De Moivre's Theorem

Question

Question 1

 

Question 2

Question 3

Answer 1

a) ( 1 - i)⁸

The complex number z = 1 - i

Write the complex number in polar form.

r = | z |

r = √(a² + b²)

r = √[(1² + (-1²)]

r = √2

The argument is θ = tan^-1 (b/a)

θ = tan^-1(-1/1)

θ = tan^-1(-1)

θ = - 45

θ = - π/4

Polar form of  1 - i is √2 [cos (- π/4) + i sin (- π/4) ].

De Moivre's theorem (cosx + i sinx )ⁿ =  cos(nx) + i sin(nx)

Now raise the complex number to given power.

zⁿ = rⁿ [cos(nθ) + i sin(nθ)]

(1 - i)⁸ = (√2)⁸ [cos(8(-π/4)) + i sin(8(- π/4))]

= 16 [cos(-2π) + i sin(-2π)]

 = 16 [cos(2π) - i sin(2π)]

= 16 [1 - 0]

(1 - i)⁸ =  16 .

Answer 2

b) ( 1 + √3 i)⁴

The complex number z = 1 + √3 i

a = 1, b = √3

Write the complex number in polar form.

r = | z |

r = √(a² + b²)

r = √[(1² + (√3²)]

r = √4

r = 2

The argument is θ = tan^-1 (b/a)

θ = tan^-1(√3/1)

θ = tan^-1(√3)

θ = 60

θ = π/3

Polar form of z = 1 + √3 i is 2 [cos ( π/3) + i sin ( π/3) ].

De Moivre's theorem (cosx + i sinx )ⁿ =  cos(nx) + i sin(nx)

Now raise the complex number to given power.

zⁿ = rⁿ [cos(nθ) + i sin(nθ)]

(1 + √3 i)⁴ = (2)⁴ [cos(4(π/3)) + i sin(4(π/3))]

= 16 [cos(4π/3) + i sin(4π/3)]

= 16 [ cos ( π + π/3) + i sin (π + π/3)]

= 16 [ - cos(π/3) - i sin(π/3) ]

= 16 [(-1/2) - i(√3/2)]

= 16 [ (-1 - i√3)/2]

( 1 + √3 i)⁴=  8 (-1 - i√3) .

Answer 3

(2).

The value of a = 2 and b = 6.