Question

Express $\frac{1}{1-\cos \theta +i\sin \theta }$ in the form $a+ib$

Answer 1

$\frac{1}{1-\cos \theta +i\sin \theta }$

$=\frac{1}{1-\cos \theta +i\sin \theta }\times \frac{(1-\cos \theta )-i\sin \theta }{(1-\cos \theta )-i\sin \theta }$

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

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

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

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

$=\frac{1}{2}-i\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{4{\sin }^2\frac{\theta }{2}}$

$=\frac{1}{2}-i\frac{\cos \frac{\theta }{2}}{2\sin \frac{\theta }{2}}$

$=\frac{1}{2}-\frac{i}{2}\cot \frac{\theta }{2}$