Q.If $x + iy = \dfrac{1}{1 + \cos\theta + i\sin\theta}$, then show that $4x^2 - 1 = 0$.

Rationalising the denominator shows $x=\dfrac12$ regardless of $\theta$, which immediately gives $4x^2-1=0$.

Step 1 — Rationalise.

$x+iy=\dfrac{1}{1+\cos\theta+i\sin\theta}$. Multiply numerator and denominator by the conjugate $1+\cos\theta-i\sin\theta$:

$x+iy = \dfrac{1+\cos\theta-i\sin\theta}{(1+\cos\theta)^2+\sin^2\theta}$

Step 2 — Simplify the denominator.

$(1+\cos\theta)^2+\sin^2\theta = 1+2\cos\theta+\cos^2\theta+\sin^2\theta = 2+2\cos\theta = 2(1+\cos\theta)$.

Step 3 — Separate real and imaginary parts. …