Math 220A HW 6 Solutions

Section 4.1
eiz
Z
12. Let I(r) = dz where γ : [0, π] → C is defined by γ(t) = reit . Show that
γ z
lim I(r) = 0.
r→∞

eiz
R R
Solution: We’ll use the bound | γ
f| ≤ γ
|f |d|z|, for f (z) = z
. Since |γ|(s) =
Rs
V (γ; [0, s]) = |reit |dt = rs,
0
Z
|I(r)| ≤ |f |d|z|
γ
Zπ it
erie
= | it |rdt
re
0
Zπ
= |er(i cos(t)−sin(t) |dt
0
Zπ
= e−r sin t dt,
0

and we are looking for a bound for this latter integral. If we could replace sin t with some
multiple of t, the integral would be easy to evaluate, so we would like e−r sin t ≤ e−rαt ; ie,
αt ≤ sin t on [0, π] , α ∈ R>0 . If we let g(t) := sin t − αt, we are looking for g(t) ≥ 0.

Observe first that g(0) = 0 for every positive α, and the derivative g 0 (t) = cos t − α
will at first be positive if α < 1. Since g(π) < 0, by continuity g will attain another
zero—cα , say—on [0, π]. This is not ideal, since the bound we want will not hold on
(cα , π]. Luckily, sin t is symmetric about π2 in this interval, so
π
Zπ Z2
e−r sin t dt = 2 e−r sin t dt
0 0

meaning we only need the bound to hold on [0, π2 ]. In other words, we would like
g( π2 ) = 1 − π2 α ≥ 0, and for there to be no other zeros of g on [0, π2 ]. In fact, the first

1
 condition implies the second. To see this, note that g 0 (t) ≥ 0 on [0, arccos α] and is
negative on (arccos α, π2 ) 1 . If there was another zero t0 inside the interval, g would have
to be increasing somewhere on (t0 , π2 ) (otherwise g(t0 ) would be > g( π2 )). But then by
our discussion of the sign of g 0 , we would have g increasing on all of (0, t0 ), which would
force g(t0 ) > g(0). Thus, we can take any α with 0 < α ≤ π2 ; we may as well take α = π2 .
Finally we have that
π
Z2
2
I(r) ≤ 2 e−r π t dt
0
2π −r
=− (e − 1).
2r
This certainly goes to 0 as r approaches ∞, so

lim I(r) = 0.
r→∞

1
z − 2 dz where:
R
13. Find γ

(a) γ is the upper half of the unit circle from +1 to −1:

Solution: Since we’re implicitly using the principal branch of the log, we should
parametrize γ as γ(t) = eit , t ∈ [0, π]. This allows us to calculate
Z Zπ
− 12 t
z dz = e−i 2 ieit dt
γ
0
Zπ
t
= iei 2 dt
0
π
= 2(ei 2 − e0 )
= 2(i − 1).
1
Since z − 2 is not actually defined at −1, we probably should be taking the limit of
the integral from 0 to s as s approaches π, but we will get the same result.
(b) γ is the lower half of the unit circle from +1 to −1.
Solution: Similarly, we find that
Z Z−π
− 12 t
z dz = e−i 2 ieit dt
γ
0
π
= 2(e−i 2 − e0 )
1
we know arccos α will be in this interval because, for example, g 0 ( π2 ) = −α < 0, so cos t = α has a
solution there.

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 = 2(−i − 1)

(z 2 − 1)−1 dz.
R
19. Let γ(t) = 1 + eit for 0 ≤ t ≤ 2π and find γ

1
Solution: We rewrite z 2 −1
as

1 1 z + 1 − (z − 1)
=
z2 −1 2 z2 − 1
1 1 1
= ( − )
2 z−1 z+1
in order to calculate (noting that log(2 + eit ) is defined on an open set containing [0, 2π])

Z2π
eit
Z
1
2
dz = i dt
γ z −1 (2 + eit )eit
0
Z2π
1 1 1
= ( it − )ieit dt
2 e 2 + eit
0
Z2π
1 ieit
= (i − )dt
2 2 + eit
0
1 2π
= (it − log(2 + eit )) 0
2
1
= (2πi − log(3) + log(3))
2
= πi.

(z 2 − 1)−1 dz.
R
20. Let γ(t) = 2eit for −π ≤ t ≤ π and find γ

Solution: Notice that T z = z−1 z+1

preserves R∞ , and therefore T z is in R≤0 ∪ {∞} if
and only if z is real and −1 ≤ z ≤ 1. This means that away from that interval—and in
z−1
particular, on an open set containing γ—log z+1 is defined. However, we can deduce from
the calculations in the previous problem that 2 log z−1
1
z+1
is a primitive of f (z) := z21−1 , so
the integral of f around the closed curve γ will be 0.

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 Section 4.2
7. Use the results of this section to evaluate the following integrals:
(d) Z
log z 1
n
dz, γ(t) = 1 + eit , 0 ≤ t ≤ 2π and n ≥ 0.
z 2
γ

Solution: Both log z and z n are analytic, and z n has no zeroes on B(1; 12 + ) for
small , so log
zn
z
is analytic there. Thus, its integral around a closed curve is 0.

9. Evaluate the following integrals:

(a)
ez − e−z
Z
dz, where n is any positive integer and γ(t) = eit , 0 ≤ t ≤ 2π.
zn
γ

Solution: If f (z) = ez − e−z , then we know that

Z
f (z) 2πi
n
dz = f (n−1) (0).
z (n − 1)!
γ

Of course, f ( n − 1)(z) = ez + (−1)n−1 e−z , so we can write this as

(
4πi
e − e−z
Z z
(n−1)!
, n even
dz =
zn 0, n odd
γ

(e)
Z 1
zm 1
m
dz, where γ(t) = 1 + eit , 0 ≤ t ≤ 2π
(z − 1) 2
γ

1
zm
Solution: If m ≤ 0, then (z−1)m
is analytic on B(1; 21 +), so its integral is 0 around
1
γ. For m positive, f (z) := z m at least is analytic there, and we can again use that
Z
f (z) 2πi
m
dz = f m−1 (1).
(z − 1) (m − 1)!
γ

We can also check that

1 1 1 1
f (m−1) (z) = ( − 1) · · · ( − (m − 2))z m −(m−1)
m m m
1 1−m 1 − (m − 2)m 1 −(m−1)
= ( )···( )z m
m m m

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 m−2
Q
(1 − jm)
j=0 1
= z m −(m−1)
mm−1
(this works even for m = 1), and we arrive at
m−2
Q
Z (1 − jm)
2πi
f (z) j=0
dz = .
(z − 1)m mm−1 (m − 1)!
γ

z2 + 1
Z
10. Evaluate dz, where γ(t) = reit , 0 ≤ t ≤ 2π, for all possible values of
z(z 2 + 4)
γ
r, 0 < r < 2, 2 < r < ∞.

2
Solution: Assume first that 0 < r < 2; then f (z) := zz2 +4 +1
is analytic on an open ball
iπ
around 0 containing γ, so the integral is just 2πi · f (0) = 2 .

Now suppose 2 < r < ∞. Then the function has 3 poles in any ball around 0 con-
taining γ, so we should use the partial fraction decomposition

z2 + 1 1 3z
2
= + 2
z(z + 4) 4z 4(z + 4)
1 3 1 1
= + ( + )
4z 8 z + 2i z − 2i
to integrate term-by-term. Each of these terms is easy: since ±2i are both at distance
2 < r from 0, we can apply Proposition 2.6 to yield
Z Z Z Z
1 3 1 1 1 1 3 1 1
+ ( + )dz = dz + ( dz + dz)
4z 8 z + 2i z − 2i 4 z 8 z + 2i z − 2i
γ γ γ γ
1 3 3
= ( + + )2πi
4 8 8
= 2πi.

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