(c) $a = \cos \theta  + i\sin \theta .$ 

$\frac{{1 + a}}{{1 – a}} = \frac{{(1 + \cos \theta ) + i\sin \theta }}{{(1 – \cos \theta ) – i\sin \theta }}.\,$

हर का परिमेयीकरण करने पर, $\frac{{1 + a}}{{1 – a}} = \frac{{(1 + \cos \theta ) + i\sin \theta }}{{(1 – \cos \theta ) – i\,\sin \theta }} \times \frac{{(1 – \cos \theta ) + i\sin \theta }}{{(1 – \cos \theta ) + i\sin \theta }}$

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

$ = \frac{{1 – {{\cos }^2}\theta  + i\sin \theta  + i\sin \theta \cos \theta  + i\sin \theta  – i\sin \theta \,\cos \theta  – {{\sin }^2}\theta }}{{1 + {{\cos }^2}\theta  – 2\cos \theta  + {{\sin }^2}\theta }}$

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

$ = \frac{{i.2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}{{2{{\sin }^2}\frac{\theta }{2}}}$$ = i\frac{{\cos \frac{\theta }{2}}}{{\sin \frac{\theta }{2}}} = i\cot \frac{\theta }{2}$