1.5 Roots of Complex Numbers

An nth root of a complex number z is a complex number ω that satisfies the equation
ωn = z

Since z = r(cos θ + i sin θ) and
$$ (\sqrt[n]{r}\ (\text{cos}\ \frac{\theta}{n} + i \ \text{sin}\ \frac{\theta}{n})) = (\sqrt[n]{r})^n (\text{cos}\ n (\frac{\theta}{n}) + i\ \text{sin}\ n (\frac{\theta}{n})) $$ $$ = r (\text{cos}\ \theta + i\ \text{sin}\ \theta) = z $$ We have an nth root of a complex number z as
$$ \sqrt[n]{r} (\text{cos}\ \frac{\theta}{n} + i\ \text{sin}\ \frac{\theta}{n}). $$ Moreover, since z can also be represented as

z = r(cos (θ + 2kπ) + i sin (θ + 2kπ))

for every integer k, we have
$$ (\sqrt[n]{r} (\text{cos}\ \frac{\theta + 2k\pi}{n} + i\ \text{sin}\ \frac{\theta + 2k\pi}{n}))^n = (\sqrt[n]{r})^n (\text{cos}\ n\ \frac{\theta + 2k\pi}{n} + i\ \text{sin}\ n\ \frac{\theta + 2k\pi}{n}) $$ $$ = r (\text{cos}\ (\theta + 2k\pi) + i\ \text{sin}\ \theta + 2k\pi )) $$ $$ = z$$ So nth roots of z are
$$ \sqrt[n]{r} (\text{cos}\ \frac{\theta + 2k\pi}{n} + i\ sin\ \frac{\theta + 2k\pi}{n}) $$ for every integer k

But for k = n, we have
cos θ + 2kπ ⁄ n + i sin θ + 2kπ ⁄ n = cos θ + 2nπ ⁄ n + i sin θ + 2nπ ⁄ n
=cos (θ ⁄ n + 2π) + i sin( θ ⁄ n + 2π)
= cos θ ⁄ n + i sin θ ⁄ n
so that root for k = n is the same as the root for k = 0. The same is also true for k > n as the root for k = n + 1 is same as the root for k = 1, and so on. Therefore the roots of z, denoted by ωk of z are
$$ \omega_k = \sqrt[n]{r} (\text{cos}\ \frac{\theta + 2k\pi}{n} + i \ \text{sin}\ \frac{\theta + 2k\pi}{n}, \ \ \ \ \ k = 1,2,..., n-1. $$
Example 10.
Find the cube roots of z = -2 - 2i.

Solution

z = -2 - 2i = (-2, -2)
r = √ 4 + 4 = 2 √ 2 ,     cos θ = - 1 ⁄ √ 2 ,     sin θ = - 1 ⁄ √ 2
θ = - 3π ⁄ 4
z = 2 √ 2 (cos (- 3π ⁄ 4) + i sin ( - 3π ⁄ 4))

x y θ r z = (-2, -2) O
Then we have
ωk = ∛ 2 √ 2 (cos - 3π ⁄ 4 + 2kπ ⁄ 3 + i sin - 3π ⁄ 4 + 2kπ ⁄ 3),       k = 0, 1, 2.
Therefore the cube roots are
ω0 = √ 2 (cos(- 3π ⁄ 12) + i sin (- 3π ⁄ 12))
= √ 2 ( √ 2 ⁄ 2 - i √ 2 ⁄ 2)
= 1 - i
ω1 = √ 2 (cos 5π ⁄ 12 + i sin 5π ⁄ 12)
= √ 2 ( √ 6 - √ 2 ⁄ 4 + i √ 6 + √ 2 ⁄ 4)
= -1 + √ 3 ⁄ 2 + 1 + √ 3 ⁄2 i
ω2 = √ 2 (cos( 13π ⁄ 12) + i sin ( 13π ⁄ 12))
= √ 2 (- √ 6 + √ 2 ⁄ 4 + i - √ 6 + √ 2 ⁄ 4)
= - 1 + √ 3 ⁄ 2 + 1 - √ 3 ⁄2 i

Example 11.

Solve z6 = 1.

Solution

1 = (1, 0) = 1(cos 0 + i sin 0) = 1 (cos (0 + 2kπ) + i sin(0 + 2kπ))
z6 = 1
26 = 1 (cos (0 + 2kπ) + i sin(0 + 2kπ))
z =1 (cos (0 + 2kπ) + i sin(0 + 2kπ))1 ⁄ 6
z = 6√ 1 (cos 0 + 2kπ ⁄ 6 + i sin 0 + 2kπ ⁄ 6),         k = 0, 1, 2, ....., 5

When k = 0, z = cos 0 + i sin 0 = 1.

When k = 1, z = cos π ⁄ 3 + i sin π ⁄ 3 = 1 ⁄ 2 + i √ 3 ⁄ 2 .

When k = 2, z = cos 2π ⁄ 3 + i sin 2π ⁄ 3 = - 1 ⁄ 2 + i √ 3 ⁄ 2 .

When k = 3, z = cos π + i sin π = 1.

When k = 4, z = cos 4π ⁄ 3 + i sin 4π ⁄ 3 = - 1 ⁄ 2 - i √ 3 ⁄ 2 .

When k = 5, z = cos 5π ⁄ 3 + i sin 5π ⁄ 3 = 1 ⁄ 2 - i √ 3 ⁄ 2 .


Exercise 1.5
  1. Find the square roots of the following complex numbers.

    2. (a) 1 + √ 3 i
    (b) i
    (c) - √ 3 + i
    (d) -1 - √ 3 i
    (e) - i
    (f) √ 3 - i

  2. Find the cube roots of the following complex numbers.

    3. (a) 1 + i
    (b) i
    (c) -1 + i
    (d) -1 - i
    (e) -i
    (f) 1 - i

  3. Solve the following equations.

    4. (a) z4 = -i
    (b) z4 = -1
    (c) z4 = -8 - 8 √ 3 i
    (d) z6 = -1


Answers

1. (a)     1 + √ 3 i = (1, √ 3 )

r = √ x2 + y2 = √ 1 +3 = 2
cos θ = x⁄r = 1⁄2
sin θ = y⁄r = √ 3 ⁄2
θ = π⁄3
z = r (cos θ + i sin θ)
  = 2(cos π⁄3 + i sin π⁄3)
ωk = √ r (cos θ + 2kπ⁄2 + i sin θ + 2kπ⁄2)
  = √ 2 (cos π⁄3 + 2kπ⁄2 + i sin π⁄3 + 2kπ⁄2)
x y θ r z = (1 , √3) O
The square roots are
ω0 = √ 2 (cos π⁄3 + 2(0)π⁄2 + i sin π⁄3 + 2(0)π⁄2)
    = √ 2 (cos π⁄6 + i sin π⁄6)
    = √ 2 (√ 3 ⁄2 + 1 ⁄2 i)
    = √ 6 ⁄2 + √ 2 ⁄2 i
ω1 = √ 2 (cos π⁄3 + 2(1)π⁄2 + i sin π⁄3 + 2(1)π⁄2)
    = √ 2 (cos π + 6π⁄3 ⁄2 + i sin π + 6π⁄3 ⁄2)
    = √ 2 (cos 7π⁄6 + i sin 7π⁄6)
    = √ 2 (cos (6+1)π⁄6 + i sin (6+1)π⁄6)
    = √ 2 [cos ( 6π⁄6 + π⁄6) + i sin ( 6π⁄6 + π⁄6)]
    = √ 2 [cos (π + π⁄6) + i sin (π + π⁄6)]
    = √ 2 [cos (π) cos ( π⁄6) - sin (π) sin π⁄6) + i (sin (π) cos π⁄6 + cos (π) sin π⁄6)]
    = √ 2 [(-1 . √ 3 ⁄2   -   0 . 1⁄2)   +   (0. √ 3 ⁄2   +   (-1 . 1⁄2 )i]
    = √ 2 [(- √ 3 ⁄2 - 0) + (0 - 1⁄2 )i]
    = √ 2 (- √ 3 ⁄2 - 1⁄2 i)
    = - √ 6 ⁄2 - √ 2 ⁄2 i

(b)     i = (0, 1)

r = √ x2 + y2 = √ 1 = 1
cos θ = x ⁄r = 0⁄1 = 0
sin θ = y⁄r = 1⁄1 = 1
θ = π⁄2
x y θ r z = (0, 1) O
z = r (cos θ + i sin θ)
  = 1 (cos π⁄2 + i sin π⁄2)
ωk = √ r (cos π⁄2 + 2kπ⁄2 + i sin π⁄2 + 2kπ⁄2)
The square roots are
ω0 = √ 1 (cos π⁄2 + 2(0)π⁄2 + i sin π⁄2 + 2(0)π⁄2)
    = 1 (cos π⁄2 + 0⁄2 + i sin π⁄2 + 0⁄2)
    = cos π⁄4 + i sin π⁄4
    = 1⁄ √ 2 + 1⁄ √ 2 i
    = √ 2 ⁄2 + √ 2 ⁄2 i

ω1 = √ 1 (cos π⁄2 + 2(1)π⁄2 + i sin π⁄2 + 2(1)π⁄2)
    = 1 (cos π + 4π⁄2 ⁄2 + i sin π + 4π⁄2 ⁄2)
    = cos 5π⁄4 + i sin 5π⁄4
    = cos (1 + 4)π⁄4 + i sin (1 + 4)π⁄4
    = cos ( π⁄4 + 4π⁄4) + i sin ( π⁄4 + 4π⁄4)
    = cos ( π⁄4 + π) + i sin ( π⁄4 + π)
    = (cos π⁄4 . cos π - sin π⁄4 . sin π) + i (sin π⁄4 . cos π + cos π⁄4 . sin π)
    = [ 1⁄ √ 2 . (-1)   -   1⁄ √ 2 . (0) ]   +   [1⁄ √ 2 . (-1)   +   1⁄ √ 2 . (0) ]i
    = (- 1⁄ √ 2 - 0) + (- 1⁄ √ 2 + 0) i
    = - 1⁄ √ 2 - 1⁄ √ 2 i
    = - √ 2 ⁄2 - √ 2 ⁄2 i

(c)     - √ 3 + i = (- √ 3 , 1)

r = √ x2 + y2 = √ 3 + 1 = 2
cos θ = x⁄r = - √ 3 ⁄2
sin θ = y⁄r = 1⁄2
θ = π - π⁄6 = 5π⁄6
x y θ r z = (-√3, 1) O
z = r (cos θ + i sin θ)
  = 2 (cos 5π⁄6 + i sin 5π⁄6)
ωk = √ r (cos θ + 2kπ⁄2 + i sin θ + 2kπ⁄2)
    = √ 2 (cos 5π⁄6 + 2kπ⁄2 + i sin 5π⁄6 + 2kπ⁄2)
The square roots are
ω0 = √ 2 (cos 5π⁄6 + 2(0)π⁄2 + i sin 5π⁄6 + 2(0)π⁄2)
    = √ 2 (cos 5π⁄12 + i sin 5π⁄12)
    = √ 2 [ cos (3 + 2)π⁄12 + i sin (3 + 2)π⁄12]
    = √ 2 [ cos (3π⁄12 + 2π⁄12) + i sin (3π⁄12 + 2π⁄12)]
    = √ 2 [cos (π⁄4 + π⁄6) + i sin (π⁄4 + π⁄6)]
    = √ 2 [(cos π⁄4.cos π⁄6 - sin π⁄4.sin π⁄6) + i (sin π⁄4.cos π⁄6 + cos π⁄4.sin π⁄6)]
    = √ 2 [(1⁄ √ 2 . √ 3 ⁄2   -   1⁄ √ 2 . 1⁄2)   +   i (1⁄ √ 2 . √ 3 ⁄2   +   1⁄ √ 2 .1⁄2)]
    = √ 2 [( √ 3 ⁄2 √ 2 - 1⁄2 √ 2 ) + ( √ 3 ⁄2 √ 2 + 1⁄2 √ 2 ) i]
    = √ 2 [ √ 3 - 1 ⁄2 √ 2 + √ 3 + 1 ⁄2 √ 2 ) i]
    = √ 3 - 1⁄2 + √ 3 + 1 ⁄2 i
ω1 = √ 2 [cos 5π⁄6 + 2(1)π⁄2 + i sin 5π⁄6 + 2(1)π⁄2]
    = √ 2 [cos 5π + 12π⁄6 ⁄2 + i sin 5π + 12π⁄6 ⁄2]
    = √ 2 [ cos 17π⁄12 + i sin 17π⁄12]
    = √ 2 [ cos (8 + 9)π⁄12 + i sin (8 + 9)π⁄12]
    = √ 2 [ cos (8π⁄12 + 9π⁄12) + i sin (8π⁄12 + 9π⁄12)]
    = √ 2 [ cos (2π⁄3 + 3π⁄4) + i sin (2π⁄3 + 3π⁄4)]
    = √ 2 [(cos 2π⁄3.cos 3π⁄4 - sin 2π⁄3.sin 3π⁄4) + i (sin 2π⁄3.cos 3π⁄4 + cos 2π⁄3.sin 3π⁄4)]
    = √ 2 [(-1⁄2. -1⁄ √ 2 - √ 3 ⁄2. 1⁄ √ 2 ) + i ( √ 3 ⁄2. -1⁄ √ 2 + -1⁄2. 1⁄ √ 2 )]
    = √ 2 [1 - √ 3 ⁄2 √ 2 - ( √ 3 + 1 ⁄2 √ 2 ) i]
    = 1 - √ 3 ⁄2 - √ 3 + 1 ⁄2 i

(d)     = -1 - √ 3 i   = (-1, - √ 3 )

r = √ x2 + y2 = √ 1 + 3   = 2
cos θ = x⁄r = -1⁄2
sin θ = y⁄r = - √ 3 ⁄2
θ = - (π - π⁄3) = - 2π⁄3
x y θ r z = (-1, -√3) O
z = r (cos θ + i sin θ)
  = 2 (cos -2π⁄3 + i sin -2π⁄3)
ωk = √ r (cos θ + 2kπ⁄2 + i sin θ + 2kπ⁄2)
    = √ 2 (cos -2π⁄3 + 2kπ⁄2 + i sin -2π⁄3 + 2kπ⁄2)
The square roots are
ω0 = √ 2 [cos -2π⁄3 + 2(0)π⁄2 + i sin -2π⁄3 + 2(0)π⁄2]
    = √ 2 (cos -2π⁄3 + 0⁄2 + i sin -2π⁄3 + 0⁄2)
    = √ 2 [cos -2π⁄6 + i sin -2π⁄6]
    = √ 2 [cos -π⁄3 + i sin -π⁄3]
    = √ 2 [1⁄2 + (- √ 3 ⁄2)i ]
    = √ 2 [1⁄2 - √ 3 ⁄2 i]
    = √ 2 ⁄2 - √ 6 ⁄2 i

ω1 = √ 2 [cos -2π⁄3 + 2(1)π⁄2 + i sin -2π⁄3 + 2(1)π⁄2]
    = √ 2 [cos -2π + 6π⁄3 ⁄2 + i sin -2π + 6π⁄3 ⁄2]
    = √ 2 [cos 4π⁄6 + i sin 4π⁄6]
    = √ 2 [cos 2π⁄3 + i sin 2π⁄3]
    = √ 2 [-1⁄2 + √ 3 ⁄2i ]
    = - √ 2 ⁄2 + √ 6 ⁄2 i

(e)     -i  = (0, -1)

r = √ x2 + y2   = √ 1   = 1
cos θ = x⁄r = 0⁄1 = 0
sin θ = y⁄r = -1⁄1 = -1
θ = - π⁄2
x y θ r z = (0, -1) O

z = r (cos θ + i sin θ)
  = 1 (cos -π⁄2 + i sin -π⁄2)
ωk = √ r (cos θ + 2kπ⁄2 + i sin θ + 2kπ⁄2)
    = √ 1 (cos -π⁄2 + 2kπ⁄2 + i sin -π⁄2 + 2kπ⁄2)
The square roots are
ω0 = √ 1 [cos -π⁄2 + 2(0)π⁄2 + i sin -π⁄2 + 2(0)π⁄2]
= 1 (cos -π⁄4 + i sin -π⁄4)
    = 1⁄ √ 2 + (-1⁄ √ 2 ) i
    = √ 2 ⁄2 - √ 2 ⁄2 i

ω1 = √ 1 [cos -π⁄2 + 2(1)π⁄2 + i sin -π⁄2 + 2(1)π⁄2]
    = 1 [cos -π + 4π⁄2 ⁄2 + i sin -π + 4π⁄2 ⁄2]
    = cos 3π⁄4 + i sin 3π⁄4
    = -1⁄ √ 2 + 1⁄ √ 2 i
    = - √ 2 ⁄2 + √ 2 ⁄2 i

(f)     √ 3   - i     = ( √ 3   , -1)

r = √ x2 + y2 = √ 3 + 1 = 2

cos θ = x⁄r = √ 3 ⁄2
sin θ = y⁄r = -1 ⁄2
θ = -π⁄6
x y θ r z = ( √ 3, -1) O
z = r (cos θ + i sin θ)
  = 2 (cos -π⁄6 + i sin -π⁄6)
ωk = √ r (cos θ + 2kπ⁄2 + i sin θ + 2kπ⁄2)
    = √ 2 (cos -π⁄6 + 2kπ⁄2 + i sin -π⁄6 + 2kπ⁄2)
The square roots are
ω0 = √ 2 [cos -π⁄6 + 2(0)π⁄2 + i sin -π⁄6 + 2(0)π⁄2]
    = √ 2 [cos -π⁄12 + i sin -π⁄12]
    = √ 2 [cos - (4-3)π⁄12 + i sin - (4-3)π⁄12]
    = √ 2 [ cos (-4π⁄12 + 3π⁄12) + i sin (-4π⁄12 + 3π⁄12)]
    = √ 2 [cos (-π⁄3 + π⁄4) + i sin (-π⁄3 + π⁄4)]
    = √ 2 [(cos -π⁄3.cos π⁄4 - sin -π⁄3.sin π⁄4) + i (sin -π⁄3.cos π⁄4 + cos -π⁄3.sin π⁄4)]
    = √ 2 [(1⁄2. 1⁄ √ 2 - (- √ 3 ⁄2.1⁄ √ 2 ) + (- √ 3 ⁄2.1⁄ √ 2 + 1⁄2.1 ⁄ √ 2 )i]
    = √ 2 [ 1 + √ 3 ⁄2 √ 2 - √ 3 - 1 ⁄2 i ]
    = 1 + √ 3 ⁄2 - √ 3 - 1 ⁄2 i

ω1 = √ 2 [cos -π⁄6 + 2(1)π⁄2 + i sin -π⁄6 + 2(1)π⁄2]
    = √ 2 [cos -π + 12π⁄6 ⁄2 + i sin -π + 12π⁄6 ⁄2]
    = √ 2 [ cos 11π⁄12 + i sin 11π⁄12]
    = √ 2 [ cos (8 + 3)π⁄12 + i sin (8 + 3)π⁄12]
    = √ 2 [ cos (8π⁄12 + 3π⁄12) + i sin (8π⁄12 + 3π⁄12)]
    = √ 2 [ cos (2π⁄3 + π⁄4) + i sin (2π⁄3 + π⁄4)]
    = √ 2 [(cos 2π⁄3.cos π⁄4 - sin 2π⁄3.sin π⁄4) + i (sin 2π⁄3.cos π⁄4 + cos 2π⁄3.sin π⁄4)]
    = √ 2 [(-1⁄2. 1⁄ √ 2 - √ 3 ⁄2. 1⁄ √ 2 ) + i ( √ 3 ⁄2. 1⁄ √ 2 + -1⁄2. 1⁄ √ 2 )]
    = √ 2 [-1 - √ 3 ⁄2 √ 2 ) + ( √ 3 - 1 ⁄2 √ 2 ) i]
    = -1 - √ 3 ⁄2 + √ 3 - 1 ⁄2 i

2.(a)     1 + i = (1, 1)

r = √ x2 + y2   = √ 1 + 1   = √ 2
cos θ = x⁄r = 1⁄ √ 2
sin θ = y⁄r = 1⁄ √ 2
θ = π⁄4
x y θ r z = (1 , 1) O
z = r (cos θ + i sin θ)
  = √ 2 (cos π⁄4 + i sin π⁄4)
ωk = √ r (cos θ + 2kπ⁄3 + i sin θ + 2kπ⁄3)
    = (∛ √ 2 ) (cos π⁄4 + 2kπ⁄3 + i sin π⁄4 + 2kπ⁄3)
The cube roots are
ω0 = (∛ √ 2 ) [cos π⁄4 + 2(0)π⁄3 + i sin π⁄4 + 2(0)π⁄3]
    = √ 2 1⁄3 [cos π⁄12 + i sin π⁄12]
    = 21⁄2.1 ⁄3 [cos (4 - 3)π⁄12 + i sin (4 - 3)π⁄12]
    = 21⁄6 [ cos (4π⁄12 - 3π⁄12) + i sin (4π⁄12 - 3π⁄12)]
    = 21⁄6 [cos (π⁄3 - π⁄4) + i sin (π⁄3 - π⁄4)]
    = 21⁄6 [(cos π⁄3.cos π⁄4 + sin π⁄3.sin π⁄4) + i (sin π⁄3.cos π⁄4 - cos π⁄3.sin π⁄4)]
    = 21⁄6 [(1⁄2. 1⁄ √ 2 + √ 3 ⁄2.1⁄ √ 2 ) + ( √ 3 ⁄2.1⁄ √ 2 - 1⁄2.1 ⁄ √ 2 )i]
    = 21⁄6 [ 1 + √ 3 ⁄2 √ 2 + √ 3 - 1 ⁄2 √ 2 i ]
    = 21⁄6 [ √ 2 + √ 6 ⁄4 + √ 6 - √ 2 ⁄4 i]

ω1 = (∛ √ 2 ) [cos π⁄4 + 2(1)π⁄3 + i sin π⁄4 + 2(1)π⁄3]
    = √ 2 1⁄3 [cos π + 8π⁄4 ⁄3 + i sin π + 8π⁄4 ⁄3]
    = 21⁄2.1 ⁄3 [cos 9π⁄12 + i sin 9π⁄12]
    = 21⁄6 [ cos 3π⁄4 + i sin 3π⁄4 ]
    = 21⁄6 [ -1 ⁄ √ 2 + 1 ⁄ √ 2 i ]
    = 21⁄6 [ - √ 2 ⁄2 + √ 2 ⁄2 i]

ω2 = (∛ √ 2 ) [cos π⁄4 + 2(2)π⁄3 + i sin π⁄4 + 2(2)π⁄3]
    = √ 2 1⁄3 [cos π + 16π⁄4 ⁄3 + i sin π + 16π⁄4 ⁄3]
    = 21⁄2.1 ⁄3 [cos 17π⁄12 + i sin 17π⁄12]
    = 21⁄2.1 ⁄3 [cos (8 + 9)π⁄12 + i sin (8 + 9)π⁄12]
    = 21⁄6 [ cos (8π⁄12 + 9π⁄12) + i sin (8π⁄12 + 9π⁄12)]
    = 21⁄6 [cos (2π⁄3 + 3π⁄4) + i sin (2π⁄3 + 3π⁄4)]
    = 21⁄6 [(cos 2π⁄3.cos 3π⁄4 - sin 2π⁄3.sin 3π⁄4) + i (sin 2π⁄3.cos 3π⁄4 - cos 2π⁄3.sin 3π⁄4)]
    = 21⁄6 [(-1⁄2. 1⁄ √ 2 - √ 3 ⁄2.1⁄ √ 2 ) + ( √ 3 ⁄2.-1⁄ √ 2 + -1⁄2.1 ⁄ √ 2 )i]
    = 21⁄6 [(-1 ⁄2 √ 2 - √ 3 ⁄2< √ 2 ) + (- √ 3 ⁄2 √ 2 - 1⁄2 √ 2 )i]
    = 21⁄6 [ -1 - √ 3 ⁄2 √ 2 - √ 3 + 1 ⁄2 √ 2 i ]
    = 21⁄6 [ - √ 2 - √ 6 ⁄4 - √ 6 + √ 2 ⁄4 i]

(b)     i   = (0, 1)

r = √ 1   = 1
cos θ = x⁄r = 0⁄1 = 0
sin θ = y⁄r = 1⁄1 = 1
θ = π⁄2
x y θ r z = (0, 1) O
z = r (cos θ + i sin θ)
  = 1 (cos π⁄2 + i sin π⁄2)
ωk = n√ r (cos θ + 2kπ⁄3 + i sin θ + 2kπ⁄3)
    = ∛ 1 (cos π⁄2 + 2kπ⁄3 + i sin π⁄2 + 2kπ⁄3)
The cube roots are
ω0 = ∛ 1 [cos π⁄2 + 2(0)π⁄3 + i sin π⁄2 + 2(0)π⁄3]
    = 1 [cos π⁄6 + i sin π⁄6]
    = √ 3 ⁄2 + 1 ⁄2 i
ω1 = ∛ 1 [cos π⁄2 + 2(1)π⁄3 + i sin π⁄2 + 2(1)π⁄3]
    = 1 [cos π + 4π⁄6 + i sin π + 4π⁄6]
    = cos 5π⁄6 + i sin 5π⁄6
    = - √ 3 ⁄2 + 1 ⁄2 i
ω2 = ∛ 1 [cos π⁄2 + 2(2)π⁄3 + i sin π⁄2 + 2(2)π⁄3]
= 1 [cos π + 8π⁄6 + i sin π + 8π⁄6
    = 1 [cos 9π⁄6 + i sin 9π⁄6]
    = cos (4 + 5)π ⁄6 + i sin (4 + 5)π ⁄6
    = cos (4π⁄6 + 5π⁄6) + i sin (4π⁄6 + 5π⁄6)
    = cos (2π⁄3 + 5π⁄6) + i sin (2π⁄3 + 5π⁄6)
    = (cos 2π⁄3.cos 5π⁄6 - sin 2π⁄3.sin 5π⁄6) + i (sin 2π⁄3.cos 5π⁄6 + cos 2π⁄3.sin 5π⁄6)
    = (-1⁄2. - √ 3 ⁄2 - √ 3 ⁄2.1⁄2 ) + ( √ 3 ⁄2.- √ 3 ⁄2 + -1⁄2.1 ⁄2)i
    = ( √ 3 ⁄4 - √ 3 ⁄4) + ( -3 ⁄4 + -1 ⁄4 ) i
    = 0 + -4⁄4 i
    = -i

(c)     -1 + i   = (-1, 1)

r = √ x2 + y2 = √ 2
cos θ = x⁄r = -1⁄ √ 2
sin θ = y⁄r = 1⁄ √ 2
θ = 3π⁄4
x y θ r z = (-1, -1) O
z = r (cos θ + i sin θ)
  = √ 2 (cos 3π⁄4 + i sin 3π⁄4)
ωk = n√ 2 (cos θ + 2kπ⁄n + i sin θ + 2kπ⁄n )
    = ∛ √ 2 ( cos 3π⁄4 + 2kπ ⁄3 + i sin 3π⁄4 + 2kπ ⁄3)
The cube roots are
ω0 = ∛ √ 2 (cos 3π⁄4 + 2(0)π⁄ 3 + i sin 3π⁄4 + 2(0)π⁄3 )
    = 2 1⁄2.1 ⁄3 (cos 3π⁄12 + i sin 3π⁄12)
    = 2 1⁄6 (cos π⁄4 + i sin π⁄4)
    = 21⁄6 (1⁄ √ 2 + 1⁄ √ 2 i)
    = 6√ 2 ( √ 2 ⁄2 + √ 2 ⁄2 i)

ω1 = ∛ √ 2 (cos 3π⁄4 + 2(1)π⁄ 3 + i sin 3π⁄4 + 2(1)π⁄3 )
∛ √ 2 (cos 3π + 8π⁄4 ⁄ 3 + i sin 3π + 8π⁄4 ⁄3 )
    = 2 1⁄2.1 ⁄3 (cos 11π⁄12 + i sin 11π⁄12)
    = 2 1⁄6 (cos (8 + 3)π⁄12 + i sin (8 + 3)π⁄12)
    = 2 1⁄6 (cos (8π⁄12 + 3π⁄12) + i sin (8π⁄12 + 3π⁄12))
    = 2 1⁄6 (cos (2π⁄3 + π⁄4) + i sin (2π⁄3 + π⁄4))
    = 21⁄6 ( (cos 2π⁄3.cos π⁄4 - sin 2π⁄3.sin π⁄4) + i (sin 2π⁄3.cos π⁄4 + cos 2π⁄3.sin π⁄4) )
    = 21⁄6 ( (-1⁄2. 1⁄ √ 2 - √ 3 ⁄2.1⁄ √ 2 ) + ( √ 3 ⁄2.1⁄ √ 2 + -1⁄2 .1⁄ √ 2 )i)
    = 21⁄6 ((-1⁄2 √ 2 - √ 3 ⁄2 √ 2 ) + ( √ 3 ⁄2 √ 2 - 1⁄2 √ 2 ) i )
    = 21⁄6 (-1 - √ 3 ⁄2 √ 2 + √ 3 - 1 ⁄2 √ 2 i)
    = 6√ 2 (- √ 2 - √ 6 ⁄4 + √ 6 - √ 2 ⁄4 i)

ω2 = ∛ √ 2 (cos 3π⁄4 + 2(2)π⁄ 3 + i sin 3π⁄4 + 2(2)π⁄3 )
    = ∛ √ 2 (cos 3π + 16π⁄4 ⁄ 3 + i sin 3π + 16π⁄4 ⁄3 )
    = 2 1⁄2.1 ⁄3 (cos 19π⁄12 + i sin 19π⁄12)
    = 2 1⁄6 (cos (10 + 9)π⁄12 + i sin (10 + 9)π⁄12)
    = 2 1⁄6 (cos (10π⁄12 + 9π⁄12) + i sin (10π⁄12 + 9π⁄12))
    = 2 1⁄6 (cos (5π⁄6 + 3π⁄4) + i sin (5π⁄6 + 3π⁄4))
    = 21⁄6 ( (cos 5π⁄6.cos 3π⁄4 - sin 5π⁄6.sin 3π⁄4) + i (sin 5π⁄6.cos 3π⁄4 + cos 5π⁄6.sin 3π⁄4) )
    = 21⁄6 ( (- √ 3 ⁄2. -1⁄ √ 2 - 1 ⁄2.1⁄ √ 2 ) + (1 ⁄2.-1⁄ √ 2 + - √ 3 ⁄2 .1⁄ √ 2 )i)
    = 21⁄6 (( √ 3 ⁄2 √ 2 - 1 ⁄2 √ 2 ) + (-1 ⁄2 √ 2 - √ 3 ⁄2 √ 2 ) i )
    = 21⁄6 ( √ 3 - 1 ⁄2 √ 2 - 1 + √ 3 ⁄2 √ 2 i)
= 6 ( 2 6 - 2 4 - 2 6 + 4 i )
(d)   -1 - i   = (-1, -1)

r = √ x2 + y2   = √ 2
cos θ = x⁄r = -1⁄ √ 2
sin θ = y⁄r = -1⁄ √ 2
θ = - (π - π⁄4 ) = -3π⁄4
x y θ r z = (-1, -1) O
z = r (cos θ + i sin θ)
  = √ 2 (cos -3π⁄4 + i sin -3π⁄4)
ωk = n√ 2 (cos θ + 2kπ⁄n + i sin θ + 2kπ⁄n )
    = ∛ √ 2 ( cos -3π⁄4 + 2kπ ⁄3 + i sin -3π⁄4 + 2kπ ⁄3)
The cube roots are
ω0 = ∛ √ 2 (cos -3π⁄4 + 2(0)π⁄ 3 + i sin -3π⁄4 + 2(0)π⁄3 )
    = 2 1⁄2.1 ⁄3 (cos -3π⁄12 + i sin -3π⁄12)
    = 2 1⁄6 (cos -π⁄4 + i sin -π⁄4)
    = 21⁄6 (1⁄ √ 2 - 1⁄ √ 2 i)
    = 6√ 2 ( √ 2 ⁄2 - √ 2 ⁄2 i)

ω1 = ∛ √ 2 (cos -3π⁄4 + 2(1)π⁄ 3 + i sin -3π⁄4 + 2(1)π⁄3 )
∛ √ 2 (cos -3π + 8π⁄4 ⁄ 3 + i sin -3π + 8π⁄4 ⁄3 )
    = 2 1⁄2.1 ⁄3 (cos 5π⁄12 + i sin 5π⁄12)
    = 2 1⁄6 (cos (2 + 3)π⁄12 + i sin (2 + 3)π⁄12)
    = 2 1⁄6 (cos (2π⁄12 + 3π⁄12) + i sin (2π⁄12 + 3π⁄12))
    = 2 1⁄6 (cos (π⁄6 + π⁄4) + i sin (π⁄6 + π⁄4))
    = 21⁄6 ( (cos π⁄6.cos π⁄4 - sin π⁄6.sin π⁄4) + i (sin π⁄6.cos π⁄4 + cos π⁄6.sin π⁄4) )
    = 21⁄6 ( ( √ 3 ⁄2. 1⁄ √ 2 - 1 ⁄2.1⁄ √ 2 ) + (1 ⁄2.1⁄ √ 2 + √ 3 ⁄2 .1⁄ √ 2 )i)
    = 21⁄6 (( √ 3 ⁄2 √ 2 - 1 ⁄2 √ 2 ) + (1 ⁄2 √ 2 + √ 3 ⁄2 √ 2 ) i )
    = 21⁄6 ( √ 3 - 1 ⁄2 √ 2 + 1 + √ 3 ⁄2 √ 2 i)
    = 6√ 2 ( √ 6 - √ 2 ⁄4 + √ 2 + √ 6 ⁄4 i)

ω2 = ∛ √ 2 (cos -3π⁄4 + 2(2)π⁄ 3 + i sin -3π⁄4 + 2(2)π⁄3 )
    = ∛ √ 2 (cos -3π + 16π⁄4 ⁄ 3 + i sin -3π + 16π⁄4 ⁄3 )
    = 2 1⁄2.1 ⁄3 (cos 13π⁄12 + i sin 13π⁄12)
    = 2 1⁄6 (cos (10 + 3)π⁄12 + i sin (10 + 3)π⁄12)
    = 2 1⁄6 (cos (10π⁄12 + 3π⁄12) + i sin (10π⁄12 + 3π⁄12))
    = 2 1⁄6 (cos (5π⁄6 + π⁄4) + i sin (5π⁄6 + π⁄4))
    = 21⁄6 ( (cos 5π⁄6.cos π⁄4 - sin 5π⁄6.sin π⁄4) + i (sin 5π⁄6.cos π⁄4 + cos 5π⁄6.sin π⁄4) )
    = 21⁄6 ( (- √ 3 ⁄2. 1⁄ √ 2 - 1 ⁄2.1⁄ √ 2 ) + (1 ⁄2.1⁄ √ 2 + - √ 3 ⁄2 .1⁄ √ 2 )i)
    = 21⁄6 ((- √ 3 ⁄2 √ 2 - 1 ⁄2 √ 2 ) + (1 ⁄2 √ 2 - √ 3 ⁄2 √ 2 ) i )
    = 21⁄6 (- √ 3 - 1 ⁄2 √ 2 + 1 - √ 3 ⁄2 √ 2 i)
= - 6 ( 2 6 - 2 4 + 2 6 - 4 i )
(e)   -i   = (0, -1)

r = √ x2 + y2   = √ 1   = 1
cos θ = x⁄r = 0⁄1 = 0
sin θ = y⁄r = -1⁄1 = -1
θ = -π⁄2
x y θ r z = (0, -1) O

z = r (cos θ + i sin θ)
  = 1 (cos -π⁄2 + i sin -π⁄2)
ωk = n√ r (cos θ + 2kπ⁄3 + i sin θ + 2kπ⁄3)
    = ∛ 1 (cos -π⁄2 + 2kπ⁄3 + i sin -π⁄2 + 2kπ⁄3)
The cube roots are
ω0 = ∛ 1 [cos -π⁄2 + 2(0)π⁄3 + i sin -π⁄2 + 2(0)π⁄3]
    = 1 [cos -π⁄6 + i sin -π⁄6]
    = √ 3 ⁄2 - 1 ⁄2 i
ω1 = ∛ 1 [cos -π⁄2 + 2(1)π⁄3 + i sin -π⁄2 + 2(1)π⁄3]
    = 1 [cos -π + 4π⁄ 6 + i sin -π + 4π⁄6]
    = cos 3π⁄6 + i sin 3π⁄6
    = cos π⁄2 + i sin π⁄2
    = 0 + 1i
    = 0 + i

ω2 = ∛ 1 [cos -π⁄2 + 2(2)π⁄3 + i sin -π⁄2 + 2(2)π⁄3]
= 1 [ cos -π + 8π⁄6 + i sin -π + 8π⁄6]
    = 1 [ cos 7π⁄6 + i sin 7π⁄6]
    = cos (4 + 3)π ⁄6 + i sin (4 + 3)π ⁄6
    = cos (4π⁄6 + 3π⁄6) + i sin (4π⁄6 + 3π⁄6)
    = cos (2π⁄3 + π⁄2) + i sin (2π⁄3 + π⁄2)
    = (cos 2π⁄3.cos π⁄2 - sin 2π⁄3.sin π⁄2) + i (sin 2π⁄3.cos π⁄2 + cos 2π⁄3.sin π⁄2)
    = (-1⁄2.0 - √ 3 ⁄2.1) + ( √ 3 ⁄2. 0 + -1⁄2. 1)i
    = - √ 3 ⁄2 - 1 ⁄2 i

(f)   1 - i   = (1, -1)

r = √ x2 + y2   = √ 2
cos θ = x⁄r = 1⁄ √ 2
sin θ = y⁄r = -1⁄ √ 2
x y θ r z = (1, -1) O
θ = -π⁄4
z = r (cos θ + i sin θ)
  = √ 2 (cos -π⁄4 + i sin -π⁄4)
ωk = n√ 2 (cos θ + 2kπ⁄n + i sin θ + 2kπ⁄n )
    = ∛ √ 2 ( cos -π⁄4 + 2kπ ⁄3 + i sin -π⁄4 + 2kπ ⁄3)
The cube roots are
ω0 = ∛ √ 2 (cos -π⁄4 + 2(0)π⁄ 3 + i sin -π⁄4 + 2(0)π⁄3 )
    = 2 1⁄2.1 ⁄3 (cos -π⁄12 + i sin -π⁄12)
    = 2 1⁄6 (cos - (4 - 3)π⁄12 + i sin - (4 - 3)π⁄12)
    = 2 1⁄6 (cos - (4π⁄12 - 3π⁄12) + i sin - (4π⁄12 - 3π⁄12 ) )
    = 2 1⁄6 (cos (-4π⁄12 + 3π⁄12) + i sin (-4π⁄12 + 3π⁄12 ) )
    = 2 1⁄6 (cos (-π⁄3 + π⁄4) + i sin (-π⁄3 + π⁄4 ) )
    = 2 1⁄6 (cos -π⁄3.cos π⁄4 - sin -π⁄3.sin π⁄4) + i (sin -π⁄3.cos π⁄4 + cos -π⁄3.sin π⁄4)
    = 21⁄6 ( (1⁄2. 1⁄ √ 2 - - √ 3 ⁄2.1⁄ √ 2 ) + (- √ 3 ⁄2.1⁄ √ 2 + 1⁄2.1 ⁄ √ 2 )i)
    = 21⁄6 ((1⁄2 √ 2 + √ 3 ⁄2 √ 2 ) + (- √ 3 ⁄2 √ 2 + 1⁄2 √ 2 ) i )
    = 21⁄6 (1 + √ 3 ⁄2 √ 2 - √ 3 - 1⁄2 √ 2 i)
= 6 ( 2 2 + 6 4 - 6 2 - 4 i )
ω1 = ∛ √ 2 (cos -π⁄4 + 2(1)π⁄ 3 + i sin -π⁄4 + 2(1)π⁄3 )
    = ∛ √ 2 (cos -π + 8π⁄4 ⁄ 3 + i sin -π + 8π⁄4 ⁄3 )
    = 2 1⁄2.1 ⁄3 (cos 7π⁄12 + i sin 7π⁄12)
    = 2 1⁄6 (cos (3 + 4)π⁄12 + i sin (3 + 4)π⁄12)
    = 2 1⁄6 (cos (3π⁄12 + 4π⁄12) + i sin (3π⁄12 + 4π⁄12))
    = 2 1⁄6 (cos (π⁄4 + π⁄3) + i sin (π⁄4 + π⁄3))
    = 21⁄6 ( (cos π⁄4.cos π⁄3 - sin π⁄4.sin π⁄3) + i (sin π⁄4.cos π⁄3 + cos π⁄4.sin π⁄3) )
    = 21⁄6 ( (1⁄ √ 2 . 1⁄2 - 1 ⁄ √ 2 . √ 3 ⁄2 ) + (1 ⁄ √ 2 .1⁄ 2 + 1⁄ √ 2 . √ 3 ⁄2)i)
    = 21⁄6 ((1 ⁄2 √ 2 - √ 3 ⁄2 √ 2 ) + (1 ⁄2 √ 2 + √ 3 ⁄2 √ 2 ) i )
    = 21⁄6 (1 - √ 3 ⁄2 √ 2 + 1 + √ 3 ⁄2 √ 2 i)
= 6 ( 2 2 - 6 4 + 2 6 + 4 i )
ω2 = ∛ √ 2 (cos -π⁄4 + 2(2)π⁄ 3 + i sin -π⁄4 + 2(2)π⁄3 )
    = ∛ √ 2 (cos -π + 16π⁄4 ⁄ 3 + i sin -π + 16π⁄4 ⁄3 )
    = 2 1⁄2.1 ⁄3 (cos 15π⁄12 + i sin 15π⁄12)
    = 2 1⁄6 (cos (6 + 9)π⁄12 + i sin (6 + 9)π⁄12)
    = 2 1⁄6 (cos (6π⁄12 + 9π⁄12) + i sin (6π⁄12 + 9π⁄12))
    = 2 1⁄6 (cos (π⁄2 + 3π⁄4) + i sin (π⁄2 + 3π⁄4))
    = 21⁄6 ( (cos π⁄2.cos 3π⁄4 - sin π⁄2.sin 3π⁄4) + i (sin π⁄2.cos 3π⁄4 + cos π⁄2.sin 3π⁄4) )
    = 21⁄6 ( ( 0 . -1⁄ √ 2 - 1 . 1 ⁄ √ 2 ) + (1 . -1 ⁄ √ 2 + 0 . 1 ⁄ √ 2 )i)
    = 21⁄6 ((0 - 1⁄ √ 2 ) + (-1 ⁄ √ 2 + 0 ) i )
    = 21⁄6 (-1⁄ √ 2 - 1 ⁄ √ 2 i)
= - 6 ( 2 2 2 - 2 2 i )
3.(a)   z4 = -i
-i = (0, -1) = 1 [cos ( -π⁄2 + 2kπ) + i sin ( -π⁄2 + 2kπ ) ]
z4 = -i
z4 = 1 [cos ( -π⁄2 + 2kπ) + i sin ( -π⁄2 + 2kπ ) ]
{z4 = 1 [cos ( -π⁄2 + 2kπ) + i sin ( -π⁄2 + 2kπ ) ] } 1⁄4
(z4)1 ⁄ 4 = 11⁄4 [cos -π⁄2 + 2kπ ⁄ 4 + i sin -π⁄2 + 2kπ ⁄4 ]
  z = ∜ 1 [cos -π⁄2 + 2kπ ⁄ 4 + i sin -π⁄2 + 2kπ ⁄4 ]   ,       k = 0, 1, 2, 3
When k = 0,
z = ∜ 1 [cos -π⁄2 + 2(0)π ⁄ 4 + i sin -π⁄2 + 2(0)π ⁄4 ]
    = 1 (cos -π⁄ 8 + i sin -π⁄8 )
    = ± √ 2 + √ 2 ⁄2   +   ( ± √ 2 - √ 2 ⁄2 i)

See how -π⁄8 is solved

    = √ 2 + √ 2 ⁄2   -   √ 2 - √ 2 ⁄2 i

When k = 1,
z = ∜ 1 [cos -π⁄2 + 2(1)π ⁄ 4 + i sin -π⁄2 + 2(1)π ⁄4 ]
z = ∜ 1 [cos -π + 4π⁄2 ⁄ 4 + i sin -π + 4π⁄2 ⁄4 ]
    = 1 (cos 3π⁄ 8 + i sin 3π⁄8 )
    = ± √ 2 - √ 2 ⁄2   +   (± √ 2 - √ 2 ⁄2 i)

See how 3π⁄8 is solved

    = √ 2 - √ 2 ⁄2   +   √ 2 - √ 2 ⁄2 i

When k = 2,
z = ∜ 1 [cos -π⁄2 + 2(2)π ⁄ 4 + i sin -π⁄2 + 2(2)π ⁄4 ]
z = ∜ 1 [cos -π + 8π⁄2 ⁄ 4 + i sin -π + 8π⁄2 ⁄4 ]
    = 1 (cos 7π⁄ 8 + i sin 7π⁄8 )
    = ± √ 2 + √ 2 ⁄2   +   ( ± √ 2 - √ 2 ⁄2 i )

See how 7π⁄8 is solved

    = - √ 2 + √ 2 ⁄2   +   √ 2 - √ 2 ⁄2 i

When k = 3,
z = ∜ 1 [cos -π⁄2 + 2(3)π ⁄ 4 + i sin -π⁄2 + 2(3)π ⁄4 ]
z = ∜ 1 [cos -π + 12π⁄2 ⁄ 4 + i sin -π + 12π⁄2 ⁄4 ]
    = 1 (cos 11π⁄ 8 + i sin 11π⁄8 )
    = ± √ 2 - √ 2 ⁄2   +   ( ± √ 2 + √ 2 ⁄2 i )

See how 11π⁄8 is solved

    = - √ 2 - √ 2 ⁄2   -   √ 2 + √ 2 ⁄2 i


(b)   z4 = -1
z4 = -1 = (-1, 0)
cos π = -1 and sin π = 0
r = √ x2 + y2   = √ 1 = 1
θ = π
z4 = r [cos θ + 2kπ + i sin θ + 2kπ]
    = 1 [ cos (π + 2kπ) + i sin (π +2kπ)]

{ z4 = 1 [cos (π + 2kπ) + i sin (π + 2kπ) ] } 1⁄4
(z4)1 ⁄ 4 = 11⁄4 [cos π + 2kπ ⁄ 4 + i sin π + 2kπ⁄4 ]
z = ∜ 1 [cos π + 2kπ ⁄ 4 + i sin π + 2kπ⁄4 ],       k = 0, 1, 2,3
When k = 0,
z = ∜ 1 [cos π + 2(0)π ⁄ 4 + i sin π + 2(0)π⁄4 ]
    = 1 [cos π ⁄ 4 + i sin π ⁄4 ]
    = 1 ⁄ √ 2 + 1 ⁄ √ 2 i
    = √ 2 ⁄2 + √ 2 ⁄2 i

When k = 1,
z = ∜ 1 [cos π + 2(1)π ⁄ 4 + i sin π + 2(1)π⁄4 ]
    = 1 [cos 3π ⁄ 4 + i sin 3π ⁄4 ]
    = -1 ⁄ √ 2 + 1 ⁄ √ 2 i
    = √ -2 ⁄2 + √ 2 ⁄2 i

When k = 2,
z = ∜ 1 [cos π + 2(2)π ⁄ 4 + i sin π + 2(2)π⁄4 ]
    = 1 [cos 5π ⁄ 4 + i sin 5π ⁄4 ]
    = [cos (4 + 1)π ⁄ 4 + i sin (4 + 1)π⁄4 ]
    = cos ( 4π ⁄ 4 + π ⁄4 ) + i sin ( 4π ⁄4 + π ⁄4 )
    = cos ( π + π ⁄ 4 ) + i sin ( π + π ⁄4 )
    = ( cos π.cos π ⁄4 - sin π.sin π⁄4 ) + ( sin π.cos π ⁄4 + cos π.sin π⁄4 ) i
    = [ -1. 1⁄ √ 2 - 0.1⁄ √ 2 ] + [ 0.1⁄ √ 2 + (-1).1⁄ √ 2 ] i
    = ( -1 ⁄ √ 2 - 0 ) + ( 0 - 1 ⁄ √ 2 ) i
    = -1 ⁄ √ 2 - 1 ⁄ √ 2 i
    = - √ 2 ⁄2 - √ 2 ⁄2 i

When k = 3,
z = ∜ 1 [cos π + 2(3)π ⁄ 4 + i sin π + 2(3)π⁄4 ]
    = 1 [cos 7π ⁄ 4 + i sin 7π ⁄4 ]
    = [cos (3 + 4)π ⁄ 4 + i sin (3 + 4)π⁄4 ]
    = cos ( 3π ⁄ 4 + 4π ⁄4 ) + i sin ( 3π ⁄4 + π ⁄4 )
    = cos ( 3π ⁄ 4 + π ) + i sin ( 3π ⁄4 + π )
    = ( cos 3π ⁄4.cos π - sin 3π⁄4.sin π ) + ( sin 3π ⁄4.cos π + cos π⁄4.sin π ) i
    = [ -1⁄ √ 2 .(-1) - 1⁄ √ 2 .0 ] + [ 1⁄ √ 2 .(-1) + (-1⁄ √ 2 ).0 ] i
    = ( 1 ⁄ √ 2 - 0 ) + ( -1 ⁄ √ 2 + 0 ) i
    = 1 ⁄ √ 2 - 1 ⁄ √ 2 i
    = √ 2 ⁄2 - √ 2 ⁄2 i

(c)   z4 = -8 - 8 √ 3 i

r = √ x2 + y2   = √ (-8)2 + (-8 √ 3 )2   = √ 64 + 64(3)   = √ 256   = 16
cos θ = x⁄r = -8⁄16 = -1⁄2
sin θ = y⁄r = -8 √ 3 ⁄16 = - √ 3 ⁄2
θ = - (π - π⁄3) = -2π⁄3
z4 = r [cos (θ + 2kπ) + i sin (θ + 2kπ) ]
    = 16 [cos ( -2π⁄3 + 2kπ) + i sin ( -2π⁄3 + 2kπ ) ]
{z4 = 16 [cos ( -2π⁄3 + 2kπ) + i sin ( -2π⁄3 + 2kπ ) ] } 1⁄4
(z4)1 ⁄ 4 = 16 1⁄4 [cos -2π⁄3 + 2kπ ⁄ 4 + i sin -2π⁄3 + 2kπ ⁄4 ]
    z = ∜ 16 [cos -2π⁄3 + 2kπ ⁄ 4 + i sin -2π⁄3 + 2kπ ⁄4 ]   ,       k = 0, 1, 2, 3
When k = 0,
z = ∜ 16 [cos -2π⁄3 + 2(0)π ⁄ 4 + i sin -2π⁄3 + 2(0)π ⁄4 ]
    = 161 ⁄4 (cos -2π⁄ 12 + i sin -2π⁄12 )
    = (24) 1⁄4 [ cos -π ⁄ 6 + i sin -π ⁄ 6 i ]
    = 2 ( √ 3 ⁄2 + -1⁄2 i )
    = √ 3   - 1 i
    = √ 3   - i

When k = 1,
z = ∜ 16 [cos -2π⁄3 + 2(1)π ⁄ 4 + i sin -2π ⁄3 + 2(1)π⁄4 ]
    = ∜ 24 [cos -2π + 6π⁄3 ⁄ 4 + i sin -2π +6π⁄3 ⁄4 ]
    = 2 [cos 4π⁄12 + i sin 4π ⁄12 ]
    = 2 [cos π⁄3 + i sin π ⁄3 ]
    = 2 ( 1 ⁄ 2 + √ 3 ⁄ 2 i )
    = 1 + √ 3   i

When k = 2,
z = ∜ 16 [cos -2π⁄3 + 2(2)π ⁄ 4 + i sin -2π⁄3 + 2(2)π⁄4 ]
    = 2 [cos -2π + 12π ⁄ 12 + i sin -2π + 12π ⁄12 ]
    = 2 [cos 10π ⁄ 12 + i sin 10π⁄12 ]
    = 2 ( cos 5π ⁄ 6 + i sin 5π ⁄6 )
    = 2 ( - √ 3 ⁄ 2   + 1 ⁄2 i )
    = - √ 3   + 1 i
    = - √ 3   + i



When k = 3,
z = ∜ 16 [ cos -2π⁄ 3 + 2(3)π⁄ 6 + i sin -2π⁄ 3 + 2(3)π⁄ 6 ]
  = ∜ 24 [ cos -2π + 18π⁄ 3⁄ 4 + i sin -2π + 18π⁄ 3 ⁄ 4 ]
  = 2 [ cos 16π⁄ 12 + i sin 16π⁄ 12 ]
  = 2 [ cos 4π⁄ 3 + i sin 4π⁄ 3 ]
  = 2 [ cos (1 + 3)π⁄ 3 + i sin (1 + 3)π⁄ 3 ]
  = 2 [ cos ( π⁄ 3 + 3π⁄ 3 ) + i sin ( π⁄ 3 + 3π⁄ 3 ) ]
  = 2 [ cos ( π⁄ 3 + π ) + i sin ( π⁄ 3 + π ) ]
  = 2 [ (cos π⁄ 3.cos π - sin π⁄ 3.sin π ) + (sin π⁄ 3.cos π + cos π⁄ 3.sin π ) i ]
  = 2 [ ( 1⁄ 2.(-1) - √ 3 ⁄ 2.0) + ( √ 3 ⁄2.(-1) + 1⁄ 2.0 ) i ]
  = 2 [ ( -1 ⁄ 2 - 0 ) + ( - √ 3 ⁄ 2 + 0 ) i ]
  = 2 ( -1 ⁄ 2 - √ 3 ⁄ 2 i )
  = -1 - √ 3 i



(d)   z = -1
sin π + i cos π = -1 + 0i = -1 = (-1, 0)
r = √ x2 + y2 = √ (-1)2 = √ 1 = 1
θ = π
z6 = r [cos (θ + 2kπ) + i sin (θ + 2kπ) ]
    = 1 [ cos (π + 2kπ) + i sin (π + 2kπ) ]
{z6 = 1 [ cos (π + 2kπ) + i sin (π + 2kπ) } 1⁄ 6
(z6) 1⁄ 6 = 1 1⁄ 6 [ cos π + 2kπ⁄ 6 + i sin π + 2kπ⁄ 6 ]
z   = 1 [ cos π + 2kπ⁄ 6 + i sin π + 2kπ⁄ 6 ] ,       k = 0, 1, 2, 3, 4, 5

When k = 0,
z   = 1 [ cos π + 2(0)π⁄ 6 + i sin π + 2(0)π⁄ 6 ]
  = ( cos π⁄ 6 + i sin π⁄ 6 )
  = √ 3 ⁄ 2 + 1⁄ 2 i

When k = 1,
z = 1 [ cos π + 2(1)π⁄ 6 + i sin π + 2(1)π⁄ 6 ]
  = ( cos 3π⁄ 6 + i sin 3π⁄ 6 )
  = ( cos π⁄ 2 + i sin π⁄ 2 )
  = 0 + 1i
  = 0 + i

When k = 2,
z = 1 [ cos π + 2(2)π⁄ 6 + i sin π + 2(2)π⁄ 6 ]
  = ( cos 5π⁄ 6 + i sin 5π⁄ 6 )
  = - √ 3 ⁄ 2 + 1⁄ 2 i

When k = 3,
z = 1 [ cos π + 2(3)π⁄ 6 + i sin π + 2(3)π⁄ 6 ]
  = [ cos 7π⁄ 6 + i sin 7π⁄ 6 ]
  = [ cos (3 + 4)π⁄ 6 + i sin (3 + 4)π⁄ 6 ]
  = [ cos ( 3π⁄ 6 + 4π⁄ 6 ) + i sin ( 3π⁄ 6 + 4π⁄ 6 ) ]
  = [ cos ( π⁄ 2 + 2π⁄ 3 ) + i sin ( π⁄ 2 + 2π⁄ 3 ) ]
  = [ (cos π⁄ 2.cos 2π⁄ 3 - sin π⁄ 2.sin 2π⁄ 3 ) + (sin π⁄ 2.cos 2π⁄ 3 + cos π⁄ 2.sin 2π⁄ 3 ) i ]
  = [0.( -1⁄ 2 ) - 1. √ 3 ⁄ 2 ] + [1.( -1⁄ 2) + 0.( -1⁄ 2) i ]
  = (0 - √ 3 ⁄ 2 ) + ( -1⁄ 2 + 0 ) i
  = - √ 3 ⁄ 2 - 1⁄ 2 i

When k = 4,
z = 1 [ cos π + 2(4)π⁄ 6 + i sin π + 2(4)π⁄ 6 ]
  = [ cos 9π⁄ 6 + i sin 9π⁄ 6 ]
  = [ cos (3 + 6)π⁄ 6 + i sin (3 + 6)π⁄ 6 ]
  = [ cos ( 3π⁄ 6 + 6π⁄ 6 ) + i sin ( 3π⁄ 6 + 6π⁄ 6 ) ]
  = [ cos ( π⁄ 2 + π ) + i sin ( π⁄ 2 + π ) ]
  = [ (cos π⁄ 2.cos π - sin π⁄ 2.sin π ) + (sin π⁄ 2.cos π + cos π⁄ 2.sin π ) i ]
  =[ 0.(-1) - 1.0] + [1.(-1) + 0.0]i
  = (0 - 0) + (-1 + 0)i
  = 0 - 1i
  = 0 - i

When k = 5,
z = 1 [ cos π + 2(5)π⁄ 6 + i sin π + 2(5)π⁄ 6 ]
  = [ cos 11π⁄ 6 + i sin 11π⁄ 6 ]
  = [ cos (5 + 6)π⁄ 6 + i sin (5 + 6)π⁄ 6 ]
  = [ cos ( 5π⁄ 6 + 6π⁄ 6 ) + i sin ( 5π⁄ 6 + 6π⁄ 6 ) ]
  = [ cos ( 5π⁄ 6 + π ) + i sin ( 5π⁄ 6 + π ) ]
  = [ (cos 5π⁄ 6.cos π - sin 5π⁄ 6.sin π ) + (sin 5π⁄ 6.cos π + cos 5π⁄ 6.sin π ) i ]
  = [ - √ 3 ⁄ 2.(-1) - 1 ⁄ 2.0 ] + [ 1⁄ 2.(-1) + ( - √ 3 ⁄ 2).0 ] i
  = ( √ 3 ⁄ 2 - 0 ) + ( -1⁄ 2 + 0 ) i
  = √ 3 ⁄ 2 - 1⁄ 2 i


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