APPM 4360/5360 Homework Assignment #5 Solutions Spring 2019

Problem #1 (15 points): Consider f n (z)g n (z) in a domain D where: |g n (z)| ≤ M ; M constant in D and f n (z)
converges to zero uniformly for all z ∈ D. Prove f n (z)g n (z) converges to zero uniformly for all z ∈ D.

Solution: By definition of uniform convergence, for any ǫ > 0 there is a number N (ǫ) such that, for all z ∈ D and
all n > N (ǫ), | f n (z)| < ǫ. For a given ǫ > 0, let ǫ1 = ǫ/M . Then, for all z ∈ D and all n > Ñ (ǫ) = N (ǫ1 ), we have

0 ≤ | f n (z)g n (z)| = | f n (z)||g n (z)| ≤ M | f n (z)| < M ǫ1 = ǫ.

This proves the statement.

Problem #2 (15 points): Find the Taylor series expansion of the following functions:

(a) z 2 /(1 + z 3 ), |z| < 1;

(b) cosh kz, k > 0 constant;

2
(c) ze i kz ; k > 0 constant.

Solution:

(a) z 2 /(1 + z 3 ), |z| < 1; using geometric series,

z2 2
∞
3 n
∞
(−1)n z 3n+2 .
X X
= z (−z ) =
1 + z3 n=0 n=0

(b) cosh kz, k > 0 constant;

X∞ (kz)2n
cosh kz = .
n=0 (2n)!

2
(c) ze i kz ; k > 0 constant.
2
∞ (i kz 2 )n ∞ i n k n z 2n+1
ze i kz = z
X X
= ,
n=0 n! n=0 n!
where i n = 1 for n = 4m, i n = i for n = 4m + 1, i n = −1 for n = 4m + 2 and i n = −i for n = 4m + 3, all m ∈ Z.

Problem #3 (20 points): Given Z∞

F (z) = f (t )e i zt d t ,
−∞

where f (t ) = e α1 t , t < 0, and f (t ) = e −α2 t , t > 0; α j > 0, j = 1, 2 constants. Find the region of the complex plane
2
where F (z) is analytic; explain. Do the same if f (t ) = t e −κt , κ > 0; explain. F (z) is referred to as the Fourier
transform of f (t ).

Solution: Let h(z, t ) = f (t )e i zt . Then, for both cases mentioned,

1) h(z, t ) is entire function of z for all t ;
2) h(z, t ) is continuous function of t for all z;
since the integral over t is improper, one has to verify condition 3) of the theorem about analyticity of such
R∞
integrals: whether there is a function G(t ) such that |h(z, t )| ≤ G(t ) and −∞ G(t )d t < ∞. Let z = x + i y.
Then
|h(z, t )| = | f (t )||e i zt | = | f (t )||e i (x+i y)t | = | f (t )|e −y t .

Then, for the first given f (t ), one gets

t <0: |h(z, t )| = e (α1 −y)t ,

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 t >0: |h(z, t )| = e −(α2 +y)t ,
R∞
so −∞ |h(z, t )|d t is finite if and only if α1 − y > 0 and α2 + y > 0. Thus, F (z) is analytic in the horizontal strip
−α2 < Imz < α1 .
R∞
For the second given f (t ), −∞ |h(z, t )|d t is finite for all z, therefore F (z) is analytic for all finite z in this case.
Formally, one can consider a region y > y 0 in C. Then
2
|h(z, t )| < G(t ) = e −κt e −y 0 t ,
R∞
and −∞ G(t )d t < ∞, so in this region one directly applies the theorem. This is true for any finite y 0 so F (z) is
analytic for all finite z.

Problem #4 (20 points): Let Z∞

F (z) = f (t )e −zt d t ,
0

where f (t ) is continuous function, | f (t )| ≤ Ae −αt , A > 0, α > 0 constants. Find the region of the complex plane
where F (z) is analytic; explain. F (z) is referred to as the Laplace transform of f (t ).

Solution: Let g (z, t ) = f (t )e −zt , then g (z, t ) is analytic in z in C and continuous in t for all t > 0. Also we have
(z = x + i y)
|g (z, t )| = | f (t )||e −zt | ≤ Ae −αt e −xt ,
R∞
and 0 Ae −(x+α)t d t is finite for x > −α. Therefore F (z) is analytic in the (infinite) vertical strip Rez > −α, by the
same theorem as used in the previous problem.

Problem #5 (20 points): Let f (z) = 1/(z 2 + α2 ), α > 0. Find the Laurent expansion in the regions

(a) |z| > α

(b) |z| < α

Solution: µ ¶
1 1 1 1
f (z) = = − .
z 2 + α2 2i α z − i α z + i α
(a) |z| > α µ ¶
1 1 1
f (z) = − =
2i αz 1 − i α/z 1 + i α/z
1 X ∞ µ (i α)n (−i α)n
¶
= − =
2i αz n=0 z n zn
X∞ (−1)m (2α)2m−2
=− .
m=1 z 2m
(b) |z| < α µ ¶
1 1 1
f (z) = + =
2α2 1 + i z/α 1 − i z/α
1 X ∞ µ (−i z)n (i z)n
¶
= + =
2α2 n=0 αn αn

1 X ∞ (−1)m z 2m
= .
α2 m=0 α2m

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Problem #6 (15 points): Given the function

2z
f (z) = ,
(z − i )(z + 2)

find the Laurent series of f (z) in the regions

(a) |z| < 1

(b) 1 < |z| < 2
(c) |z| > 2

Solution: Using partial fractions, we see that

2z 2/5 + 4i /5 8/5 − 4i /5
= + .
(z − i )(z + 2) z −i z +2

(a) For |z| < 1,

2/5 + 4i /5 4/5 − 2i /5
f (z) = i +
1+iz 1 + z/2
µ ¶ ∞ µ µ ¶n ¶
4 − 2i X n −1
= −(−i ) + zn .
5 n=0 2

(b) For 1 < |z| < 2,

2/5 + 4i /5 4/5 − 2i /5
f (z) = +
z(1 − i /z) 1 + z/2
¶ ∞ µ ¶n
in
µ ¶ ∞ µ
2(1 + 2i ) 2(2 − i ) X −1
zn
X
= n+1
+
5 n=0 z 5 n=0 2

(c) For |z| > 2,

µ ¶
2(1/5 + 2i /5) 2(4/5 − 2i /5) 1
f (z) = +
z(1 − i /z) z 1 + 2/z
2(1 + 2i ) X i n ∞ (−2)n
µ ¶ ∞
2(4/5 − 2i /5) X
= n+1
+ n
5 n=0 z z n=0 z
µ ¶ ∞
2(2 − i ) X ¡ n+1 ¢ 1
= i − (−2)n+1 n+1 .
5 n=0 z

Problem #7 (45 points): Discuss all singularities of the following functions including the type of singularity:
removable, pole – include order, essential, branch point, cluster, . . . , that each of these functions has in the finite
z-plane. For parts a,b,c,d, if the functions have a Laurent series around any of the singularities find the first two
nonzero terms.

(a) secz
(b) e z1−1
log z
(c) z(z−2)
(d) sin(1/z 2 )
(e) coth(1/z)

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Solution:

(a) secz = 1/ cos z. Since cos z is an entire function, the only singular points are those where cos(z) = 0,
i.e. z = z k = π/2 + πk, k ∈ Z. Around such a point, let z = z k + (z − z k ) = z k + u, then
1 1 1
= =− =
cos(z) cos(z k + u) sin(z k ) sin(u)

(−1)k (−1)k
= = =
sin(u) u − u 3 /6 + . . .
(−1)k (−1)k (−1)k u
= (1 + u 2 /6 + . . . ) = + +...,
u u 6
where . . . correspond to positive powers of u = z − z k greater than 1. I.e. z = z k = π/2 + πk is simple pole
with strength (−1)k .
(b) e z1−1 . Since the denominator is an entire function, the only singular points are those where e z − 1 = 0,
i.e. z = z n = 2i πn, n ∈ Z. They are isolated. Around such a point, let z = z n + u, then
1 1 1
= z u = u =
ez −1 e e −1 e −1
n

1 1
= = =
1 + u + u 2 /2 + · · · − 1 u(1 + u/2 + . . . )
1 − u/2 + . . . 1 1
= = − +...,
u u 2
i.e. z = z n = 2i πn is simple pole with residue 1 (for every n).
log z
(c) z(z−2) . Due to log z, there are two branch points, z = 0 and z = ∞. A branch cut must connect them, and a
branch of log z is analytic everywhere outside the cut. All points on the cut are nonisolated (jump) s.p. of
the function. If the cut passes through the point z = 2, then z = 2 is a nonisolated s.p. If it does not, e.g. if
the cut connects 0 and ∞ on the negative real axis, then Laurent expansion around z = 2 is

log z log 2 + log(1 + (z − 2)/2) (log 2 + (z − 2)/2 + . . . )(1 − (z − 2)/2 + . . . )

= = =
z(z − 2) (2 + (z − 2))(z − 2) 2(z − 2)
log 2 1 − log 2
= + +...,
2(z − 2) 4
which means that, for every branch of log, z = 2 is a simple pole with residue log 2/2.
(d) sin(1/z 2 ). Since sin ζ is an entire function of ζ, the only singular point in the finite C is z = 0, being the only
one, it is isolated. For all finite z 6= 0, sin(1/z 2 ) is equal to the convergent series,
∞ (−1)n ∞ (−1)n
sin(1/z 2 ) =
X X
2 2n+1 (2n + 1)!
= 4n+2 (2n + 1)!
.
n=0 (z ) n=0 z

This is the Laurent series around z = 0, which shows that z = 0 is essential singular point. The first two
terms of the series are
1 1
sin(1/z 2 ) = 2 − 6 + . . .
z 6z
(e) coth(1/z). The only s.p. are points where sinh(1/z) = 0 i.e. 1/z = i πk, k ∈ Z, and z = 0. They are isolated
except for z = 0, which is the limit of the sequence of the other points, so z = 0 is a cluster point. Around a
point z = z k = −i /(πk), let 1/z = u + i πk, then

cosh(i πk + u) cosh(i πk) cosh u

coth(1/z) = = =
sinh(i πk + u) cosh(i πk) sinh u

1 + u 2 /2 + . . . 1 u
= = + +··· =
u + u 3 /6 + . . . u 3

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 z 1 − i πkz −i /(πk) + (z + i /(πk)) i πk(z + i /(πk))
= + +··· = − + +··· =
1 − i πkz 3z i πk(z + i /(πk)) 3i /(πk)
1 i π2 k 2 (z + i /(πk))
= + + +...
π2 k 2 (z + i /(πk)) πk 3
i.e. each s.p. is simple pole of strength 1/(π2 k 2 ) for k 6= 0. (k = 0 corresponds to z = ∞ which should be
considered separately, then z = 1/u, expand around u = 0 and get a simple pole again.)

Problem #8 (30 points): Evaluate the integral

1
I
I= f (z) d z,
2πi C

where C is the unit circle centered at the origin, and f (z) is given below:
z2
(a) f (z) = ,0<a <1
z2 + a2
(b) f (z) = cot(2z)
log(z + a)
(c) f (z) = , a > 1, principal branch
z + 1/a

Solution:

(a) There are singular points at z = i a and z = −i a; both are inside the unit circle. We have

z2 a2
µ ¶
ia 1 1
f (z) = = 1 − = 1 + − ,
z2 + a2 z2 + a2 2 z −ia z +ia

therefore
1 ia
I
I= f (z) d z = 0 + (1 − 1) = 0.
2πi C 2

(b) The singular points are those where sin(2z) = 0, i.e. z = z k = πk/2, k ∈ Z. Only one such point, z = 0, is
inside C . Using that

1 − (2z)2 /2 + . . . (2z)2
µ ¶
cos(2z) 1 1 2z
cot(2z) = = 3
= 1− +... = − +...,
sin(2z) 2z − (2z) /6 + . . . 2z 3 2z 3

where . . . stand for higher powers of z. Thus, integrating powers of z, we get

1 1 1 1
I I
I= f (z) d z = dz +0 = .
2πi C 2πi C 2z 2

(c) This f has a branch point at z = −a, make the branch cut on (−∞, −a] and, for the principal branch, when
z = x > a, log(z − a) = log |x − a|. Then z = −1/a is a simple pole inside C . Expanding the function in the
Laurent series around z = −1/a, we get

log(z + a) log(a − 1/a + (z + 1/a)) log(a − 1/a) + (z + 1/a)/(a − 1/a) + . . . log(a − 1/a) 1
= = = + +...,
z + 1/a z + 1/a z + 1/a z + 1/a a − 1/a
where dots stand for positive powers of z + 1/a. Thus, deforming the contour to the small circle around
z = −1/a, we get

1 1 log(a − 1/a)
I I
I= f (z) d z = d z + 0 = log(a − 1/a) = log |a − 1/a|,
2πi C 2πi C z + 1/a
since a − 1/a > 0 > −a.

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Problem #9 (20 points):

(a) Let f (z) = 1 + z 2 + z 4 + . . . , |z| < 1. Find a function, call it g (z), that analytically continues f (z) to |z| > 1;
what can be said about g (z) on |z| = 1 and for |z| < 1; explain.

(b) Consider f (z) = log(2(z − 1)); z − 1 = r e i θ . Discuss/explain the analytic continuation of the function from
R 1 → R 2 → R 3 where r > 0 and θ is in the regions: R 1 : 0 ≤ θ ≤ π/2; R 2 : π/3 ≤ θ ≤ 4π/3; R 3 : π ≤ θ ≤ 7π/3.

Solution:

(a) For |z| < 1, f (z) is a convergent geometric series and its sum is 1/(1 − z 2 ). Thus, if we define
g (z) = 1/(1 − z 2 ), we get g (z) = f (z) for |z| < 1, and g (z) is analytic for all z ∈ C except for z = ±1. Thus, g (z)
is the analytic continuation of f (z) to |z| > 1. On |z| = 1, g (z) is analytic (and continuous) except for
z = ±1.

(b) f (z) = log(2(z − 1)) = log 2 + log(z − 1). We have to find such branches of log(z − 1), f 1 (z) in R 1 , f 2 (z) in R 2
and f 3 (z) in R 3 that an analytic function f (z) can be defined in R 1 ∪ R 2 ∪ R 3 by

 f 1 (z), z ∈ R1
f (z) = f (z), z ∈ R2
 2
f 3 (z), z ∈ R3

This is possible if f 1 (z) = f 2 (z) in R 1 ∪ R 2 and f 2 (z) = f 3 (z) in R 2 ∪ R 3 .

First, we have to choose f 1 (z) as a branch analytic in R 1 . Define f 1 (z) = log(z − 1) for 0 < r < ∞,
0 ≤ θ ≤ π/2, as the principal branch of log(z − 1) with the branch cut on (−∞, 1]. Then f 2 (z) can be taken
e.g. as the principal branch of log(z − 1) for the cut on [1, +∞), then f (z) = f 1 (z) = f 2 (z) for π/3 ≤ θ ≤ π/2.
Now, to continue f 2 from R 2 to R 3 , one can take f 3 (z) as e.g. the branch of log(z − 1) for the cut outside of
R 3 e.g. on the ray [1, +∞ · e 2πi /3 ), such that the range of θ is 2π/3 ≤ θ < 8π/3 for this branch. Then
f (z) = f 2 (z) = f 3 (z) for π ≤ θ ≤ 4π/3. The whole range of θ for f (z) here exceeds 2π therefore such f (z) is
defined on the Riemann surface of log(z − 1) but cannot be defined in C.

Extra-Credit Problem #10 (10 points): Given the function

e −1/t
Z∞
A(z) = dt,
z t2
Find a Laurent expansion in powers of z for |z| > R, R > 0. Why will the same procedure fail if we consider

e −t
Z∞
E (z) = dt
z t

Solution: In fact ¯∞
A(z) = e −1/t ¯z = 1 − e −1/z ,
therefore the Laurent expansion in powers of z for |z| > R, R > 0, is

X (−1)n X (−1)n+1
A(z) = 1 − n
= n
.
n=0 z n! n=1 z n!

Consider the ratio test for the series:

|a n+1 (z)| n!|z|n 1
= n+1
= →n→∞ 0,
|a n (z)| (n + 1)!|z| (n + 1)|z|

for all z s.t. |z| > R > 0, so the series converges.

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As for E (z), iterating the identity
e −t e −z e −t
Z∞ Z∞
dt = −k dt,
z tk zk z t k+1
starting with k = 1 up to k = n, one obtains the series expansion with the remainder term R n (z). Consider the
ratio test for the series:
|a n+1 (z)| (n + 1)!|z|n+1
= = (n + 1)z →n→∞ ∞,
|a n (z)| n!|z|n
for all z 6= 0, so the series diverges.