Question

The variable complex number z is given by z=1+cos 2θ+isin2θ,where θ takes all values in the interval −1/2π<θ<1/2π

(i) Show that the modulus of z is 2 cos θ and the argument of z is θ
(ii) Prove that the real part of 1/z is constant.

Answer 1

The given complex number is
z = 1 + cos(2θ) + i sin(2θ), for -1/2π < θ < 1/2π

Part (i)
Let V = the modulus of z.
Then
V² = [1 + cos(2θ)]² + sin²(2θ)
= 1 + 2 cos(2θ) + cos²2θ + sin²2θ

Because sin²x + cos²x = 1, therefore
V² = 2(1 + cos2θ)

Because cos(2x) = 2 cos²x - 1, therefore
V² = 2(1 + 2cos²θ - 1) = 4 cos²θ
Because -1/2π < θ < 1/2π,
V = 2 cosθ PROVEN

Part ii.
1/z = 1/[1 + cos2θ + i sin 2θ]

`(1)/(z) = ((1+cos2\theta - i\, sin2\theta))/((1 + cos 2\theta + i\, sin 2\theta)(1+cos2\theta - i \,sin2\theta))\\ = (1+cos2\theta - i \,sin 2\theta)/((1+cos2\theta)^(2) + sin^(2)2\theta)`

The denominator is

`(1+cos2\theta)^(2)+sin^(2)2\theta \\ = 1+2cos2\theta+cos^(2)2\theta+sin^(2)2\theta \\ =2cos2\theta+2 \\ = 2(1+cos2\theta)`

Therefore

`(1)/(z) = (1)/(2) -i (sin2\theta)/(2(1+cos2\theta))`

The real part of 1/ = 1/ (constant).