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

Rationalise by multiplying numerator and denominator by the conjugate of the denominator; the real part $x$ comes out to be the constant $\tfrac12$ regardless of $\theta$.

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

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

Simplify the denominator:

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

So

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