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The University of Sydney

School of Mathematics and Statistics

Solutions to Tutorial 12: Complex integrals

MATH3068 Analysis Semester 1, 2007

Web Page: http://www.maths.usyd.edu.au:8000/u/UG/SM/MATH3068/Lecturer: Donald Cartwright

1. Evaluate the integral

C

|z| dz along the curve C, whereC is(a) the line segment from 0 to 1 i;

Solution: The curveC is parametrized by z (t) = (1 i)t, 0 t 1. On this segmentC,|z| = t2 +t2 = 2 t (since t 0), andz (t) = (1 i). So

C

|z| dz = 10

|z(t)|z(t)dt = 10

2 t(1 i)dt =

2(1 i) t

2

2

10

=

2

2 (1 i).

(b) the circleC1(0) = {z C : |z| = 1}.Solution: OnC, |z| = 1 and so

C|z| dz =

C1 dz = 0 by Cauchys Theorem, because the

constant function 1 is differentiable. Alternatively,Cis parametrized byz = eit, 0 t 2,and z (t) =ieit, so

C

|z| dz = 20

|z(t)|z(t)dt = 20

ieit dt= eit20

= 0.

2. Evaluate

C

ez dz, whereC is

(a) the line segment from 1 toi;

Solution: Becauseez

=F

(z) for allz, whereF(z) =ez

, the integral ofez

along a curveCisF() F() =e e, where and are the start and finish points ofC.In both parts (a) and (b), C is a curve from 1 to i. The integral therefore has the samevalue in parts (a) and (b):

C

ez dz = ei e= cos 1 e+i sin1.

(b) the circular arc from 1 toi along the circle|z| = 1;Solution: See the solution of the previous part.

(c) the circleC2(0) = {z C : |z| = 2}, taken once anti-clockwise.Solution: The integral is around a closed curve, starting at = 2 and ending at = 2.So it equals e e = 0.

3. Evaluate

C

(2z+i)dz whereC is

(a) the circular arc from 3 + 4ito5 along the circle|z| = 5;Solution: Since 2z + i= F(z) forF(z) =z2 + iz, the integral of 2z + ialong a curveC isF()F(), where and are the start and finish points ofC. So the value of

C2z + i dz

depends only on the end-points ofC.

C

2z+i dz = F(5) F(3 + 4i) =

z2 +iz

5

3+4i

= 36 32i.

This method would work with 2z+i replaced by any polynomial.

Copyright c 2007 The University of Sydney 1

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(b) the straight line segment from5 to 3 + 4i;Solution: The curve has the same end-points as in part (a) but is directed from5 to3 + 4i, so that and are the reverse of what they were in part (a). The value of theintegral is therefore36 + 32i.

(c) the circleC5(0) = {z C : |z| = 5}, taken once anti-clockwise.Solution: The curve is closed, i.e., = , and so the integral is zero.

4. EvaluateC

f(z)dz , wheref(z) = ez

z andCis the circle C2(1) = {z C : |z 1| = 2}, traversed

in the anti-clockwise direction.

Solution: Sinceez differentiable on and insideC, we can puta = 0 in Cauchys Integral Formula,obtaining

C

ez

z dz= 2ie0 = 2i.

5. Evaluate the following integralsC

f(z)dz . All the circles are traversed once in the anticlockwisedirection.

(a) f(z) = 2zz2+1

, Cis the circle C3(0) = {z C : |z| = 3}.Solution:

1z+i

+ 1z i = z i+x+i(z+i)(z i) = 2zz2 + 1 .

SoC

2z

z2 + 1 dz =

C

1

z+i+

1

z i

dz =

C

1

z+idz+

C

1

z i dz= 2i+ 2i = 4i

since C

1

z a dz = 2iwhenever Cwinds once around a, as shown in lectures.

(b) f(z) = 1z2+5z+6

, C is the unit circleC1(0) = {z C : |z| = 1}.Solution: Becausez 2 + 5z+ 6 = (z+ 2)(z+ 3), the only problem points are

2 and

3,

which are outside C. So the integrand is differentiable on and inside C, and thereforeC

dzz2+5z+6

= 0 by Cauchys theorem.

(c) f(z) = cos zz(z2+1)

, Cis the circle C13

(0) = {z C : |z| = 13}.

Solution: The only problem points for this function are at 0 and i. Of these, only z = 0lies inside C. So we can apply the Cauchy Integral Formula to g(z) = cos z

z2+1 and a = 0,

getting C

cos z dz

z(z2 + 1) dz=

C

g(z)

z dz = 2ig(0) = 2i.

(d) f(z) = z3+2z

(z1)3, Cis the circle C1

2

(0) = {z C : |z| = 12}.Solution: The integral is 0 by Cauchys Theorem, since f(z) is differentiable on andinsideC.

6. Evaluate the integral

1

x2 + 2x+ 2 dx

by integrating 1z2+2z+2

around the closed curve consisting of the line segment from Rto R(whereR >

2), followed by the semicircle above the x-axis going from R around toR.

Solution:

1+i

R

2

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The denominator z2 + 2z+ 2 factors into (z )(z ), where =1 +i and =1 i. Sowe can think of the integrand as f(z)

z, wheref(z) = 1

z, which is differentiable on and inside C.

So the integral around C is

2if() = 2i 1

=.

The horizontal part of the curve is parametrized by z (t) =t,R t R. So the integral alongthat part is simply RR

1

t2 + 2t+ 2dt.

On the semicircular partC ofC, z2 + 2z+ 2 is dominated by the z2 term when R is large. Oneway to express this is by writing

|z2 + 2z+ 2| =R21 +2

z+

2

z2

R21 2R 2

R2

R2

1 2

32

9

=

R2

9 onceR 3.

Hence the integrand is in modulus at most 9R2

on C. NowC has length R, and so

C

1

z2 + 2z+ 2 dz

9

R2 R = 9

R,

which tends to 0 as R . Therefore the second term on the right in the equation

=

C

1

z2 + 2z+ 2dz =

RR

1

t2 + 2t+ 2 dt+

C

1

z2 + 2z+ 2 dz

is thought of as an error term E(R), and has modulus at most 9/R. Hence

RR

1

t2 + 2t+ 2dt= E(R) as R .

That is,

1t2 + 2t + 2

dt= .

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