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

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]
$\frac{1}{z} = \frac{(1+cos2\theta - i\, sin2\theta)}{(1 + cos 2\theta + i\, sin 2\theta)(1+cos2\theta - i \,sin2\theta)}\\ = \frac{1+cos2\theta - i \,sin 2\theta}{(1+cos2\theta)^{2} + sin^{2}2\theta}$

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
$\frac{1}{z} = \frac{1}{2} -i \frac{sin2\theta}{2(1+cos2\theta)}$

The real part of 1/ = 1/  (constant).