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MATH 185: COMPLEX ANALYSISFALL 2009/10

PROBLEM SET 10 SOLUTIONS

1. (a) Show that if f has a pole or an essential singularity at a, then ef has an essential singularityat a.Solution. If f has a pole of order m at a, then there exists > 0 and g : D(a, ) Canalytic, g(a) = 0, such that

f (z) =g(z)

(z a)m

for all z D (a, ). Let the power series representation of g on D (a, ) be

g(z) =

n =0an (z a)n .

Leth(z) :=

n =0an + m (z a)n .

Then 1

f (z) = h(z) +m 1

n =0

an(z a)m n

and so

ef (z ) = eh (z )m 1

n =0e

a n

( z a ) m n =: eh (z )F (z).

Note that eh (z ) is analytic and non-zero. If ef (z ) has a pole or a removable singularity at a,then F (z) = ef (z )e h (z ) will have a pole or a removable singularity at a. So since F (z) hasan essential singularity at a, ef (z ) must have an essential singularity at a.If f has an essential singularity at a, then f (D (a, )) is dense in C for all > 0, ie.f (D (a, )) = C (here S denotes the closure of the set S ). Recall that if g is any continuousfunction, then g(S ) g(S ). Since exp : C C is a continuous function, let S = f (D (a, ))and we have

exp(S ) exp(S ) = exp( C ) = C .Hence, (exp f )(D (a, )) is dense in C . By part (a), a is an essential singularity of exp f .

(b) Let C be a region. Let a and f : \{ a} C be a function with an isolatedsingularity at a. Suppose for some m N and > 0,

Re f (z) m log|z a |for all z D (a, ). Show that a is a removable singularity of f .Solution. Note that the condition implies that

|ef (z ) | = eRe f (z ) e m log|z a | =1

|z a |m.

Date : December 5, 2009.1Now that you have learnt about Laurent series and its relation with poles, you could of course just write this

down without going through the preceding arguments.

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and so f 0 on D (0, 1).(b) Let f : C C be analytic on C with a pole of order 1 at 0. Show that if f (z) R for

all |z | = 1, then for some C and R ,

f (z) = z + 1z

+

for all z C .Solution. Since f has a pole of order 1 at 0 and is analytic otherwise, the Laurentexpansion of f takes the form

f (z) =

n = 1an zn . (2.1)

Let be the closed curve z : [0, 2] C , z() = ei . Since f (ei ) R for all [0, 2], wemust have

f (ei ) = f (ei ).

The integral formula for Laurent coefficients yields, for any n Z ,

an = 12i f (z)zn +1 dz=

12

2

0f (ei )e in dz

=1

2 2

0f (ei )e in dz

=1

2 2

0f (ei )ein dz.

Note that

a

n =1

2 2

0f (ei )ein dz

and so we getan = a n

for all n Z . By (2.1), a n = 0 for all n > 1. So an = 0 for all n > 1. Let a1 = and soa 1 = a1 = . Let a0 = . Then (2.1) becomes

f (z) = z + 1z

+ .

We know that = 0 since f has a simple pole at 0. We also know that R since = a 0 = a0 = .

3. Evaluate the integral

i f ifor i = a, b.(a) f a : C C is given by

f a (z) = ee1z

and a is the boundary D (0, 2) traversed once counter-clockwise.

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Solution. f a is analytic in C and its Laurent expansion about z = 0 may be obtainedas follows:

ee1z = 1 + e

1z +

12!

e2z + +

1n!

enz +

= 1 + 1 +1z

+12!

1z2

+ +12!

1 +2z

+12!

2z

2

+

+ + 1n!

1 + nz

+ 12!

nz

2 + + .

Observe that the coefficient of the term z 1 is simply

0 + 1 +22!

+33!

+ +nn!

+ = 1 +11!

+12!

+ +1

(n 1)!+

= e.

By the residue theorem

a f a = 2 i Res( f a ;0)Ind( a ; 0) = 2ei.(b) f b : D (0, ) C is given by

f b(z) =1

(sin z)3

and b is the boundary D (0, 1) traversed once counter-clockwise.Solution. f b is analytic in D (0, ) and its Laurant expansion about z = 0 may beobtained as follows:

1(sin z)3

= z 13!

z3 +15!

z5 3

=1z3

1 13!

z2 15!

z4 + 3

= 1z3

1 + 3 13!

z2 15!

z4 + + 6 13!

z2 15!

z4 + 2 + .

Now observe that the terms enclosed in the second parentheses onwards would all havepowers at least 4 and so will not contribute to the z 1 term. The only term in the rstparentheses that contribute to the z 1 term is the z2 term, which has coefficient 3 / 3! = 1/ 2.By the residue theorem,

b f b = 2 i Res( f b;0)Ind( b; 0) = i.4. (a) Let the Laurent expansion of cot( z ) on A(0;1, 2) be

cot( z ) =

n = an zn .

Compute an for n < 0.Solution. Let be the circle D (0, r ) traversed once counter-clockwise and 1 < r < 2.Note that Ind(; z) = 1 for all z D (0, r ), ie. the bounded component of . By the integralformula for Laurent coefficients,

a k =1

2i cot( z )z k+1 dz = 12i zk 1 cot( z ) dz4

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for k N . For k = 1,

zk 1 cot( z ) = cot( z ) =cos(z )sin(z )

has three isolated (non-removable) singularities in the bounded component of , namely, 1, 0, 1. So by the residue theorem 2,

a 1 =1

2i cot( z ) dz= Res(cot( z ); 1) + Res(cot( z ); 0) + Res(cot( z );1)=

cos(z ) cos(z ) z = 1

+cos(z )

cos(z ) z=0+

cos(z ) cos(z ) z =1

=3

.

For k 2,

limz 0

z[zk 1 cot( z )] = limz0

zsin(z )

[zk 1 cos(z )]

=1

limz0

z

sin(z ) lim

z0zk 1 cos(z )

= 0

In other words, 0 is a removable singularity of zk 1 cot( z ) for k 2. Note that the residueabout any removable singularity is 0. So by the residue theorem 1,

a k =1

2i zk 1 cot( z ) dz= Res( zk 1 cot( z ); 1) + Res( zk 1 cot( z ); 0) + Res( zk 1 cot( z );1)

=zk 1 cos(z )

ddz sin(z ) z = 1

+ 0 +zk 1 cos(z )

ddz cos(z ) z =1

= ( 1)k 1

+ 1

=0 if k is odd,2

if k is even,

for k 2.(b) For n = 0 , 1, 2, . . . , compute

12i n dzz3 sin z

where n is the circle D (0, r n ) traversed once counter-clockwise and r n = ( n + 12 ).Solution. Note that for m = 0,

limz m

(z m )1

z3 sin z= lim

z m

z mz3 sin(z m )

=1

m33lim

z m

z msin(z m )

=1

m33

2We use result that Res( / ; a) = (a)/ (a) if (a) = 0, (a) = 0 and (a) = 0.

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while for m = 0

limz 0

z41

z3 sin z= lim

z0

zsin z

= 1 .

So the integrand z 3 csc z has a pole of order 3 at 0 and simple poles at m for all m Z ,m = 0. Since D (0, r n ), the bounded component of n , contains {m | m = n, , 1, 0, 1, . . . , n }, the residue theorem yields

12i n dzz3 sin z =n

m = nRes 1z3 sin z ; m .

Now for m = 0,

Res1

z3 sin z; m =

1/z 3d

dz sin z z= m=

1/m 33

cos(m )=

( 1)m

m33,

and for m = 0,

Res1

z3 sin z; 0 = 0 .

For the latter, observe that the Laurent expansion of z 3 csc z contains only even powers,

and so a 1 = 0. Hence1

2i n dzz3 sin z =n

m =1

( 1)m

m33

n

m =1

( 1)m

m33= 0 .

5. (a) Does the following function have an antiderivative on A(0;4, )?z

(z 1)(z 2)(z 3)Solution. Let f : A(0;4, ) C be

f (z) :=z

(z 1)(z 2)(z 3).

Let A(0;4, ) be the boundary of a rectangle R traversed once counter-clockwise.Since A(0;4, ), we must have 1 , 2, 3 R, the bounded component of . So thewinding numbers Ind(; n) = 1 for n = 1 , 2, 3. By the residue theorem

12i f (z) dz = Res( f ; 1) + Res( f ; 2) + Res( f ; 3).

Since f has simple poles at 1 , 2, 3, the required residues may be evaluated by

Res( f ;1) = limz1

(z 1)f (z) = limz 1

z(z 2)(z 3)

=12

,

Res( f ;2) = limz2

(z 2)f (z) = limz 2

z(z 1)(z 3)

= 2,

Res( f ;3) = limz3

(z 3)f (z) = limz 3

z(z 1)(z 2)

= 32

.

Hence,1

2i f (z) dz = 0 .Note that this holds for arbitrary and thus all A(0;4, ), = R for some R. Henceby Problem Set 9, Problem 3(a) (which is really Moreras Theorem), f has an antiderivativeon A(0;4, ).

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