INSTRUCTOR’S SOLUTIONS MANUAL
1.
APPENDIX I. (PAGE A-10)
APPENDICES
15.
Appendix I. Complex Numbers
(page A-10)
16.
z = −5 + 2i,
Re(z) = −5,
z = −5 + 2i
Im(z) = 2
y
z-plane
z = −6
17.
z = 4−i
x
z = −πi
4π
4π
+ 3i sin
5
5
4π
|z| = 3, Arg (z) =
5
π
3π
and Arg (w) = , then
If Arg (z) =
4
2
3π
π
5π
arg (zw) =
+
=
, so
4
2
4
5π
−3π
Arg (zw) =
− 2π =
.
4
4
π
5π
and Arg (w) = , then
If Arg (z) = −
6
4
5π
π
13π
arg (z/w) = −
− =−
, so
6
4
12
13π
11π
Arg (z/w) = −
+ 2π =
.
12
12
z = 3 cos
18. |z| = 2, arg (z) = π ⇒ z = 2(cos π + i sin π ) = −2
Fig. .1
2.
z = 4 − i,
Re(z) = 4,
Im(z) = −1
3.
z = −πi,
Re(z) = 0,
Im(z) = −π
4.
z = −6,
5.
6.
Re(z) = −6, Im(z) = 0
√
z = −1 + i, |z| = 2, Arg (z) = 3π/4
√
z = 2 (cos(3π/4) + i sin(3π/4))
z = −2, |z| = 2, Arg (z) = π
z = 2(cos π + i sin π )
7.
z = 3i, |z| = 3, Arg (z) = π/2
z = 3(cos(π/2) + i sin(π/2))
8.
z = −5i, |z| = 5, Arg (z) = −π/2
z = 5(cos(−π/2) + i sin(−π/2))
√
z = 1 + 2i, |z| = 5, θ = Arg (z) = tan−1 2
√
z = 5(cos θ + i sin θ )
√
z = −2 + i, |z| = 5, θ = Arg (z) = π − tan−1 (1/2)
√
z = 5(cos θ + i sin θ )
9.
10.
θ = Arg (z) = −π + tan−1 (4/3)
11.
z = −3 − 4i, |z| = 5,
z = 5(cos θ + i sin θ )
12.
z = 3 − 4i, |z| = 5, θ = Arg (z) = −tan−1 (4/3)
z = 5(cos θ + i sin θ )
√
z = 3 − i, |z| = 2, Arg (z) = −π/6
z = 2(cos(−π/6) + i sin(−π/6))
√
√
z = − 3 − 3i, |z| = 2 3, Arg (z) = −2π/3
√
z = 2 3(cos(−2π/3) + i sin(−2π/3))
13.
14.
19.
20.
21.
|z| = 5, θ = arg (z) = π ⇒ sin θ = 3/5, cos θ = 4/5
z = 4 + 3i
3π
3π
3π
⇒ z = cos
+ i sin
|z| = 1, arg (z) =
4
4
4
1
1
⇒ z = −√ + √ i
2
2
π
π
π
⇒ z = π cos + i sin
|z| = π, arg (z) =
6
6
6
√
π 3 π
⇒ z=
+ i
2
2
22. |z| = 0 ⇒ z = 0 for any value of arg (z)
23.
π
1
π
1
π
, arg (z) = −
⇒ z=
cos − i sin
2
3
2
3
3
√
3
1
i
⇒ z= −
4
4
|z| =
24. 5 + 3i = 5 − 3i
25. −3 − 5i = −3 + 5i
26. 4i = −4i
27. 2 − i = 2 + i
28. |z| = 2 represents all points on the circle of radius 2
centred at the origin.
29. |z| ≤ 2 represents all points in the closed disk of radius 2
centred at the origin.
30. |z − 2i | ≤ 3 represents all points in the closed disk of
radius 3 centred at the point 2i .
31. |z − 3 + 4i | ≤ 5 represents all points in the closed disk of
radius 5 centred at the point 3 − 4i .
32. arg (z) = π/3 represents all points on the ray from the
origin in the first quadrant, making angle 60◦ with the
positive direction of the real axis.
687
APPENDIX I. (PAGE A-10)
33. π ≤ arg (z) ≤ 7π/4 represents the closed wedge-shaped
R. A. ADAMS: CALCULUS
48.
region in the third and fourth quadrants bounded by the
ray from the origin to −∞ on the real axis and the ray
from the origin making angle −45◦ with the positive
direction of the real axis.
cos(3θ ) + i sin(3θ ) = (cos θ + i sin θ )3
= cos3 θ + 3i cos2 θ sin θ − 3 cos θ sin2 θ − i sin3 θ
Thus
cos(3θ ) = cos3 θ − 3 cos θ sin2 θ = 4 cos3 θ − 3 cos θ
34. (2 + 5i ) + (3 − i ) = 5 + 4i
sin(3θ ) = 3 cos2 θ sin θ − sin3 θ = 3 sin θ − 4 sin3 θ.
35. i − (3 − 2i ) + (7 − 3i ) = −3 + 7 + i + 2i − 3i = 4
36. (4 + i )(4 − i ) = 16 − i 2 = 17
49.
37. (1 + i )(2 − 3i ) = 2 + 2i − 3i − 3i 2 = 5 − i
38.
(a + bi )(2a − bi) = (a + bi )(2a + bi ) = 2a 2 − b2 + 3abi
39. (2 + i )3 = 8 + 12i + 6i 2 + i 3 = 2 + 11i
40.
(2 − i )2
3 − 4i
2−i
=
=
2
2+i
4−i
5
41.
(1 + 3i )(2 + i )
1 + 3i
−1 + 7i
=
=
2−i
4 − i2
5
42.
1+i
(1 + i )(−3 − 2i )
−1 − 5i
1+i
=
=
=
i (2 + 3i )
−3 + 2i
9+4
13
43.
8+i
(1 + 2i )(2 − 3i )
=
=1
(2 − i )(3 + 2i )
8+i
50.
form cos θ +i sin θ where θ = π/3, θ = π , and θ = 5π/3.
Thus they are
√
1
3
+i
,
2
2
real, then
52.
45. Using the fact that |zw| = |z||w|, we have
z
w
46.
47.
=
zw
|w|2
−1,
√
1
3
−i
.
2
2
3π
3π
+ i sin
The three cube roots of −8i = 8 cos
2
2
are of the form 2(cos θ + i sin θ ) where θ = π/2,
θ = 7π/6, and
θ = 11π/6. Thus they are
2i,
zw
zw
z
= .
=
=
ww
w
|w|2
√
√ π
π
z = 3 + i 3 = 2 3 cos + i sin
6
6 √
2π
2π
+ i sin
w = −1 + i 3 = 2 cos
3
3
√
5π
5π
zw = 4 3 cos
+ i sin
6
6
√
√
z
−π
−π
= 3 cos
+ i sin
= −i 3
w
2
2
√ 3π
3π
+ i sin
z = −1 + i = 2 cos
4
4
π
π
w = 3i = 3 cos + i sin
2
2 √
5π
5π
zw = 3 2 cos
+ i sin
= −3 − 3i
4
4
√ z
π 1 1
2
π
=
cos + i sin
= + i
w
3
4
4
3 3
688
b) z = −2/z can be rewritten |z|2 = zz = −2, which
has no solutions since the square of |z| is nonnegative for all complex z.
√
If z = w =√−1, then zw = 1, so zw = 1. √But if we
√
use
and the same value for w, then
√ √ z = 2 −1 = i √
z w = i = −1 = zw.
51. The three cube roots of −1 = cos π + i sin π are of the
44. If z = x + yi and w = u + vi , where x, y, u, and v are
z + w = x + u + (y + v)i
= x + u − (y + v)i = x − yi + u − vi = z + w.
a) z = 2/z can be rewritten |z|2 = zz = 2, so is √
satisfied by all numbers z on the circle of radius 2
centred at the origin.
√
− 3 − i,
√
3 − i.
√
3π
3π
+ i sin
are of the
roots of −1 + i = 2 cos
4
4
1/6
form 2 (cos θ + i sin θ ) where θ = π/4, θ = 11π/12,
and θ = 19π/12.
53. The three cube
54. The four
√ fourth roots of 4 = 4(cos 0 + i sin 0) are of the
55.
form 2(cos θ + i sin θ ) where
θ = 0,√θ = π/2, π√
, and
√ √
θ = 3π/2. Thus they are 2, i 2, − 2, and −i 2.
√
The equation z 4 + 1 − i 3 = 0 hassolutions that are the
√
2π
2π
+ i sin
.
four fourth roots of −1 + i 3 = 2 cos
3
3
Thus they are of the form 21/4 (cos θ + i sin θ ), where
θ = π/6, 2π/3, 7π/6, and 5π/3. They are the complex
numbers
√
√ 3 1
3
1/4
1/4 1
+ i , ±2
−
i .
±2
2
2
2
2
INSTRUCTOR’S SOLUTIONS MANUAL
APPENDIX II. (PAGE A-19)
56. The equation z 5 + a 5 = 0 (a > 0) has solutions that are
8. The function w = z 2 = x 2 − y 2 + 2x yi transforms the
57. The n nth roots of unity are
9. The function w = z 2 = x 2 − y 2 + 2x yi transforms the
−a 5
the five fifth roots of
= a (cos π + i sin π ); they are
of the form a(cos θ + i sin θ ), where θ = π/5, 3π/5, π ,
7π/5, and 9π/5.
line x = 1 to u = 1 − y 2 , v = 2y, which is the parabola
v 2 = 4 − 4u with vertex at w = 1, opening to the left.
line y = 1 to u = x 2 − 1, v = 2x, which is the parabola
v 2 = 4u + 4 with vertex at w = −1 and opening to the
right.
ω1 = 1
2π
2π
+ i sin
n
n
4π
4π
ω3 = cos
+ i sin
= ω22
n
n
6π
6π
+ i sin
= ω23
ω4 = cos
n
n
..
.
2(n − 1)π
2(n − 1)π
ωn = cos
+ i sin
= ω2n−1 .
n
n
ω2 = cos
10. The function w = 1/z = (x − yi )/(x 2 + y 2 ) transforms
the line x = 1 to the curve given parametrically by
u=
Appendix II. Complex Functions
(page A-19)
u 2 + v2 =
the horizontal strip −∞ < x < ∞, π/4 ≤ y ≤ π/2 to
the wedge π/4 ≤ arg (w) ≤ π/2, or, equivalently, u ≥ 0,
v ≥ u.
12. The function w = ei z = e−y (cos x + i sin x) transforms
the vertical half-strip 0 < x < π/2, 0 < y < ∞ to the
first-quadrant part of the unit open disk |w| = e−y < 1,
0 < arg (w) = x < π/2, that is u > 0, v > 0, u 2 + v 2 < 1.
13.
line u − v = 1.
3. The function w = z 2 transforms the closed annular sector
14.
circular disk 0 ≤ |z| ≤ 2, 0 ≤ arg (z) ≤ π/2 to the closed
three-quarter disk 0 ≤ |w| ≤ 8, 0 ≤ arg (w) ≤ 3π/2.
5. The function w = 1/z = z/|z|2 transforms the closed
7.
forms the wedge π/4 ≤ arg (z) ≤ π/3 to the wedge
−π/4 ≤ arg (z) ≤ −π/6.
√
The function w = z transforms the ray arg (z) = −π/3
(that is, Arg (z) = 5π/3) to the ray arg (w) = 5π/6.
f (z) = z 3 = (x + yi )3 = x 3 − 3x y 2 + (3x 2 y − y 3 )i
v = 3x 2 y − y 3
u = x 3 − 3x y 2 ,
∂u
∂v
∂v
∂u
= 3(x 2 − y 2 ) =
,
= −6x y = −
∂x
∂y
∂y
∂x
∂v
∂u
+i
= 3(x 2 − y 2 + 2x yi ) = 3z 2 .
f (z) =
∂x
∂x
4. The function w = z 3 transforms the closed quarter-
6. The function w = −i z rotates the z-plane −90◦ , so trans-
f (z) = z 2 = (x + yi )2 = x 2 − y 2 + 2x yi
v = 2x y
u = x 2 − y 2,
∂u
∂v
∂u
∂v
= 2x =
,
= −2y = −
∂x
∂y
∂y
∂x
∂v
∂u
+i
= 2x + 2yi = 2z.
f (z) =
∂x
∂x
2. The function w = z transforms the line x + y = 1 to the
quarter-circular disk 0 ≤ |z| ≤ 2, 0 ≤ arg (z) ≤ π/2
to the closed region lying on or outside the circle
|w| = 1/2 and in the fourth quadrant, that is, having
−π/2 ≤ arg (w) ≤ 0.
1 + y2
= u,
(1 + y 2 )2
11. The function w = e z = e x cos y + i e x sin y transforms
0 ≤ x ≤ 1, 0 ≤ y ≤ 2 to the closed rectangle 0 ≤ u ≤ 1,
−2 ≤ v ≤ 0.
1 ≤ |z| ≤ 2, π/2 ≤ arg (z) ≤ 3π/4 to the closed annular
sector 1 ≤ |w| ≤ 4, π ≤ arg (w) ≤ 3π/2.
−y
.
1 + y2
with centre w = 1/2 and radius 1/2.
In Solutions 1–12, z = x + yi and w = u + vi , where x,
y, u, and v are real.
1. The function w = z transforms the closed rectangle
v=
This curve is, in fact, a circle,
Hence
ω1 + ω2 + ω3 + · · · + ωn = 1 + ω2 + ω22 + · · · + ω2n−1
1 − ω2n
0
=
=
= 0.
1 − ω2
1 − ω2
1
,
1 + y2
15.
x − yi
1
= 2
z
x + y2
x
−y
u= 2
,
v= 2
2
x +y
x + y2
2
2
y −x
∂u
−2x y
∂v
∂v
∂u
= 2
,
= 2
=
=−
∂x
(x + y 2 )2
∂y
∂y
(x + y 2 )2
∂x
∂v
−(x 2 − y 2 ) + 2x yi
∂u
−(z)2
1
+i
=
=
= − 2.
f (z) =
2
2
2
2
∂x
∂x
(x + y )
(zz)
z
f (z) =
689