1.4 Trigonometric Form

As a complex number z = x + yi is an ordered pair (x, y) of real numbers, we can place z in xy-coordinate plane as usual. If the length of the line segment from O to z is r, we say that the absolute value of z is r and is denoted by |z|. Here angle θ, measure in radians, is an angle with positive x-axis and the line segment. From trigonometry, we have known that

x y θ r z = (x, y) = x + yi O

r = √ x2 + y2       x = r cos θ       y = r sin θ


So we have z = (x, y) = x + yi = r cos θ + i r sin θ.
z = r(cos θ + i sin θ)

Which is called the trigonometric form of a complex number. z = (r, θ) is called the polar form of z.

Example 5.
Find the trigonometric form with -π < θ ≤ π.
(a) z = 1 + √ 3 i
(b) z = -1 + i
(c) z = - √ 3 - i
(d) z = -1

Solution

(a) z = 1 + √ 3 i = (1, √ 3 )
r = √ 1 + 3 = 2,       cos θ = 1⁄2       sin θ = √ 3 ⁄ 2
θ = π⁄3
z = 2(cos π⁄3 + i sin π⁄3)
x y θ r z = 1 + √ 3 O
(b) z = -1 + i = (-1, 1)
r = √ 1 + 1 = √ 2 ,       cos θ = - 1 ⁄ √ 2       sin θ = 1⁄ √ 2
θ = 3π⁄4
z = √ 2 (cos 3π⁄4 + i sin 3π⁄4 )
x y θ r z = (-1, 1) O
(c) z = -√ 3 - i = ( -√ 3 - 1)
r = √ 3 + 1 = 2,       cos θ = - √ 3 ⁄2,       sin θ = - 1⁄ 2
θ = - 5π ⁄ 6
z = 2(cos (- 5π ⁄ 6 ) + i sin(- 5π ⁄ 6 ))

x y θ r z = (-√ 3, -1) O

(d) z = -1 = (-1, 0)
r = 1,       cos θ = -1,       sin θ = 0
θ = π
z = cos π + i sin π
x y θ r z = (-1, 0) O

ပံ့ပိုးကူညီသူ

Product in Trigonometric Form
Let z1 = r1(cos θ1 + i sin θ1) and z2 = r2(cos θ2 + i sin θ2). Then their product is

z1z2 = r1(cos θ1 + i sin θ1) . r2(cos θ2 + i sin θ2)
= r1r2((cos θ1 cos θ2 - sin θ1 sin θ2) + i(sin θ1 cos θ2 + cos θ1 sin θ2))
= r1r2(cos(θ1 + θ2) + i sin((θ1 + θ2))

z1z2 = r1r2(cos(θ1 + θ2) + i sin((θ1 + θ2))


Example 6.
Given that z1 = 1 + √ 3 i, z2 = -1 + i, find z1z2 by using trigonometric forms. Check your answer by direct multiplication.

Solution

z1 = 1 + √ 3 i = 2(cos π ⁄ 3 + i sin π ⁄ 3)                (See Example 5(a))
z2 = -1 + i = √ 2 (cos 3π ⁄ 4 + i sin 3π ⁄ 4)                (See Example 5(b))
z1z2 = 2√ 2 [ cos ( π ⁄ 3 + 3π ⁄ 4) + i sin ( π ⁄ 3 + 3π ⁄ 4)]
= -1 - √ 3 + (1 - √ 3 )i

cos(θ1 + θ2) + i sin((θ1 + θ2) = (cos θ1 cos θ2   +   sin θ1 sin θ2)   +   i(sin θ1 cos θ2 + cos θ1 sin θ2)

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

checking
z1z2 = (1 + √ 3 i)(-1 + i)
= -1 + i - √ 3 i + √ 3 i2
= -1 - √ 3 + (1 - √ 3 )i

Multiplicative Inverse in Trigonometric Form
Since z-1 = x ⁄ x2 + y2 - y ⁄ x2 + y2 i for z = x + yi,

z-1 = 1 ⁄ x2 + y2 (x - yi)
= 1 ⁄ r2 r (cos θ - i sin θ)
= 1 ⁄ r(cos θ - i sin θ)

z-1 = 1 ⁄ r(cos (-θ) + i sin (-θ))


Example 7.
Given that z = -√ 3 - i, using trigonometric form of z, find z-1. Check your answer by showing that zz-1 = 1.

Solution

z = -√ 3 - i
z = 2(cos (-5π ⁄ 6) + i sin (- 5π ⁄ 6))                   (See Example 5(c))
z-1 = 1 ⁄ 2(cos 5π ⁄ 6 + i sin (5π ⁄ 6))
= 1 ⁄ 2 ( -√ 3 ⁄ 2 + 1 ⁄ 2 i)
= - √ 3 ⁄ 4 + 1 ⁄ 4 i
zz-1 = (-√ 3 - i) (- √ 3 ⁄ 4 + 1 ⁄ 4 i) = 3 ⁄ 4 - √ 3 ⁄ 4 i + √ 3 ⁄ 4 i - 1 ⁄ 4 i2 = 3 ⁄ 4 + 1 ⁄ 4 = 1

Division in Trigonometric Form
Let z1 = r1(cos θ1 + i sin θ1) and z2 = r2(cos θ2 + i sin θ2). Then

z1 ⁄ z2 = z1z2-1
= r1(cos θ1 + i sin θ1) 1 ⁄ r2(cos(-θ2) + i sin(- θ2) + i sin(-θ2))
z1 ⁄ z2 =r1 ⁄ r2 (cos(θ1 - θ2) + i sin(θ1 - θ2))


Example 8.

Given that z1 = 1 +√ 3 i, z2 = -1 + i, find z1 ⁄ z2 by using trigonometric form. Check your answer by direct solution.

Solution

z1 = 1 +√ 3 i = 2(cos π ⁄ 3 + i sin π ⁄ 3)
z2 = -1 + i = √ 2 (cos 3π ⁄ 4 + i sin 3π ⁄ 4)
z1 ⁄ z2 = 2 ⁄ √ 2 (cos ( π ⁄ 3 - 3π ⁄ 4) + i sin ( π ⁄ 3 - 3π ⁄ 4))
= -1 + √ 3 ⁄ 2 - 1 + √ 3 ⁄ 2 i
z1 ⁄ z2 = 1 + √ 3 i ⁄ -1 + i -1 - i⁄ -1 - i
= -1 - i - √ 3 i - √ 3 i2 ⁄ 1 + 1
= -1 + √ 3 - (1 + √ 3 ) i ⁄ 2
= -1 + √ 3 ⁄ 2 - 1 + √ 3 ⁄ 2 i

Powers of Complex Numbers
The power of a complex number z = r(cos θ + i sin θ) is given by
zn = rn (cos nθ +i sin nθ),     n in an integer


Example 9.
Given that z = 1 + √ 3 i, find (a) z10       (b) z-10

Solution

z = 1 + √ 3 i = 2(cos π ⁄ 3 + i sin π ⁄ 3)               (See Example 5(a))
(a) z10 = 210 (cos 10 π ⁄ 3 + i sin 10 π ⁄ 3) = 1024 (- 1 ⁄ 2 - √ 3 ⁄ 2 i) = -512 - 512 √ 3 i

(b) z-10 = 1 ⁄ 210 (cos (- 10 π ⁄ 3) + i sin (- 10 π ⁄ 3)) = 1 ⁄ 1024 (- 1 ⁄ 2 + √ 3 ⁄ 2 i) = - 1 ⁄ 2048 + √ 3 ⁄ 2048 i

Exercise 1.4
  1. Find the trigonometric form with -π < θ ≤ π.
    2. (a) z = 1 - √ 3 i
    (b) z = - √ 2 + √ 2 i
    (c) z = -2 - 2i
    (d) z = √ 3 - i
    (e) z = i
    (f) z = -3i
  2. Given that z1 = 2 - 2√ 3 i, z2 = -1 - i, find the following complex numbers by using trigonometric forms. Check your answer by direct calculation.
    3. (a) z1z2
    (b) z1-1
    (c) z2-1
    (d) z1 ⁄ z2
    (e) z2 ⁄ z1
  3. Given that z = -2 √ 3 - 2i, find (a) z5       (b) z-5

Answer
1. (a)     z = 1 - √ 3 i
Solution
z = 1 - √ 3 i = (1, - √ 3 )

r = √ x2 + y2 = √ 1 + 3 = 2
x = r cos θ
x⁄r = cos θ
1⁄2 = cos θ
cos θ = 1⁄2
r sin θ = y
sin θ = y⁄r = - √ 3 ⁄2
θ = - π⁄3
z = 2 ( cos - π⁄ 3 + i sin - π⁄3)
x y θ r z = (1, - √ 3) O

(b)     z = - √ 2   + √ 2 i = ( - √ 2 , √ 2 )

r = √ x2 + y2 = √ 2 + 2 = 2
cos θ = x⁄ r = - √ 2 ⁄2
sin θ = y⁄r = √ 2 ⁄2
θ = 3π⁄4
z = r (cos θ + i sin θ)
= 2 (cos 3π⁄4 + i sin 3π⁄4)
x y θ r z = (-√2, √2) O
(c)     z = -2 - 2i = (-2, -2)

r = √ x2 + y2   =   √ (-2)2 + (-2)2   =   √ 8   =   2 √ 2
cos θ = x⁄r = -2⁄2 √ 2 = -1 ⁄ √ 2
sin θ = y ⁄ r = -2 ⁄ 2 √ 2 = -1 ⁄ √ 2
θ = -3π ⁄ 4
z = r (cos θ + i sin θ)
    = 2 √ 2 ( cos -3π ⁄ 4 + i sin -3π ⁄ 4 )
x y θ r z = (-2, -2) O

(d)     z   =  √ 3 - i  =   (√ 3 , -1)

r = √ x2 + y2   =   √ 4   =  2
cos θ = x ⁄ r = √ 3 ⁄ 2
sin θ = y⁄r = -1⁄ 2
θ = -π⁄6
z = r ( cos θ + i sin θ)
= 2 (cos -π⁄6 + i sin -π⁄6 )
x y θ r z = ( √ 3, -1) O

(e)     z = 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
z = r ( cos θ + i sin θ )
  = 1 (cos π⁄2 + i sin π⁄2)
cos π⁄2 + i sin π⁄2
x y θ r z = (0, 1) O
(f)     z = -3i = (0, -3)

r = √ x2 + y2   =   √ (-3)2 = 3
cos θ = x⁄r = 0⁄3 = 0
sin θ = y⁄r = -3⁄3 = -1
θ = -π⁄2
z = ( cos θ + i sin θ )
  = 3 ( cos -π⁄2 + i sin -π⁄2)
x y θ r z = (0, -3) O

2.    z1 = 2 - 2 √ 3 i,     z2 = -1 - i

z1 = 2 - 2 √ 3 i = (2, - 2 √ 3 )

r1 = √ x2 + y2   =   √ 22 + (-2 √3)2   =   = √ 4 + 12   =   4
cos θ = x⁄r = 2⁄4 = 1⁄2
sin θ = y⁄r = -2 √3⁄4 = -√3⁄2
θ1 = -π⁄3
z1 = r ( cos θ + i sin θ )
    = 4 ( cos -π⁄3 + i sin -π⁄3)
x y θ r z = (2, -2 √ 3) O

z2 = -1 - i = (-1, -1)

r2 = √ x2 + y2   =   √ (-1)2 + (-1)2   =   √ 2
cos θ = x⁄r = -1⁄ √ 2
sin θ = y⁄r = -1⁄ √ 2
x y θ r z = (-1, -1) O
θ2 = -3π⁄ 4
z2 = r2 ( cos θ2 + i sin θ2)
    = √ 2 ( cos -3π⁄4 + i sin -3π⁄4 )

(a)     z1z2 = r1r2 [ cos (θ1 + θ2) + i sin (θ1 + θ2 ) ]

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

    = 4 √ 2   [ { cos -π⁄3 - 3π⁄4} + i { sin -π⁄3 - 3π⁄4 } ]

    = 4 √ 2   [ { ( cos -π⁄3 . cos -3π⁄ 4 ) - ( sin -π⁄3 . sin -3π⁄4 ) } + i { sin -π⁄3 . cos -3π⁄ 4 ) + ( cos -π⁄3 . sin -3π⁄4 ) } ]

    = 4 √ 2   [ { (1⁄2 . -1⁄ √ 2 ) - (- √ 3 ⁄2 . -1⁄ √ 2 ) } + i { ( - √ 3 ⁄2 . -1 ⁄ √ 2 ) + ( 1⁄ 2 . -1⁄ √ 2 ) } ]

    = 4 √ 2   [ ( -1⁄ 2 √ 2 - √ 3 ⁄ 2 √ 2 ) + i ( √ 3 ⁄2 √ 2 - 1⁄ 2 √ 2 ) ]

    = 4 √ 2   [ (-1 - √ 3 ⁄2 √ 2 ) + i ( √ 3   - 1⁄2 √ 2 ) ]

    = 4 √ 2 (-1 - √ 3 ⁄2 √ 2 )  +   4 √ 2 ( √ 3   - 1⁄2 √ 2 ) i

    = -2 - 2√ 3   +   (2√ 3   - 2) i

Checking
z1z2 = (2 - 2√ 3 i) (-1 - i)

    = -2 - 2i + 2√ 3 i + 2√ 3 i2

    = -2 - 2i + 2√ 3 i + 2√ 3 (-1)

    = -2 - 2√ 3   +   2√ 3 i - 2i

    = -2 - 2√ 3   +   (2√ 3 - 2) i

(b)     z1-1
z1 = 4 ( cos -π⁄3 + i sin -π⁄3 )

z1-1 = 1⁄4 ( cos π⁄3 + i sin π⁄3 )
    = 1⁄4 ( 1⁄2 + √ 3 ⁄2 i )
    = 1⁄8 + √ 3 ⁄8 i

(c)     z2-1
z2 = √ 2 (cos -3π⁄4 + i sin -3π⁄4 )
z2-1 = 1⁄ √ 2 ( cos 3π⁄4 + i sin 3π⁄4 )
      = 1⁄ √ 2 ( -1⁄ √ 2 + 1⁄ √ 2 i )
      = -1⁄ √ 2 . √ 2   +   1⁄ √ 2 . √ 2 i
      = -1⁄2 + 1⁄2 i

(d)     z1⁄z2
z1⁄z2 = r1⁄r2 [ cos (θ1 - θ2) + i sin (θ1 - θ2) ]
      = 4⁄ √ 2 [ cos ( -π⁄ 3 - -3π⁄4) + i sin ( -π⁄3 - -3π⁄ 4 ) ]
      = 4⁄ √ 2 [ cos (-π⁄3 + 3π⁄4 ) + i sin ( -π⁄3 + 3π⁄4 ) ]
      = 4⁄ √ 2 [ ( cos -π⁄3 . cos 3π⁄4   -   sin -π⁄3 . sin 3π ⁄4 )  +   i ( sin -π⁄3 . cos 3π⁄4 + cos -π ⁄3 . sin 3π⁄4 ) ]
      = 4⁄ √ 2 [ ( 1⁄2 . -1⁄ √ 2   -   - √ 3 ⁄2 . 1⁄ √ 2 ) + i ( - √ 3 ⁄2 . -1⁄√ 2   +   1⁄2 . 1⁄ √ 2 ) ]
      = 4⁄ √ 2 [ -1⁄2 √ 2 + √ 3 ⁄2 √ 2   +   ( √ 3 ⁄2 √ 2 + 1⁄2 √ 2 ) i ]
      = 4⁄ √ 2 [ -1 + √ 3 ⁄2 √ 2   +   √ 3   + 1⁄2 √ 2 i ]
      = 4⁄ √ 2 ( -1 + √ 3 ⁄2 √ 2 )   +   4⁄ √ 2 ( √ 3   + 1⁄2 √ 2 ) i
      = -1 + √ 3   +   ( √ 3   + 1) i

Checking
z1⁄z2 = 2 - 2√ 3 i⁄-1 - i   ×   -1 + i⁄-1 + i
      = -2 + 2i + 2√ 3 i - 2√ 3 i2 ⁄1 - i + i - i2
      = -2 + (2 + 2√ 3 ) i - 2√ 3 (-1)⁄1 - (-1)
      = -2 + 2√ 3   +   (2√ 3   + 2) i⁄2
      = 2 [-1 + √ 3   +   (√ 3   + 1) i ]⁄2
      = -1 + √3   +   (√3 + 1) i

(e)     = z2⁄z1
z2⁄z1 = r2⁄r1 [ cos (θ2 - θ1) + i sin (θ2 - θ1) ]
      = √ 2 ⁄4 [ cos ( -3π⁄ 4 - -π⁄3) + i sin ( -3π⁄4 - -π⁄ 3 ) ]
      = √ 2 ⁄4 [ cos (-3π⁄4 + π⁄3 ) + i sin ( -3π⁄4 + π⁄3 ) ]
      = √ 2 ⁄4 [ ( cos -3π⁄4 . cos π⁄3   -   sin -3π⁄4 . sin π ⁄3 )  +   i ( sin -3π⁄4 . cos π⁄3 + cos -3π ⁄4 . sin π⁄3 ) ]
      = √ 2 ⁄4 [ ( -1⁄ √ 2 . 1⁄2   -   -1⁄ √ 2 . √ 3 ⁄2 ) + i ( -1⁄√ 2 . 1⁄2   +   -1⁄ √ 2 . √ 3 ⁄2 ) ]
      = √ 2 ⁄4 [ -1⁄2 √ 2 + √ 3 ⁄2 √ 2   +   ( -1⁄2 √ 2 + - √ 3 ⁄2 √ 2 ) i ]
      = √ 2 ⁄4 [ -1 + √ 3 ⁄2 √ 2   -   1 + √ 3 ⁄2 √ 2 i ]
      = √ 2 ⁄4 ( -1 + √ 3 ⁄2 √ 2 )   -   √ 2 ⁄4 ( 1 + √ 3 ⁄2 √ 2 ) i
      = -1 + √ 3 ⁄8   -   ( 1 + √ 3 ⁄8) i

Checking
z2⁄z1 = -1 - i⁄2 - 2√ 3 i   ×   2 + 2√ 3 i⁄2 + 2√ 3 i
      = -2 - 2√ 3 i - 2i - 2√ 3 i2 ⁄4 + 4√ 3 i - 4√ 3 i - 4(3) i2

      = -2 + ( -2√ 3   - 2 ) i - 2√ 3 (-1)⁄4 - 12(-1)

      = -2 + 2√ 3   -   (2√ 3   + 2) i⁄16

      = 2 [-1 + √ 3   -   (√ 3   + 1) i ]⁄16

      = -1 + √ 3 ⁄8   -   ( √ 3   + 1⁄8 ) i

3.    z = -2√ 3 - 2i = (-2√ 3 , -2)

r = √ x2 + y2 = √ (- 2√ 3 )2 + (-2)2 = √ 16   = 4

cos θ = x⁄r = - 2√ 3 ⁄ 4   = - √ 3 ⁄2
sin θ = y⁄x = -2⁄4 = -1⁄2
      θ = - 5π⁄6
z = r (cos θ + i sin θ )
  = 4 ( cos -5π⁄6  +   i sin -5π⁄6 )
zn = rn ( cos n θ + i sin n θ)
(a)  z5 = 45 ( cos 5 . -5π⁄6 + i sin 5 . -5π⁄6 )
      = 1024 ( cos -25π⁄6   +   i sin -25π⁄6 )
      = 1024 [ cos -(24 + 1)π⁄6   +   i sin -(24 + 1)π⁄6 ]
      = 1024 [ cos - (24π⁄6 + π⁄6)   +   i sin - ( 24π⁄6 + π⁄6 ) ]
      = 1024 [ cos - ( 4π + π⁄6)   +   i sin - (4π + π⁄6 ) ]
      = 1024 [ cos - ( 2π + 2π + π⁄6)   +   i sin - (2π + 2π + π⁄6 ) ]
      = 1024 [ cos -π⁄6   +   i sin -π⁄6 ]
(Just take π⁄6 and the two full cycles of 2π is ignored)
      = 1024 [ √ 3 ⁄2 + (- 1⁄2)i ]
      = 512 √ 3   -   512 i

(b)   z-5 = 1⁄45 [ cos 5(-5 π⁄6)   +   i sin 5 (-5π⁄6 ) ]
      = 1⁄1024 [ cos -25π⁄6   +   i sin -25π⁄6 ) ]
      = 1⁄1024 [ cos -(24 + 1)π⁄6   +   i sin -(24 + 1)π⁄6 ]
      = 1⁄1024 [ cos - (24π⁄6 + π⁄6)   +   i sin - ( 24π⁄6 + π⁄6 ) ]
      = 1⁄1024 [ cos - ( 4π + π⁄6)   +   i sin - (4π + π⁄6 ) ]
      = 1⁄1024 [ cos - ( 2π + 2π + π⁄6)   +   i sin - (2π + 2π + π⁄6 ) ]
      = 1⁄1024 [ cos -π⁄6   +   i sin -π⁄6 ]
(Just take π⁄6 and the two full cycles of 2π is ignored)
      = 1⁄1024 [ √ 3 ⁄2 + (- 1⁄2)i ]
      = √ 3 ⁄2048 - 1⁄2048i

photo of exercise 1.4, No.2 (d)
Image of exercise 1.4, No.2 (d)