wished to show. We could also write this as
π cot(πλ) =
∞
X
n=−∞
3
λ2
λ
.
− k2
(1)
Remarks: Replacing λ by z, we have now seen three similar-looking expansions
of analytic functions on C − Z, with pole set Z.
∞
X
(−1)k
,
π csc(πz) =
z
+
k
k=−∞
2
∞
X
1
(z + k)2
k=−∞
2
π csc (πz) =
∞
X
π cot(πz) =
n=−∞
z2
z
.
− k2
In contrast to the power series expansions, which involved Bernoulli numbers,
these expansions are simple, make the periodicity of csc(πz) and cot(πz) evident.
This expansion of cot πz is particularly useful. We’ll use it next semester to
get the product formula for
P sin πz and thereby vindicate Euler’s first, wild and
crazy, way of computing 1/n2 .
Instead of summing over Z, one can sum over two-dimensional lattices in C,
leading to elliptic functions, which are doubly periodic, and are connected to many
areas of mathematics, such as elliptic curves.
6. Let α be a fixed complex number. Define
(1 + z)α = exp(αLog(1 + z)),
where Log is the principal branch of the logarithm. Find the power series expansion of (1 + z)α about 0, and determine the radius of convergence.
Applying the chain rule to the right side, one proves the familar formula
d
(1 + z)α = α(1 + z)α−1 .
dz
Then, for f (z) = (1 + z)α , one proves by induction that
f (n) (z) = α(α − 1) · · · (α − n + 1)(1 + z)α ,
which gives the famous Binomial Series
(1 + z)α =
∞
X
α(α − 1) · · · (α − n + 1)
n!
n=0
4
zn.
If α is an integer ≥ 0 then the coefficients are zero for n > α, so (1 + z)α is
a polynomial of degree n (as we knew!) with infinite radius of convergence. If
α is not an integer ≥ 0 then the series converges up to the nearest point of nonanalyticity of Log(1 + z), which is z = −1. Hence the radius of convergence is
≥ 1. In fact the ratio test shows the radius of convergence is exactly one.
7. LetRγ(t) = eit , 0 ≤ t ≤ 2π. Calculate the following integrals.
a) γ sin( z1 ) dz
Since
1
1
1
1
sin
= −
+
− ··· ,
3
z
z 3!z
5!z 5
we have Res[sin(z −1 ), 0] = 1, so
Z
sin(z −1 ) dz = 2πi.
γ
b)
R
γ
sin2 (z −1 ) dz.
The quick answer is that the integral is zero, since the integrand is even, hence
has no odd powers in its Laurent expansion, so its residue at 0 is zero. To see this
explicitly, use the identity
sin2 z =
1 − cos 2z
,
2
which we know holds for real z, hence for all z ∈ C since both sides are entire
functions. This gives the power series
1
(2z)2 (2z)4 (2z)6
2
sin z =
−
+
− +··· ,
1−1+
2
2!
4!
6!
which gives the Laurent expansion
1
2
23
25
2
sin
−
+
− +··· .
=
z
2!z 2 4!z 4 6!z 6
in which the coefficient of 1/z is zero.
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8. Compute
Z
∞
−∞
dz
.
1 + z + z2 + z3 + z4
In class, we derived the formula
Z ∞
m
X
P (ai )
P (z)
dz = 2πi
,
Q0 (ai )
−∞ Q(z)
i=1
(2)
where P and Q are polynomials with deg P ≤ deg Q − 2, all poles of P/Q are
simple and nonreal, and a1 , . . . , am are the poles of P/Q in the upper half-plane.
In the present case there are two such poles:
a1 = α = e2πi/5 ,
a2 = α2 = e4πi/5 .
Formula (2), with P = 1 and Q(z) = 1 + z + z 2 + z 3 + z 4 , gives
Z ∞
dz
1
1
= 2πi
+
.
2
3
4
1 + 2α + 3α2 + 4α3 1 + 2α2 + 3α4 + 4α6
−∞ 1 + z + z + z + z
Let’s call this number I, for “integral”. There are various ways to simplify the
above expression for I. Mathematica can be coaxed to write it as
q
q
√
√
√
√
π
I = √ (3 + 5) 5 − 5 + (3 − 5) 5 + 5 .
10 2
Using the values
√
5+ 5
,
sin
=
8
√
3
−
5
2π
,
cos2
=
5
8
2
2π
5
√
4π
5− 5
sin
,
=
5
8
√
4π
3
+
5
cos2
,
=
5
8
2
we can also write it as
4π
I=
sin
5
4π
5
4π
2π
2
2 cos
+ cos
.
5
5
This simplifies even more, because 2 cos(2π/5) satisfies the equation x2 = 1 − x,
and this implies that
2π
4π
2π
2
2 cos
2 cos
+ cos
= 1.
5
5
5
6
So in fact,
p
√
5+ 5
√
= 2π ·
.
5 2
In hindsight, we could have arrived at this more easily if we had written the integrand as
z−1
.
z5 − 1
4π
sin
I=
5
2π
5
Then, with P = z − 1 and Q = z 5 − 1 and α = e2πi/5 , formula (2) gives
2πi 2
4π
2π
α − 1 α2 − 1
−1
2
+
=
(α − α + α − α ) =
sin
,
I = 2πi ·
4
8
5α
5α
5
5
5
since α5 = 1. The same method shows that
Z ∞
dx
4π X
2kπ
=
sin
,
2m
2m + 1 k
2m + 1
−∞ 1 + x + · · · + x
where the sum runs through all odd integers k such that 1 ≤ k ≤ m.
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