Math 520a - Homework 3

1. For each of the following four functions find all the singularities and for
each singularity identify its nature (removable, pole, essential). For poles
find the order and principal part.

Solution: z cos(z −1 ) : The only singularity is at 0. Using the power series

expansion of cos(z), you get the Laurent series of cos(z −1 ) about 0. It is an
essential singularty. So z cos(z −1 ) has an essential singularity at 0.
z −2 log(z + 1) : The only singularity in the plane with (−∞, −1] removed
is at 0. We have
z2 z3
log(z + 1) = z − + ···
2 3
So
1 z
z −2 log(z + 1) = z −1 − + ···
2 3
So at 0 there is a simple pole with principal part 1/z.
z −1 (cos(z) − 1) The only singularity is at 0. The power series expansion
of cos(z) − 1 about 0 is z 2 /2 − z 4 /4! · · ·, and so the singularity is removable.
cos(z)
The singularities are at the zeroes of sin(z) and of ez − 1, i.e.,
sin(z)(ez −1)
at πn and i2πn for integral n. These zeroes are all simple, so for n 6= 0 we
get simple poles and at z = 0 we get a pole of order 2. For n 6= 0, the residue
of the simple pole at πn is

cos(z) cos(πn) 1
lim (z − πn) z
= nπ
= nπ
z→πn sin(z)(e − 1) cos(πn)(e − 1) e −1

For n 6= 0, the residue of the simple pole at 2πni is

cos(z) cos(2πni)
lim (z − 2πni) z
= = −i coth(2πn)
z→2πni sin(z)(e − 1) sin(2πni)

For the pole of order 2 at z = 0 you can get the principal part by plugging
in power series for the various functions and doing enough of the division to
get the z −2 and z −1 terms. The principal part is z −2 − 21 z −1 .

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2. In class we showed that the gamma function Γ(z) can be analytically
continued to the complex plane minus the points 0, −1, −2, −3, · · ·. Show
that this function has simple poles at 0, −1, −2, −3, · · · and the residue of
the pole at −n is (−1)n /n!.

Solution: What we did in class showed that Γ(z) can be analytically contin-
ued to the complex plane minus the points 0, −1, −2, −3, · · ·, and it satisfies
Γ(z) = Γ(z + 1)/z on this domain. By a straightforward induction agrument
it satisfies
Γ(z + n + 1)
Γ(z) =
z(z + 1)(z + 2) · · · (z + n)

on this domain. Now consider this equation for z in U = {z : 0 < |z+n| < 1},
a deleted neighborhood of −n. On U, Re(z + n + 1) > 0, so Γ(z + n + 1)
is analytic on U. Clearly 1/(z + k) is analytic on U. So the above equation
says Γ(z) = g(z)/(z + n) where g(z) is analytic on U. Thus the pole at −n
is simple and the residue is g(−n). Since Γ(1) = 1, the residue is

1 (−1)n
g(−n) = =
(−n)(−n + 1)(−n + 2) · · · (−n + n − 1) n!

3. Problem 7 on p. 104 of the book.

Solution: I just give the highlights of the computation. Let z = eiθ . So

dz = ieiθ dθ, i.e., dθ = −idz/z. Then the given integral becomes

zdz
Z
−4i 2 2
γ (z + 2az + 1)
√
where γ is the unit circle.
√ There are poles at −a ± a2 − 1. Since a > 1,
2
only the pole at −a + a − 1 is inside circle. The zero in the denominator
is of order 2, so the pole is second order. So its residue is given by
√
√ 2 − 1)2 z
 
d (z + a − a
Res(−a + a2 − 1) =
dz (z 2 + 2az + 1)2 √
2
  z=a− a −1
d z a
= √ =
dz (z + a + a − 1) z=a− a2 −1 4(a − 1)3/2
2 2 √ 2

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4. (a) If f (z) has an isolated singularity at z0 , prove that exp(f (z)) cannot
have a pole there.

Solution: If f (z) has an isolated singularity at z0 , then clearly exp(f (z))

does too.
If f (z) has an essential singularity, consider any non-zero c ∈ C. By
the Casorati-Weierstrass theorem, there is a sequence zn → z0 such that
f (zn ) → log(c). So exp(f (zn )) → c. Since this is true for all non-zero c,
exp(f (z)) must have an essential singularity at z0 .
Finally we will show that if f has a pole at z0 , then exp(f (z)) has an
essential singularity there. Let n be the order of the pole of f . Then
g(z)
f (z) =
(z − z0 )n
where g(z) is analytic and non-zero on a deleted neighborhood of z0 . Let
g(z0 ) = reiθ . Consider the sequence
exp(iθ/n)
zk = z0 +
n
Then f (zk ) = g(zk ) exp(−iθ)nn . Since g(zk ) → reiθ , exp(f (zk )) converges to
∞. So the singularity for exp(f ) is not removable. If we consider
exp(i(π + θ)/n)
wk = z0 +
n
then we see that exp(f (zk )) converges to zero. Thus the singularity for exp(f )
is essential.
(b) Use (a) to show that if f has an isolated singularity at z0 and for some
positive constant c,
Ref (z) ≤ −c log(|z − z0 |)
in a deleted neighborhood of z0 then the singularity in f is removable.

Solution: The bound on f implies

| exp(f (z))| = exp(Re(f (z)) ≤ exp(−c log(|z − z0 |)) = |z − z0 |−c
So for large enough integers n,
lim (z − z0 )n exp(f (z)) = 0
z→z0

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This implies that the singularity of exp(f (z)) is either a pole or removable.
By (a) it cannot be a pole, so it is removable. From part (a) we know that
exp(f (z)) can have a removable singularity only if f (z) has a removable
singularity.

5. This is problem 14 on p. 105 in the book and you can find a hint there.
Prove that if f (z) is entire and injective (one to one), then there are complex
constants a 6= 0, b such that f (z) = az + b.

Solution: Since f is entire it has a power series about the origin that con-
verges on the entire complex plane. First suppose this power series is a
polynomial P (z). If P (z) has two or more distince roots, then it is not in-
jective since is sends two different numbers to 0. So it must have a single
root of order n where n is the degree of P . So P (z) = c(z − z0 )n . But then
z0 + exp(ik2π/n) for k = 0, 1, · · · , n − 1 all get mapped to the same image.
So n can only be 1. So P (z) is linear az + b and clearly a cannot be zero.
Now suppose f is not a polynomial. Then g(z) = f (1/z) has an essential
singularity at 0. By Casorati-Weierstrass theorem g maps {z : 0 < |z| < r}
to a dense subset of C for all r > 0. Now pick a point w0 such that that
f (w0 ) 6= 0. Then take r = |f (w0 )|/2. By the density, we can find a sequence
zn with 0 < |zn | < 1 such that g(zn ) converges to f (w0 ). Since the sequence
zn is bounded, it has a convergent subsequence znk converges to some w.
Note that |w| ≤ r/2. By continuity g(w) = f (w0 ). So f (1/w) = f (w0 ).
Since f is injective, 1/w = w0 . But |w0 | = r and |w| ≤ r/2, a contradiction.

6. In our proof of Runge’s theorem we used the following proposition: Fix

a compact subset of the complex plane. Let A be a collection of continuous
functions on K such that if f, g ∈ A and c ∈ C, then cf, f g, f + g ∈ A.
Suppose that a continuous function f can be uniformly approximated on K
by functions in A. Then any polynomial in f can be uniformly approximated
on K by functions in A.

Solution: Let P (z) be a polynomial. We want to show that P (f (z)) can be

uniformly approximated on K by functions in A. Let ǫ > 0.
Since f is continuous and K is compact, f (K) is compact. So P (z) is
uniformly continuous on f (K). So there is a δ > 0 so that z, w ∈ f (K)
and |z − w| < δ implies |P (z) − P (w)| < ǫ. Let g ∈ A be such that
|f (z) − g(z)| < δ for z ∈ K. Then since f (z), g(z) ∈ f (K), this implies

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|P (f (z)) − P (g(z))| < ǫ. The properties on A imply that P (g(z)) ∈ A. So
this completes the proof.

7. Fix w = reiθ with w 6= 0. Let γ be a curve in C \ {0} from 1 to w. Show

that there is an integer k such that
dz
Z
= log(r) + iθ + 2πik
γ z

Solution: I assume r ≥ 1. The changes for the other case of r < 1 are minor.
Let γ1 Rbe the contour that is the line segment from 1 to r: γ1 (t) = t, 1 ≤ t ≤ r.
Then γ1 dz/z is log(r).
Let γ2 be the contour that is the subarc of the circle of radius r from r
to reiθ . So γ2 (t) = reit , 0 ≤ t ≤ θ. Then
θ θ
dz γ2′ (t)
Z Z Z
= dt = idt = iθ
γ2 z 0 γ2 (t) 0

Let γ − γ2 − γ1 be the contour that follows γ from 1 to w, then follows

γ2 backwards from w to r and then follows γ1 backwards from r to 1. This
is a closed contour, so the integral of 1/z around this contour is 2πik, where
k is the winding number of this contour. So
dz dz dz
Z Z Z
= 2πik + + = 2πik + log(r) + iθ
γ z γ1 z γ2 z