Caushys Theorem

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Homework #7 Solutions

Math 128, Fall 2013Instructor: Dr. Doreen De Leon

1 HW #7(a)

1.1 p. 135: 1, 3

Use parametric representation of C to evaluate C f (z) dz1. f (z) = (z2 )z and C is

(a) the semicircle z = 2 ei (0 )Solution:

C f (z) dz =

0f [z()]z () d

=

0(2e

i+ 2)

2e i 2ie i d

= 2 i

0(ei + 1) d

= 2 i1i

e i + |0

= 2 i1i

e i + 1i

ei (0) + 0

= 2 i 1i +

1i

= 2 i 2i + = 4 + 2 i .

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(b) the semicircle z = 2 ei ( 2 )Solution:

C f (z) dz = 2

f [z()]z () d

= 2

(2ei + 2)

2ei 2ie i d

= 2 i 2

(e i + 1) d

= 2 i1i

ei + |2= 2 i

1i

ei2 + 2 1i

ei ( ) +

= 2 i1i + 2

1i +

= 2 i 2i +

= 4 + 2 i .

(c) the circle z = 2e i (0 2 )Solution: Let C 1 be the contour from part (a) and C 2 be the contour from part (b).Then C = C 1 + C 2, so

C f (z) dz = C 1 f (z ) dz + C 2 f (z) dz= (

4 + 2 i ) + (4 + 2 i )

= 4 i .

3. f (z) = exp(z ) and C is the boundary of the square with vertices at 0 , 1, 1 + i, and i, theorientation of C being counterclockwise.Solution:

1

11 + iC 3

C 2

C 1

C 4

So, C = C 1 + C 2 + C 3 + C 4, and a parameterization for the curves is

C 1 : z(t ) = t, 0 t 1 = z (t ) = 1

C 2 : z(t ) = 1 + ( t 1)i, 1 t 2 = z (t) = i

C 3 : z(t ) = (3 t ) + i, 2 t 3 = z (t ) = 1

C 4 : z(t ) = (4 t )i, 3 t 4 = z (t ) = i

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Integrate f (z) along each contour.

C 1 f (z) dz = 1

0f [z(t )]z (t ) dt

= 1

0 et

1 dt= et 10= e 1.

C 2 f (z) dz = 2

1f [z(t )]z (t ) dt

= 2

1 exp( (1 (t 1)i)) i dt

= 2

1ie e (t 1) i dt

= e 21 ie ( t 1) i dt= e e ( t

1) i 2

1

= e 1 ei

= 2 e .

C 3 f (z) dz = 3

2f [z(t )]z (t ) dt

=

3

2

exp( ((3

t )

i))(

1) dt

= 3

2e i e (3 t ) dt

= e i e (3 t )3

2

= 1 (1 e )= e 1.

C 4 f (z) dz = 4

3f [z(t )]z (t ) dt

=

4

3 exp(

(4

t )i)(

i) dt

= 4

3 ie ( t 4) i dt

= e ( t 4) i 4

3

= 1 ei= 2.

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So,

C f (z) dz = C 1 f (z) dz + C 2 f (z) dz + C 3 f (z) dz + C 4 f (z) dz= e

1 + 2 e + ( e

1)

2

= 4 e 4 = 4 ( e 1) .

1.2 Problem 2

Evaluate C 1(z 1) dz , where C is the circle of radius 2 centered at 1, traversed counterclockwise.Solution: Parameterization of C : z() = 1 + 2 e i , 0 2 . Then, z () = 2 ie i . Therefore,

C

f (z) dz =

2

0

f [z()]z () d

= 2

0

1(1 + 2 ei ) 1

2ie i d

= 2

0i d

= i|20= 2 i .

1.3 Problem 3

Evaluate C z2 dz along two paths joining 0 to 1 + i as follows:

(a) C is the straight line segment joining 0 to 1 + i.

(b) C is the union of the line segment joining 0 to 1, then joining 1 to 1 + i.

What do you observe?Solution: Let f (z) = z2.

(a)

1

11 + i

A parameterization of C is

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z (t ) = (1 + i)t, 0 t 1= z

(t ) = 1 + i.

Then,

C f (z) dz = 1

0f [z(t )]z (t ) dt

= 1

0[(1 + i)t]2(1 + i) dt

= (1 + i) 1

0(1 i)2t 2 dt

= 2i(1 + i) 1

0t 2 dt

= 2i(1 + i) t331

0

= 2i(1 + i)13

= 23

23

i .

(b)

C 1

C 21

C 1 : z(t) = t, 0 t 1 = z (t ) = 1C 2 : z(t) = 1 + ( t 1)i, 1 t 2 = z (t) = i Then,

C 1 f (z) dz = 1

0[t]2 1 dt

= 1

0t 2 dt

= t3

3

1

0

= 13

.

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C 2 f (z) dz = 2

1[(1 + ( t 1)i)]2 i dt

= i 2

1(1 + (1 t )i)2 dt

= i 2

1(1 + 2(1 t )i (1 t )2) dt

= i t (1 t )2i + 13

(1 t )32

1

= i 2 i 13 1

= i23 i

= 1 + 23

i.

Then,

C f (z) dz = C 1 f (z) dz + C 2 f (z) dz=

13

+ 1 + 23

i

= 43

+ 23

i .

The integrals in (a) and (b) yield different results. Therefore, the integral is not path-independent.

2 HW #7(b)

2.1 Problem 1

Find a number M such that

C dzz 2 + 2 M,where C is the upper half of the unit circle.Solution:

1z2 + 2

1

|z2 + 2 |

1

|z2 (2)|

1

||z|2 | 2|| =

1

||1|2 2| = 1 .

Therefore,

C dz

z2 + 2 1(length of C ) = = M.

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2.2 p. 141: 4

Let C R denote the upper half of the circle |z| = R (R > 2), taken in the counterclockwise direction.Show that

C R2z2

1

z4 + 5 z2 + 4 dz R (2R 2 + 1)(R 2 1)(R 2 4) .

Then, show that the value of the integral tends to 0 as R .Solution:

Find an upper bound on 2 z2 1.

|2z2 1| 2|z |2 + 1= 2 R 2 + 1 .

Find a lower bound on z4

+ 5z2

+ 4.

|z4 + 5 z2 + 4 | = |(z2 1)(z2 4)|= |z2 1||z2 4| ||z|2 1|||z|2 4| = ( R 2 1)(R 2 4).

Therefore,2z2 1

z4 + 5 z2 + 4 2R 2 + 1

(R 2 1)(R 2 4).

So,

C R 2z2 1

z4 + 5 z 2 + 4 dz

2R 2 + 1(R 2 1)(R 2 4)

(length of C R )

= R (2R 2 + 1)(R 2 1)(R 2 4)

.

As R ,

0 limR C R 2z2 1

z4 + 5 z2 + 4 dz limR

R (2R 2 + 1)(R 2 1)(R 2 4)

= 0

=

limR

C R

2z2 1z4 + 5 z2 + 4

dz = 0

2.3 p. 149: 1, 2

1. Use an antiderivative to show that for every contour C extending from a point z1 to a pointz2,

C z n dz = 1n + 1 z n +12 z n +11 (n = 0 , 1, 2, . . . ).

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Solution:

C z n dz = z2

z 1z n dz

= zn +1

n + 1

z 2

z 1

= zn +12n + 1

zn +11n + 1

= 1n + 1

z n +12 z n +11 .

2. By nding an antiderivative, evaluate each of these integrals, where the path is any contourbetween the indicated limits of integration.

(a)

i2

iez dz

Solution:

i

2

iez dz =

1

ezi2

i

= 1

ei(2 ) ei

= 1

(i + 1) .

(b) +2 i

0cos

z2

dz

Solution:

+2 i

0cos

z2

dz = 2sinz2

+2 i

0

= 2sin + 2 i

2 2sin0= 2sin

2

+ i

= 2 sin2

cos i + cos2

sin i

= 2 cos i

= 2e i

2

e i

2

2

= e 1 + e.

(c) 3

1 (z 2)3 dzSolution:

3

1(z 2)3 dz =

14

(z 2)43

1

= 14

(3 2)4 14

(1 2)4= 0 .

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3 HW #7(c)

3.1 p. 160: 1(b), (c), (f)

Apply the Cauchy-Goursat Theorem to show that C f (z) dz = 0, where the contour C is the unitcircle |z| = 1, in either direction, and when(b) f (z) = ze z

Solution: f (z) = ze z is entire = f is analytic on and inside the contour C . Therefore,

C ze z dz = 0.(c) f (z) =

1(z2 + 2 z + 2)

Solution: f (z) is analytic for all z except where z2 + 2 z + 2 = 0 =

z =

1 + i,

1

i .

Neither of these points is on or inside C . Therefore, f is analytic on and inside the contourC

= C z(z2 + 2 z + 2) dz = 0 .(f) f (z) = Log ( z + 2)

Solution: Log (z) is analytic everywhere except the set {(x, y )|x 0 and y = 0}, andw = z + 2 is entire. Note that w = ( x +2)+ iy . Therefore, Log ( z + 2) is analytic everywhereexcept the set

{(x, y )|x + 2 0 and y = 0}= {(x, y )|x 2 and y = 0}.The set where f is not analytic is not on or inside C =

f is analytic on and inside C

= C Log (z + 2) dz = 0 .3.2 p. 161: 2(a), (c)

Let C 1 denote the positively oriented boundary of the square whose sides lie along the line x =

1, y = 1 and let C 2 be the positively oriented circle |z| = 4. Show that

C 1 f (z) dz = C 2 f (z) dz,where

(a) f (z) = 1

3z2 + 1Solution:

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1

-1

-4

4C 2

C 1

f (z) = 1

3z2 + 1 is analytic for all z except where 3 z2 + 1 = 0 = f is analytic for all

z= 1 3.

So, f is analtyic on C 1 and C 2 and between C 1 and C 2. Therefore, by the Path DeformationPrinciple,

C 1 13z2 + 1 dz = C 2 13z2 + 1 dz.(c) f (z) =

z1 ez

Solution: f (z) = z1 ez

is analytic for all z such that

1 ez = 0=

ez

= 1

But,

ez = 1

= ex eiy = 1 e i0

= x = ln1 and y = 0 + 2 n, n Z .

Therefore, f is analytic for

z= 2 ni, n Z .

So, f is analytic for z = 0 , 2i, 4 i , . . . = f is anlytic on C 1 and C 2 and between C 1and C 2. Therefore, C 1 z1 ez dz = C 2

z1 ez

dz.

3.3 Problem 3

Evaluate the following

(a) C (z3

+ 3) dz , where C is the upper half of the unit circle traversed counterclockwise.

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C 1

-1 1 -1 1

So,

C (z3 + 3) dz = 1

11(z3 + 3) dz

= z4

4 + 3 z

1

1

= 6 .

(b) C (z3 + 3) dz , where C is the unit circle traversed clockwise.Solution: f (z) = z3 + 3 is entire, and so is analytic on and inside C . Therefore, by theCauchy-Goursat theorem,

C (z3 + 3) dz = 0 .(c)

C

e1z dz , where C is the circle of radius 3 centered at 1 + 5 i traversed counterclockwise.

e1z is analytic for all z= 0, because

1z

is analytic for all z= 0 and ez is entire. z0 = 0 is not on

or inside C , so e1z is analytic on and inside C . Therefore,

C e 1z dz = 0by the Cauchy-Goursat theorem.

(d) C cos 3 + 1z 3 dz , where C is the unit square with corners at 0, 1, 1 + i, and i, traversedclockwise.Solution:

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1

1 C

32

f (z) = cos 3 + 1z 3

is analytic for all z= 3 = f is analytic on and inside C . So, by the

Cauchy-Goursat theorem,

C cos 3 + 1z 3 dz = 0 .

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Caushys Theorem

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