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The University of Sydney

School of Mathematics and Statistics

Solutions to Tutorial 11: Complex integrals

MATH3068 Analysis Semester 2, 2009

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 , where C is

(a) the line segment from 0 to 1− i;

Solution: The curve C is parametrized by z (t) = (1 − i)t, 0 ≤ t ≤ 1. On this segment C ,|z| =

√ t2 + t2 =

√ 2 t (since t ≥ 0), and z ′(t) = (1 − i). So

C

|z| dz =

10

|z(t)|z′(t) dt =

10

√ 2 t(1− i) dt =

√ 2(1− i)

t2

2

10

=

√ 2

2 (1− i).

(b) the circle C 1(0) = {z ∈ C : |z| = 1}.

Solution: On C , |z| = 1 and so C |z| dz =

C

1 dz = 0 by Cauchy’s Theorem, because theconstant function 1 is differentiable. Alternatively, C is parametrized by z = eit, 0 ≤ t ≤ 2π,and z ′(t) = ieit, so

C

|z| dz =

2π0

|z(t)|z′(t) dt =

2π0

ieit dt = eit2π0

= 0.

2. Evaluate

C

ez dz, where C is

(a) the line segment from 1 to i;

Solution: Because ez

= F ′

(z) for all z, where F (z) = ez

, the integral of ez

along a curve C is F (β ) − F (α) = eβ − eα, where α and β are the start and finish points of C .

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 to i along the circle |z| = 1;

Solution: See the solution of the previous part.

(c) the circle C 2(0) = {z ∈ C : |z| = 2}, taken once anti-clockwise.

Solution: The integral is around a closed curve, starting at α = 2 and ending at β = 2.So it equals eβ − eα = 0.

3. Evaluate

C

(2z + i) dz where C is

(a) the circular arc from 3 + 4i to −5 along the circle |z| = 5;

Solution: Since 2z + i = F ′(z) for F (z) = z2 + iz, the integral of 2z + i along a curve C isF (β )−F (α), where α and β are the start and finish points of C . So the value of

C

2z + i dzdepends only on the end-points of C .

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 2009 The University of Sydney 1

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(b) the straight line segment from −5 to 3 + 4i;

Solution: The curve has the same end-points as in part (a) but is directed from −5 to3 + 4i, so that α and β are the reverse of what they were in part (a). The value of theintegral is therefore −36 + 32i.

(c) the circle C 5(0) = {z ∈ C : |z| = 5}, taken once anti-clockwise.

Solution: The curve is closed, i.e., α = β , and so the integral is zero.

4. Evaluate C

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

z and C is the circle C 2(1) = {z ∈ C : |z− 1| = 2}, traversed

in the anti-clockwise direction.

Solution: Since ez differentiable on and inside C , we can put a = 0 in Cauchy’s Integral Formula,obtaining

C

ez

z dz = 2πie0 = 2πi.

5. Evaluate the following integrals C

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

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

, C is the circle C 3(0) = {z ∈ C : |z| = 3}.

Solution:

1z + i

+ 1z − i

= z − i + x + i(z + i)(z − i)

= 2zz2 + 1

.

So C

2z

z2 + 1 dz =

C

1

z + i +

1

z − i

dz =

C

1

z + i dz +

C

1

z − i dz = 2πi + 2πi = 4πi

since C

1

z − a dz = 2πi

whenever C winds once around a, as shown in lectures.

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

, C is the unit circle C 1(0) = {z ∈ C : |z| = 1}.

Solution: Because z 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 therefore C

dzz2+5z+6

= 0 by Cauchy’s theorem.

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

, C is the circle C 13

(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 = 2πig(0) = 2πi.

(d) f (z) = z3+2z(z−1)3

, C is the circle C 12

(0) = {z ∈ C : |z| = 12}.

Solution: The integral is 0 by Cauchy’s Theorem, since f (z) is differentiable on andinside C .

6. Evaluate the integral ∞

−∞

1

x2 + 2x + 2 dx

by integrating 1z2+2z+2

around the closed curve consisting of the line segment from −R to R (where

R >√

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

Solution:

•

−

1+i

R

2

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The denominator z2 + 2z + 2 factors into (z − α)(z − β ), where α = −1 + i and β = −1 − i. So

we can think of the integrand as f (z)z−α

, where f (z) = 1z−β

, which is differentiable on and inside C .So the integral around C is

2πif (α) = 2πi 1

α − β = π.

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

1

t2 + 2t + 2 dt.

On the semicircular part C ′ of C , 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

≥ R2

1− 2

R − 2

R2

≥ R2

1 − 2

3 − 2

9

=

R2

9 once R ≥ 3.

Hence the integrand is in modulus at most 9R2 on C ′. Now C ′ has length π R, and so

C ′

1

z2 + 2z + 2 dz ≤

9

R2 · πR =

9π

R ,

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

π =

C

1

z2 + 2z + 2 dz =

R−R

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 9π/R. Hence

R−R

1

t2 + 2t + 2 dt = π − E (R) → π as R →∞.

That is, ∞−∞

1t2 + 2t + 2

dt = π.

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