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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.
4 Advanced Engineering Mathematics with MATLAB
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
6 Advanced Engineering Mathematics with MATLAB
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,
8 Advanced Engineering Mathematics with MATLAB
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) =
10 Advanced Engineering Mathematics with MATLAB
−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!
12 Advanced Engineering Mathematics with MATLAB
%
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!+ · · ·
14 Advanced Engineering Mathematics with MATLAB
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.
16 Advanced Engineering Mathematics with MATLAB
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.
The final result follows by substitution into the first equation.
3. ∫ ∞
−∞
x dx
(x2 + 1)(x2 + 2x+ 2)=
∮
C
z dz
(z2 + 1)(z2 + 2z + 2),
18 Advanced Engineering Mathematics with MATLAB
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.
The final result follows by substitution into the first equation.
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.
The final result follows by substitution into the first equation.
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.
The final result follows by substitution into the first equation.
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.
The final result follows by substitution into the first equation.
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
].
20 Advanced Engineering Mathematics with MATLAB
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.