Question

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

Answer 1

$x+iy=\frac{1}{1+\cos \theta +i\sin \theta }$

$\therefore x+iy=\frac{1}{1+e^{i\theta }}$

$=\frac{1}{1+e^{i\theta }}\times \frac{1+e^{-i\theta }}{1+e^{-i\theta }}$

$=\frac{1+e^{-i\theta }}{1+e^{i\theta }+e^{-i\theta }+1}$

$\therefore x+iy=\frac{1+e^{-i\theta }}{2+2\cos \theta }$

Separating the real part, we have

$x=\frac{1+\cos \theta }{2+2\cos \theta }=\frac{1}{2}$

$\therefore 4x^2-1=0$

Hence proved