Solutions Manual of Complex Analysis by Gamelin | 1st edition

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IV 1 2 3 4 5 6 7 8

1 X X X X X X X X

2 X X X X X X X X

3 X X X X X X X X

4 X X X X X X X X

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 X X X X X X X X X X X X X X

X X X X X X X X X

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X

X

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I.4.4 Describe how to construct the following functions p Riemann surfaces for the 2=5 (a) w = z 1=4 ; (b) w = z i; (c) w = (z 1) : Remark. To describe the Riemann surface of a multivalued function, begin with one sheet for each branch of the function, make branch cuts so that the branches are de…ned continuously on each sheet, and identify each edge of a cut on one sheet to another appropriate edge so that the function values match up continuously.

Solution (a) Use four sheets, can make branch cuts along real axis from 1 to 0. (b) Use two sheets, can make branch cuts along horizontal line from 1 + i to i. (c) Use …ve sheets, can make branch cuts along real axis from 1 to 1.

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II.1.11 At what points does the function Arg z have a limit? Where is Arg z continuous? Justify your answer.

Solution Arg z have a limit at each point of Cn ( 1; 0]. It is continuous in Cn ( 1; 0].

It is discontinuous at each point of ( 1; 0]. Values of function ! f g at points of ( 1; 0], the function may take any value in intervall [ ; ] as z ! 0. The value is depending in that direktion we approach z = 0.

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III.2.5 Let 0 (t) and 1 (t), 0 t 1, be paths in the slit annulus fa < jzj < bg n ( b; a) from A to B. Write down explicitly a family of paths s (t) from A to B in the slit annulus that deforms continuously to 1 .

Suggestion. Deform separately the modulus and the principal value of the argument.

Solution Write 0 (t) = r0 ei 0 (t) , 0 6 t 6 1 , when continuous. (Use the fact that (t) = Arg annulus) Also 1 (t) = r1 ei 1 (t) .

Then consider s

s) r0 (t)] ei[s

(t) = [sr1 (t) + (1

< 0 (t) < , r0 (t), 0 (t) (t) is continuous on the slit 0

1 (t)+(1

s)

0 (t)]

;

0

s

1:

This does the trick, it deform 0 to 1 continuously in D, and each s is a path in D from A to B. We see that s (t) belongs to the slit annulus for 0 s 1 and 0 t 1 since for every 0 5 1 we have a < min (r0 (t) ; r1 (t))

sr1 (t) + (1

s) r0 (t)

max (r0 (t) ; r1 (t)) < b

and < min (

0

(t) ;

1

(t))

s

1

(t) + (1

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s)

0

(t)

max (

0

(t) ;

1

(t)) <

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V.1.3 1 2 3 P L K P1

Show that if p > 1, then the series

k=1

n X 1 S < p k (p k=1 Rk Hint. Use the estimate k1p < k 1 dx . xp

1=k p converges to S, where

1 1) np

1

:

Solution Let k be a nonnegative integer. Then since the function x 7! 1=x is decreasing on the positive reals, we have for all x 2 [k 1; k] that 1=k 1=x. Becauce p > 1, it follows that 1=k p 1=xp . Hence locking att the undersum in the interval [k 1; k] Z k 1 1 1 (k (k 1)) = p dx p p k k k 1 x Therefore n X 1 kp k=1

n Z X k=1

k

k 1

Z n dx x p+1 dx = 1+ = 1+ p xp p+1 1 x

n

= 1+ 1

1 p

n 1 p

p+1

n p

p+1

1

Hence the serie converge to S, where S

1+

1 p

1

=

p p

1

;

since its terms are > 0.

Now we have that Z N N X 1 dx 6 =1+ p k xp n k=n+1

x p+1 p+1

as N ! 1.

Thus we have

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N

= n

n

p+1

p

N 1

p+1

!

1

1+

1 p

1

:

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(a) Double poles at 1 , principal parts (1=4) (z 1)2 . z 1 ez ez + z+1 . Principal part (b) Singularity at z = 1, simple poles zze 2 1 = 2 z 1 1=2 1=2 at z = 1, z+1 at z = 1. z 1 (d) tan z = sin z= cos z, poles at z = =2 + n , n = 0; 1; 2; : : :. cos z = sin ( =2 + n ) (z =2 n ) + O (z =2 n )3 sin( =2+n )+O((z =2+n )2 ) tan z = sin( =2+n )(z =2 n )+O (:::)3 = z =21 n + O (z =2 n ) Poles ( ) are simple, principal part 1= (z =2 n ) .

2 2 (f) cos z= (z =4), singularities at =2 are both removable. (g) De…ned and analytic for 1 1=z 2 Cn [ 1; 0), 1=z 1 2 Cn [0; 1), 1=z 2 Cn [1; 1), z 2 Cn [0; 1]. No isolated singularities.

Rz R2 dw RP (h) Isolated singularity at z = 1, Log z = dw = = ( 1)n (w 1)n dw = w 1+w 1 1 P

1

n

( 1) (z 1) n+1

n=0 Log z (z 1)3

=

n+1

1 (z 1)2

Log z = (z 1 z(z 1)

1)

(z 1) 2

2

+

(z 1) 3

3

:::

+ analytic

2 Double pole, principal part (z 11)2 z(z1 2) at z = 1. (i) e1=(z +1) , singularities at z = i. Values of 1= (z 2 + 1) approach 1 from all directions as z ! i, 2 so values of e1=(z +1) do not converge, and singularities are essential.?????

2 2 check e1(1 t ) = f (it) ! +1 as t " 1 or t # 1, e1(1 t ) = f (it) ! 0 as t # 1 or t " 1, so no limit at i.

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VII.3.5 Show using residue theory that Z d 1 r2 = 1; 2 1 2r cos + r 2

0

r < 1:

Solution

VII.3.5 (r = 1/2)

|z | = 1 z2

z1

Set

I=

Z

1

1 r2 d = t = = 2 d = dt 2r cos + r 2 Z 0 1 r2 dt = = 2 2r cos ( t) + r 2 2 1 Z 2 dt 1 r2 = 2 1 + 2r cos t + r 2 0

A change of variables as in book page 203-204 gives

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VIII.7.1 1 2 3 P L K Let f (z) be an entire function, and suppose g ( ) is analytic for in the open upper hand lower half-planes and across the interval ( 1; 1) on the real line.

Suppose that Z 1 f (x) dx = g ( ) 1 z for in the upper half-plane. What is the value of the integral when the lower half-plane? Justify your answer carefully.

Solution

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is in