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SOLUTIONS MANUAL FOR

by

Advanced Engineering Mathematics with MATLAB®

Third Edition

Dean G. Duffy

SOLUTIONS MANUAL FOR

by

Advanced Engineering Mathematics with MATLAB®

Third Edition

Dean G. Duffy

Taylor & Francis Group, an informa business

Chapter 0

Solution Manual

Section 1.1

1.5i

2 + i=

5i

2 + i× 2 − i

2 − i=

5 + 10i

4 + 1= 1 + 2i

2.5 + 5i

3 − 4i+

20

4 + 3i=

5 + 5i

3 − 4i× 3 + 4i

3 + 4i+

20

4 + 3i× 4 − 3i

4 − 3i

=−1 + 7i

5+

16 − 12i

5= 3 − i

3.

1 + 2i

3 − 4i+

2 − i

5i=

1 + 2i

3 − 4i× 3 + 4i

3 + 4i+

2 − i

5i× −5i

−5i=

−1 + 2i

5− 1 + 2i

5= −2

5

4.(1 − i)4 = (1 − i)2(1 − i)2 = (−2i)2 = −4

5.i(1 − i

√3 )(

√3 + i) = i(

√3 − 3i+ i+

√3) = 2 + 2i

√3

6.r =

√02 + (−1)2 = 1

θ = tan−1(−1/0) = π/2 or 3π/2

Because −i lies below the real axis, θ = 3π/2 and z = e3πi/2.

7.r =

√42 + 02 = 4 and θ = tan−1(0/− 4) = 0 or π

Because −4 lies in the left side of the complex plane, θ = π and z = 4eπi.

8.r =

√4 + 12 = 4

1

2 Advanced Engineering Mathematics with MATLAB

θ = tan−1(2√

3/2) = tan−1(√

3) = π/3 or 4π/3

Because 2 + 2√

3i lies in the first quadrant, z = 4eπi/3.

9.r =

√25 + 25 = 5

√2

θ = tan−1[5/(−5)] = tan−1(−1) = 3π/4 or 7π/4

Because −5 + 5i lies in the second quadrant, z = 5√

2e3πi/4.

10.r =

√4 + 4 = 2

√2

θ = tan−1(−2/2) = tan−1(−1) = 3π/4 or 7π/4

Because 2 − 2i lies in the fourth quadrant, z = 2√

2e7πi/4.

11.r =

√1 + 3 = 2

θ = tan−1[√

3/(−1)] = tan−1(−√

3) = 2π/3 or 5π/3

Because −1 +√

3i lies in the second quadrant, z = 2e2πi/3.

12.

e(α+β)i = eαieβi

cos(α+ β) + i sin(α+ β) = [cos(α) + i sin(α)][cos(β) + i sin(β)]

= cos(α) cos(β) − sin(α) sin(β)

+ i cos(α) sin(β) + i sin(α) cos(β)

Taking the real and imaginary parts,

cos(α+ β) = cos(α) cos(β) − sin(α) sin(β)

andsin(α+ β) = cos(α) sin(β) + sin(α) cos(β).

13.N∑

n=0

eint =1 − exp[i(N + 1)t]

1 − exp(it)

= exp(iNt/2)exp[i(N + 1)t/2] − exp[−i(N + 1)t/2]

exp(it/2) − exp(−it/2)

= eiNt/2sin[(N + 1)t/2]

sin(t/2).

Worked Solutions 3

To obtain the final answer, take the real and imaginary parts of the lastequation.

14. (a)∞∑

n=0

ǫneint =1

1 − ǫ exp(it)=

1

1 − ǫ cos(t) − iǫ sin(t)

=1 − ǫ cos(t) + iǫ sin(t)

1 + ǫ2 − 2ǫ cos(t).

To obtain the final answer, take the real and imaginary parts of the lastequation.

(b)

∞∑

n=1

e−na sin(nt) =e−a sin(t)

1 + e−2a − 2e−a cos(t)=

sin(t)

2 cosh(a) − 2 cos(t).

Now multiply both sides of the equation by 2.

Section 1.2

1.8 = 23e0i+2kπi.

Therefore,zk =

√2ekπi/3, k = 0, 1, 2, 3, 4, 5

or

z0 =√

2, z1 =√

2[cos(π

3

)+ i sin

(π3

)]=

√2

[1

2+

√3i

2

],

z2 =√

2

[cos

(2π

3

)+ i sin

(2π

3

)]=

√2

[−1

2+

√3i

2

],

z3 =√

2eπi = −√

2,

z4 =√

2

[cos

(4π

3

)+ i sin

(4π

3

)]=

√2

[−1

2−

√3i

2

]

and

z5 =√

2

[cos

(5π

3

)+ i sin

(5π

3

)]=

√2

[1

2−

√3i

2

].

2.−1 = eπi+2kπi.


Therefore,zk = e(π+2kπ)i/3, k = 0, 1, 2

or

z0 = eπi/3 = cos(π

3

)+ i sin

(π3

)=

1

2+

√3

2i, z1 = eπi = −1

and

z2 = e5πi/3 = cos

(5π

3

)+ i sin

(5π

3

)=

1

2−

√3

2i.

3.−i = e3πi/2+2kπi.

Therefore,zk = eπi/2+2kπi/3, k = 0, 1, 2

orz0 = eπi/2 = i,

z1 = e7πi/6 = cos

(7π

6

)+ i sin

(7π

6

)= −

√3

2− i

2

and

z2 = e11πi/6 = cos

(11π

6

)+ i sin

(11π

6

)=

√3

2− i

2.

4.−27i = 33e3πi/2+2kπi.

Therefore,zk =

√3eπi/4+kπi/3, k = 0, 1, 2, 3, 4, 5

or

z0 =√

3[cos(π

4

)+ i sin

(π4

)], z1 =

√3

[cos

(7π

12

)+ i sin

(7π

12

)],

z2 =√

3

[cos

(11π

12

)+ i sin

(11π

12

)],

z3 =√

3

[cos

(5π

4

)+ i sin

(5π

4

)],

z4 =√

3

[cos

(19π

12

)+ i sin

(19π

12

)],

and

z5 =√

3

[cos

(23π

12

)+ i sin

(23π

12

)].

Worked Solutions 5

5. If we find one root w1, then the other root is −w1. Let z = reθi, wherer =

√a2 + b2 and θ = tan−1(−b/a). The value of θ lies between 3π/2 and 2π

because z lies in the fourth quadrant. We know that

w1 =√reθi/2 =

√r [cos(θ/2) + i sin(θ/2)].

But cos(θ/2) = ±√

[1 + cos(θ)]/2 and sin(θ/2) = ±√

[1 − cos(θ)]/2, where

cos(θ) = a/√a2 + b2. Therefore,

w1 =(a2 + b2)1/4√

2

−

√√a2 + b2 + a√a2 + b2

+ i

√√a2 + b2 − a√a2 + b2

=1√2

(−√√

a2 + b2 + a+ i

√√a2 + b2 + a

).

Our choice of signs for cos(θ/2) and sin(θ/2) is determined by the fact that3π/4 < θ/2 < π.

6. From the quadratic formula, z2 = (3i±√−9 + 8)/2 = 2i, i. Therefore,

z1,2 = ±√

2eπi/4 = ±(1 + i) and z3,4 = ±eπi/4 = ±(

1√2

+i√2

).

7. From the quadratic formula,

z2 =−6i±

√−36 − 64

2= −8i, 2i.

Therefore,

z1,2 = ±√

2eπi/4 = ±(1 + i) and z3,4 = ±2√

2e−πi/4 = ±2(1 − i).

Section 1.3

1.w = u+ iv = i(x+ iy) + 2 = 2 − y + xi

Therefore, u = 2 − y and v = x. Then, ux = 0, vy = 0, vx = 1 and uy = −1.Thus, ux = vy and vx = −uy for all x and y. Therefore, iz + 2 is an entirefunction because the derivatives are continuous.

2. w = u + iv = e−(x+iy) = e−x[cos(y) − i sin(y)]. Therefore, u = e−x cos(y)and v = −e−x sin(y). Then, vx = e−x sin(y), ux = −e−x cos(y), vy =−e−x cos(y) and uy = −e−x sin(y). Thus, ux = vy and vx = −uy for all


x and y. Therefore, e−z is an entire function because the derivatives arecontinuous.

3.

w = u+ iv = (x+ iy)3 = x3 − 3xy2 + i(3x2y − y3)

Therefore, u = x3 − 3xy2 and v = 3x2y − y3. Then, ux = 3x2 − 3y2, vy =3x2 − 3y2, vx = 6xy and uy = −6xy. Thus, ux = vy and vx = −uy forall x and y. Therefore, z3 is an entire function because the derivatives arecontinuous.

4. w = u + iv = cosh(x + iy) = cosh(x) cos(y) + i sinh(x) sin(y). Therefore,u = cosh(x) cos(y) and v = sinh(x) sin(y). Then, ux = sinh(x) cos(y), vy =sinh(x) cos(y), vx = cosh(x) sin(y) and uy = − cosh(x) sin(y). Thus, ux = vyand vx = −uy for all x and y. Therefore, cosh(z) is an entire function becausethe derivatives are continuous.

5.

f ′(z) =3

2(1 + z2)1/2(2z) = 3z(1 + z2)1/2

6.

f ′(z) = 13 (z + 2z1/2)−2/3(1 + z−1/2)

7.

f ′(z) = 2(1 + 4i)z − 3

8.

f ′(z) =2

z + 2i− 2z − i

(z + 2i)2=

5i

(z + 2i)2

9.

f ′(z) = −3i(iz − 1)−4

10.

limz→i

z2 − 2iz − 1

z4 + 2z2 + 1= limz→i

2z − 2i

4z3 + 4z= limz→i

2

12z2 + 4= −1

4

11.

limz→0

z − sin(z)

z3= limz→0

1 − cos(z)

3z2= limz→0

sin(z)

6z= limz→0

cos(z)

6=

1

6

Worked Solutions 7

12. Because u = x and v = −y, we have ux = 1 and vy = −1. Therefore,ux 6= vy for any x and y and f(z) is not differentiable anywhere on the complexplane.

13. Because uxx = 2 and uyy = −2, we have uxx + uyy = 0 and u(x, y) isharmonic. To find v(x, y), we use the Cauchy-Riemann equations: vy = ux =2x or v(x, y) = 2xy + f(x). To find f(x) we use vx = 2y + f ′(x) = −uy = 2yor f ′(x) = 0. Therefore, the final answer is v(x, y) = 2xy + constant.

14. Because uxx = 12x2 − 12y2 and uyy = −12x2 + 12y2, we have that uxx +uyy = 0 and u(x, y) is harmonic. To find v(x, y), we use the Cauchy-Riemannequations: vy = ux = 4x3 − 12xy2 +1 or v(x, y) = 4x3y− 4xy3 + y+ f(x). Tofind f(x) we use vx = 12x2y− 4y3 + f ′(x) = −uy = 12x2y− 4y3 or f ′(x) = 0.Therefore, the final answer is v(x, y) = 4x3y − 4xy3 + y + constant.

15. Because

uxx = [−2 sin(x) − x cos(x)]e−y + ye−y sin(x)

anduyy = x cos(x)e−y + 2e−y sin(x) − ye−y sin(x),

we have uxx + uyy = 0 and u(x, y) is harmonic. To find v(x, y), we use theCauchy-Riemann equations:

vy = ux = [cos(x) − x sin(x)]e−y − ye−y cos(x)

orv(x, y) = x sin(x)e−y + ye−y cos(x) + f(x).

To find f(x) we use

vx = sin(x)e−y + x cos(x)e−y − sin(x)ye−y + f ′(x)

= −uy = x cos(x)e−y + sin(x)[e−y − ye−y]

or f ′(x) = 0. Therefore, the final answer is

v(x, y) = x sin(x)e−y + ye−y cos(x) + constant.

16. Because

uxx = 2 cos(y)ex+4x cos(y)ex+(x2−y2) cos(y)ex−4y sin(y)ex−2xy cos(y)ex

and

uyy = −2 cos(y)ex+4y sin(y)ex−(x2−y2) cos(y)ex−4x cos(y)ex+2xy sin(y)ex,


we have that uxx + uyy = 0 and u(x, y) is harmonic. To find v(x, y), we usethe Cauchy-Riemann equations:

vy = ux = 2x cos(y)ex + (x2 − y2) cos(y)ex − 2y sin(y)ex − 2xy sin(y)ex

orv(x, y) = 2xy cos(y)ex + (x2 − y2) sin(y)ex + f(x).

To find f(x) we use

vx = 2y cos(y)ex + 2xy cos(y)ex + 2x sin(y)ex + (x2 − y2) sin(y)ex + f ′(x)

= −uy = 2y cos(y)ex + (x2 − y2) sin(y)ex + 2x sin(y)ex + 2xy cos(y)ex

or f ′(x) = 0. Therefore, the final answer is

v(x, y) = 2xy cos(y)ex + (x2 − y2) sin(y)ex + constant.

Section 1.4

1. Because z = eθi, z∗ = e−θi and dz = ieθidθ. Then

∮

C

(z∗)2dz =

∫ 2π

0

e−2θiieθi dθ =

∫ 2π

0

e−θii dθ = −e−θi∣∣∣∣2π

0

= 0.

2. ∮

C

|z|2dz =

∮

C

(x2 + y2)(dx+ i dy)

=

∫

C1

|z|2dz +

∫

C2

|z|2dz +

∫

C3

|z|2dz +

∫

C4

|z|2dz

Then

∫

C1

|z|2dz =

∫ 1

0

(x2 + 02) dx = 13 ,

∫

C2

|z|2dz =

∫ 1

0

(12 + y2)i dy = i(1 + 13 ),

∫

C3

|z|2dz =

∫ 0

1

(x2 + 12) dx = − 13 − 1,

∫

C4

|z|2dz =

∫ 0

1

(02 + y2)i dy = − i3 .

Finally, ∮

C

|z|2dz = −1 + i.

3. We have z = eθi and dz = ieθidθ with −π/2 < θ < π/2. Then

∫

C

|z| dz =

∫ π/2

−π/21 × ieθidθ = eθi

∣∣π/2−π/2 = 2i.

Worked Solutions 9

4. Because y = x, z = x+ ix and dz = dx+ i dx. Then

∫

C

ezdz =

∫ 1

−1

e(1+i)x(1 + i) dx = e(1+i)x∣∣∣1

−1

= e1+i − e−1−i = 2 sinh(1) cos(1) + 2i cosh(1) sin(1).

5. Because y = x2, z = x+ x2i and dz = (1 + 2ix) dx. Then

∫

C

(z∗)2dz =

∫ 1

0

(x− ix2)2(1 + 2ix) dx

=

∫ 1

0

(x2 − x4 − 2ix3)(1 + 2ix) dx

=

∫ 1

0

(x2 + 3x4 − 2ix5) dx = 1415 − i

3 .

6. For (a), z = eθi and dz = ieθidθ. Then

∫

C

dz√z

=

∫ π

0

i exp(θi)

exp(θi/2)dθ =

∫ π

0

ieθi/2dθ = 2eθi/2∣∣∣π

0= −2 + 2i.

For (b) ∫

C

dz√z

=

∫ 0

−πieθi/2dθ = 2eθi/2

∣∣∣0

−π= 2 + 2i.

Note the jump in eθi/2 as we move across the negative real axis. Just abovethe negative real axis, eθi/2 = i while below this axis we have eθi/2 = −i. Thisjump occurs because of the presence of the branch cut along the negative realaxis associated with our multivalued, complex square root function.

Section 1.5

1. Because u = e−2x cos(2y) and v = −e−2x sin(2y), ux = −2e−2x cos(2y) =vy and vx = 2e−2x sin(2y) = −uy for all x and y. Therefore, any integrationbetween two points is path independent. Thus,

∫ 2+3πi

1−πie−2zdz = − 1

2e−2z∣∣2+3πi

1−πi = − 12

[e−4−6πi − e−2+2πi

]= 1

2

[e−2 − e−4

].

2. Because u = ex cos(y) − cos(x) cosh(y) and v = ex sin(y) + sin(x) sinh(y),ux = ex cos(y) + sin(x) cosh(y) = vy and vx = ex sin(y) + cos(x) sinh(y) =


−uy for all x and y. Therefore, any integration between two points is pathindependent. Thus,

∫ 2π

0

[ez − cos(z)] dz = [ez − sin(z)]∣∣2π0

= e2π − 1.

3. Because u = 12 − 1

2 cos(2x) cosh(2y) and v = 12 sin(2x) sinh(2y), ux =

sin(2x) cosh(2y) = vy and vx = cos(2x) sinh(2y) = −uy for all x and y.Therefore, any integration between two points is path independent. Thus,

∫ π

0

sin2(z) dz = 12

∫ π0

[1 − cos(2z)] dz = 12 [z − 1

2 sin(2z)]∣∣π0

= π/2.

4. Because u = x+ 1 and v = y, ux = 1 = vy and vx = 0 = −uy for all x andy. Therefore, any integration between two points is path independent. Thus,

∫ 2i

−i(z + 1) dz =

(12z

2 + z)∣∣2i

−i = − 32 + 3i.

Section 1.6

1. ∮

|z|=1

sin6(z)

z − π/6dz = 2πi sin6(π/6) =

πi

32

2. ∮

|z|=1

sin6(z)

(z − π/6)3dz =

2πi

2!

d2[sin6(z)]

dz2

∣∣∣∣z=π/6

= πi[30 sin4(π/6) cos2(π/6) − 6 sin6(π/6)

]

= 21πi/16

3. ∮

|z|=1

dz

z(z2 + 4)=

∮

|z|=1

1/(z2 + 4)

zdz = 2πi

1

z2 + 4

∣∣∣∣z=0

=πi

2

4. ∮

|z|=1

tan(z)

zdz = 2πi tan(z)

∣∣z=0

= 0

5.∮

|z−1|=1/2

dz

(z − 1)(z − 2)=

∮

|z−1|=1/2

1/(z − 2)

z − 1dz = 2πi

1

z − 2

∣∣∣∣z=1

= −2πi

Worked Solutions 11

6.

∮

|z|=5

exp(z2)

z3dz =

2πi

2!

d2(ez2

)

dz2

∣∣∣∣z=0

= πi(4z2ez

2

+ 2ez2)∣∣∣z=0

= 2πi

7.

∮

|z−1|=1

z2 + 1

z2 − 1dz =

∮

|z−1|=1

(z2 + 1)/(z + 1)

z − 1dz = 2πi

z2 + 1

z + 1

∣∣∣∣z=1

= 2πi

8. ∮

|z|=2

z2

(z − 1)4dz =

2πi

3!

d3(z2)

dz3

∣∣∣∣z=1

= 0

9. ∮

|z|=2

z3

(z + i)3dz =

2πi

2!

d2(z3)

dz2

∣∣∣∣z=i

= 6πiz|z=i = −6π

10. ∮

|z|=1

cos(z)

z2n+1dz =

2πi

(2n)!

d2n[cos(z)]

dz2n

∣∣∣∣z=0

=2πi

(2n)!(−1)n

MATLAB code used in the complex variables project

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

% MATLAB Code for Complex Variables Project

%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

% initialize parameters

clear

n = 11; R = 1; x 0 = 2; y 0 = 0;

%

% load in values for Gaussian-Legendre quadrature

%

x gauss(1) = -0.906179845938664; A(1) = 0.236926885056189;

x gauss(2) = -0.538469310105683; A(2) = 0.478628670499366;

x gauss(3) = 0.000000000000000; A(3) = 0.568888888888889;

x gauss(4) = 0.538469310105683; A(4) = 0.478628670499366;

x gauss(5) = 0.906179845938664; A(5) = 0.236926885056189;

%

% compute n!


%

factorial = 1; radius = 1; answer ave = 0;

for j = 1:n

factorial = j * factorial; radius = R * radius;

end

%

% test out various resolutions

%

for m = 1:85

%

M = m+15; m plot(m) = M; dtheta = 2 * pi / M;

answer1 = 0; answer2 = 0;

%

% now do the integration the circle

%

for j = 1:M

%

a = (j-1)*dtheta; b = j*dtheta; h = 0.5*(b-a); ave = 0.5*(b+a);

%

for k = 1:5

theta = ave + h * x gauss(k);

x = x 0 + R * cos(theta);

y = y 0 + R * sin(theta);

z = x + i*y;

f = 8 * z / (z * z+4);

u = real(f);

v = imag(f);

cosine = cos(n * theta);

sine = sin(n * theta);

integrand1 = u * cosine + v * sine;

integrand2 = v * cosine - u * sine;

answer1 = answer1 + h * A(k) * integrand1;

answer2 = answer2 + h * A(k) * integrand2;

end

end

%

% nth derivative = answer1 + i * answer2

%

format long

answer1 = factorial * answer1 / (2 * pi * radius);

answer2 = factorial * answer2 / (2 * pi * radius);

derivative plot(m) = answer1;

answer ave = answer ave + derivative plot(m);

end

%

Worked Solutions 13

% plot difference in the answer as a function of order of scheme

%

answer ave = answer ave / m;

plot(m plot,derivative plot-answer ave)

xlabel(’number of Gaussian intervals’,’Fontsize’,20)

ylabel(’the eleventh derivative of f(z)’,’Fontsize’,20)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Section 1.7

1.1

(1 − z)2=

∞∑

n=0

f (n)(0)

n!zn

Because f (n)(z) = (n+ 1)!/(1 − z)n+2, f (n)(0) = (n+ 1)! Then

1

(1 − z)2=

∞∑

n=0

(n+ 1)!

n!zn =

∞∑

n=0

(n+ 1)zn.

2.f(z) = e(z − 1)e(z−1)

= e(z − 1)

[1 +

z − 1

1!+

(z − 1)2

2!+

(z − 1)3

3!+ · · ·

]

= e(z − 1) + e(z − 1)2 + 12e(z − 1)3 + 1

6e(z − 1)4 + · · ·

3.

f(z) = z10

(1 − 1

z+

1

2z2− 1

6z3+ · · · − 1

11!z11+ · · ·

)

= z10 − z9 + 12z

8 − 16z

7 + · · · − 111!z + · · ·

We have an essential singularity and the residue equals −1/11!

4.f(z) = z−3 sin2(z) = 1

2z−3[1 − cos(2z)]

= 12z

−3(1 − 1 + (2z)2

2! − (2z)4

4! + (2z)6

6! − · · ·)

=1

z− 23z

4!+

25z3

6!− · · ·

We have a simple pole with a residue that equals 1.

5.

f(z) =cosh(z) − 1

z2=

1

z2

(1 +

z2

2!+z4

4!+z6

6!+ · · · − 1

)

=1

2!+z2

4!+z4

6!+ · · ·


We have a removable singularity where the value of the residue equals zero.

6.

f(z) =z + 2 − 2

(z + 2)2= − 2

(z + 2)2+

1

z + 2

We have a second-order pole where the residue equals one.

7.ez + 1

e−z − 1=

2 + z + 12z

2 + 16z

3 + · · ·1 − z + 1

2z2 − 1

6z3 + · · · − 1

= − 1z (2 + z + 1

2z2 + 1

6z3 + · · ·)(1 + 1

2z + 112z

2 + · · ·)= − 2

z − 2 − 76z − 1

2z2 − · · ·

We have a simple pole and the residue equals −2.

8.

eiz

z2 + b2=

ei(z−bi)e−b

(z − bi)[2bi+ (z − bi)]

=e−b

2bi

1

z − bi

[1 − z − bi

2bi+

(z − bi)2

(2bi)2− · · ·

]

×[1 + i(z − bi) + 1

2 i2(z − bi)2 + · · ·

]

=e−b

2bi

1

z − bi+e−b

4b2(1 + 2b) − e−b

8b3i(1 + 2b+ 2b2)(z − bi) + · · ·

We have a simple pole and the residue equals e−b/(2bi).

9.

f(z) =1

[2 + (z − 2)](z − 2)=

1

2

1

z − 2

[1 − z − 2

2+

(z − 2)2

4− · · ·

]

=1

2

1

z − 2− 1

4+

1

8(z − 2) − · · ·

We have a simple pole and the residue equals 1/2.

10.

f(z) =1

z4

(1 + z2 +

z4

2!+z6

3!+ · · ·

)=

1

z4+

1

z2+

1

2+z2

6+ · · ·

We have a fourth-order pole and the residue equals zero.

Section 1.8

Worked Solutions 15

1. ∮

|z|=1

z + 1

z4 − 2z3dz = 2πi Res

(z + 1

z4 − 2z3; 0

)= −3πi

4,

because

Res

(z + 1

z4 − 2z3; 0

)= limz→0

1

2!

d2

dz2

[z3 z + 1

z3(z − 2)

]

= limz→0

[− 1

(z − 2)2+

z + 1

(z − 2)3

]= −3

8.

2.

∮

|z|=1

(z + 4)3

z4 + 5z3 + 6z2dz = 2πi Res

[(z + 4)3

z4 + 5z3 + 6z2; 0

]= −16πi

9

because

Res

[(z + 4)3

z4 + 5z3 + 6z2; 0

]= limz→0

d

dz

[z2 (z + 4)3

z2(z2 + 5z + 6)

]

= limz→0

[3(z + 4)2

z2 + 5z + 6− (z + 4)3(2z + 5)

(z2 + 5z + 6)2

]= −8

9.

3. Because

1

1 − ez=

1

1 − 1 − z − 12z

2 − · · · = −1

z

(1

1 + 12z + · · ·

)

= −1

z

(1 − 1

2z − · · ·

),

we have a simple pole at z = 0 with Res = −1. Therefore,

∮

|z|=1

1

1 − ezdz = −2πi.

4. ∮

|z|=2

z2 − 4

(z − 1)4dz = 2πi Res

[z2 − 4

(z − 1)4; 1

]= 0

because

Res

[z2 − 4

(z − 1)4; 1

]=

1

3!limz→1

d3

dz3

[(z − 1)4

z2 − 4

(z − 1)4

]= 0.


5. The singularities are located where z4 = 1 or zn = ±1 and ±i. Thecorresponding residues are

Res

(z3

z4 − 1; 1

)= limz→1

(z − 1)z3

z4 − 1=(

limz→1

z3)(

limz→1

1

4z3

)=

1

4,

Res

(z3

z4 − 1;−1

)= limz→−1

(z + 1)z3

z4 − 1=

(limz→−1

z3

)(limz→−1

1

4z3

)=

1

4,

Res

(z3

z4 − 1; i

)= limz→i

(z − i)z3

z4 − 1=(limz→i

z3)(

limz→i

1

4z3

)=

1

4

and

Res

(z3

z4 − 1;−i)

= limz→−i

(z + i)z3

z4 − 1=

(limz→−i

z3

)(limz→−i

1

4z3

)=

1

4.

Therefore, ∮

|z|=2

z3

z4 − 1dz = 2πi.

6. Because the Laurent expansion of the integrand is

f(z) = zn + 2zn−1 + · · · + 1

(n+ 1)!

2n+1

z+ · · · ,

we have an essential singularity and the residue equals 2n+1/(n+1)! Therefore,

∮

|z|=1

zne2/z dz = 4πi2n

(n+ 1)!.

7. Because

e1/z cos(1/z) =1

2

[e(1+i)/z + e(1−i)/z

]= 1 +

1

z+

0

z2+ · · · ,

we have an essential singularity and the residue equals one. Then

∮

|z|=1

e1/z cos(1/z) dz = 2πi.

8. ∮

|z|=2

2 + 4 cos(πz)

z(z − 1)2dz = 2πi

{Res

[2 + 4 cos(πz)

z(z − 1)2; 0

]

+ Res

[2 + 4 cos(πz)

z(z − 1)2; 1

]}= 16πi

Worked Solutions 17

because

Res

[2 + 4 cos(πz)

z(z − 1)2; 0

]= limz→0

z2 + 4 cos(πz)

z(z − 1)2= 6

and

Res

[2 + 4 cos(πz)

z(z − 1)2; 1

]= limz→1

d

dz

[(z − 1)2

2 + 4 cos(πz)

z(z − 1)2

]= 2.

Section 1.9

1. ∫ ∞

0

dx

x4 + 1=

1

2

∫ ∞

−∞

dx

x4 + 1=

1

2

∮

C

dz

z4 + 1,

where C denotes a semicircle of infinite radius in the upper half-plane. Thiscontour encloses two simple poles at z = eπi/4 and z = e3πi/4. Therefore,

∮

C

dz

z4 + 1= 2πi Res

[1

z4 + 1; eπi/4

]+ 2πi Res

[1

z4 + 1; e3πi/4

]

= 2πi limz→eπi/4

z − eπi/4

z4 + 1+ 2πi lim

z→e3πi/4

z − e3πi/4

z4 + 1

=πi

2

[e−3πi/4 + e−9πi/4

]=π√

2

2.

The final result follows by substitution into the first equation.

2. ∫ ∞

−∞

dx

(x2 + 4x+ 5)2=

∮

C

dz

(z2 + 4z + 5)2,

where C denotes a semicircle of infinite radius in the upper half-plane. Thiscontour encloses a second-order pole at z = −2 + i. Therefore,

∮

C

dz

(z2 + 4z + 5)2= 2πi Res

[1

(z2 + 4z + 5)2;−2 + i

]

= 2πi limz→−2+i

d

dz

[(z + 2 − i)2

(z + 2 − i)2(z + 2 + i)2

]

= 2πi limz=−2+i

−2

(z + 2 + i)3=π

2.


3. ∫ ∞

−∞

x dx

(x2 + 1)(x2 + 2x+ 2)=

∮

C

z dz

(z2 + 1)(z2 + 2z + 2),


where C denotes a semicircle of infinite radius in the upper half-plane. Thiscontour encloses simple poles at z = i and z = −1 + i. Therefore,

∮

C

z dz

(z2 + 1)(z2 + 2z + 2)= 2πi Res

[z

(z2 + 1)(z2 + 2z + 2); i

]

+ 2πi Res

[z

(z2 + 1)(z2 + 2z + 2);−1 + i

]

= 2πi limz→i

(z − i)z

(z − i)(z + i)(z2 + 2z + 2)

+ 2πi limz→−1+i

(z + 1 − i)z

(z2 + 1)(z + 1 − i)(z + 1 + i)

= 2πi

[i

(2i)(1 + 2i)+

−1 + i

(2i)(1 − 2i)

]= −π

5.


4. ∫ ∞

0

x2 dx

x6 + 1=

1

2

∫ ∞

−∞

x2 dx

x6 + 1=

1

2

∮

C

z2 dz

z6 + 1,

where C denotes a semicircle of infinite radius in the upper half-plane. Thiscontour encloses three simple poles at z = eπi/6, z = eπi/2 and z = e5πi/6.Therefore,

∮

C

z2 dz

z6 + 1= 2πi Res

(z2

z6 + 1; eπi/6

)+ 2πi Res

(z2

z6 + 1; eπi/2

)

+ 2πi Res

(z2

z6 + 1; e5πi/6

)

= 2πi limz→eπi/6

(z − eπi/6)z2

z6 + 1+ 2πi lim

z→eπi/2

(z − eπi/2)z2

z6 + 1

+ 2πi limz→e5πi/6

(z − e5πi/6)z2

z6 + 1

= 2πi

[1

6i+

1

−6i+

1

6i

]=π

3.


5. ∫ ∞

0

dx

(x2 + 1)2=

1

2

∫ ∞

−∞

dx

(x2 + 1)2=

1

2

∮

C

dz

(z2 + 1)2,

where C denotes a semicircle of infinite radius in the upper half-plane. This

Worked Solutions 19

contour encloses a second-order pole at z = i. Therefore,

∮

C

dz

(z2 + 1)2= 2πi Res

[1

(z2 + 1)2; i

]

= 2πi limz→i

d

dz

[(z − i)2

(z − i)2(z + i)2

]

= 2πi limz→i

−2

(z + i)3=π

2.


6.∫ ∞

0

dx

(x2 + 1)(x2 + 4)2=

1

2

∫ ∞

−∞

dx

(x2 + 1)(x2 + 4)2=

1

2

∮

C

dz

(z2 + 1)(z2 + 4)2,

where C denotes a semicircle of infinite radius in the upper half-plane. Thiscontour encloses simple poles at z = i and a second-order pole at z = 2i.Therefore,

∮

C

dz

(z2 + 1)(z2 + 4)2= 2πi Res

[1

(z2 + 1)(z2 + 4)2; i

]

+ 2πi Res

[1

(z2 + 1)(z2 + 4)2; 2i

]

= 2πi limz→i

z − i

(z + i)(z − i)(z2 + 4)2

+ 2πi limz→2i

d

dz

[(z − 2i)2

(z2 + 1)(z − 2i)2(z + 2i)2

]

=2πi

(2i)(9)+

(2πi)(−2)

(4i)3(−3)+

(2πi)(−1)(4i)

(4i)2(−3)2=

5π

144.


7. ∫ ∞

−∞

x2

(x2 + a2)(x2 + b2)2dx =

∮

C

z2

(z2 + a2)(z2 + b2)2dz,

where C denotes a semicircle of infinite radius in the upper half-plane. Thiscontour encloses a simple pole at z = ai and a second-order pole at z = bi.Therefore,

∮

C

z2

(z2 + a2)(z2 + b2)2dz = 2πi Res

[z2

(z2 + a2)(z2 + b2)2; ai

]

+ 2πi Res

[z2

(z2 + a2)(z2 + b2)2; bi

].


At z = ai,

Res

[z2

(z2 + a2)(z2 + b2)2; ai

]= limz→ai

(z − ai)

(z − ai)(z + ai)limz→ai

z2

(z2 + b2)2

=ai

2(a2 − b2)2.

At z = bi,

Res

[z2

(z2 + a2)(z2 + b2)2; bi

]= limz→bi

d

dz

[z2

(z2 + a2)(z + bi)2

]

= − i

2b(a2 − b2)− bi

2(a2 − b2)2+

i

4b(a2 − b2).

Therefore,

∫ ∞

−∞

x2

(x2 + a2)(x2 + b2)2dx = 2πi

[ai

2(a2 − b2)2− i

2b(a2 − b2)

− bi

2(a2 − b2)2+

i

4b(a2 − b2)

]

=π

2b(a+ b)2.

8. ∫ ∞

0

t2

(t2 + 1)[t2(a/h+ 1) + (a/h− 1)]dt

=1

2

∫ ∞

−∞

t2

(t2 + 1)[t2(a/h+ 1) + (a/h− 1)]dt

=1

2(1 + a/h)

∮

C

z2

(z2 + 1)(z2 + b2)dz

where b2 = [(1− h/a)/(1 + h/a)]. We have a simple pole at z = i and z = bi.Then∮

C

z2

(z2 + 1)(z2 + b2)dz = 2πi Res

[z2

(z2 + 1)(z2 + b2); i

]

+ 2πi Res

[z2

(z2 + 1)(z2 + b2); bi

]

= 2πi limz→i

(z − i)z2

(z − i)(z + i)(z2 + b2)

+ 2πi limz→bi

(z − bi)z2

(z2 + 1)(z − bi)(z + bi)

= 2πi−1

(2i)(b2 − 1)+ 2πi

−b2(2bi)(1 − b2)

=π(1 − b)

1 − b2

Worked Solutions 21

and

∮

C

z2

(z2 + 1)(z2 + b2)dz =

π

1 − [(1 − h/a)/(1 + h/a)]

(1 −

√1 − h/a

1 + h/a

).

Substitution into the first equation yields the desired result.

9. We begin by converting the real integral into a closed contour integration:

∫ π/2

0

dθ

a+ sin2(θ)=

1

4

∫ 2π

0

dθ

a+ sin2(θ)= i

∮

|z|=1

z

(z2 − 1)2 − 4az2dz,

where z = eθi. The integrand has four poles: z = ±√a ±

√1 + a. Only two

are located inside the contour:

z1 = −√a+

√1 + a and z2 =

√a−

√1 + a.

The corresponding residues are

Res

[z

(z2 − 1)2 − 4az2; z1

]= limz→z1

(z − z1)z

(z2 − 1)2 − 4az2= − 1

8√a+ a2

and

Res

[z

(z2 − 1)2 − 4az2; z2

]= limz→z2

(z − z2)z

(z2 − 1)2 − 4az2; z1 = − 1

8√a+ a2

.

Employing the residue theorem and substituting into the top line, we havethat ∫ π/2

0

dθ

a+ sin2(θ)=

π

2√a+ a2

.

10. We begin by converting the real integral into a closed contour integration:

∫ π/2

0

dθ

a2 cos2(θ) + b2 sin2(θ)=

1

4

∫ 2π

0

dθ

a2 cos2(θ) + b2 sin2(θ)

= −i∮

|z|=1

z

a2(z2 + 1)2 − b2(z2 − 1)2dz,

where z = eθi. The integrand has four simple poles located at

z2+ =

b+ a

b− a, and z2

− =b− a

b+ a.

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