Question

Represent $\sin \alpha -i\cos \alpha$ in polar form.

Answer 1

Real part $>0$; Imaginary part $<0$

Argument of $\sin \alpha –i\cos \alpha$ is in the nature of a negative acute angle.

$\therefore \sin \alpha -i\cos \alpha =\cos (\alpha -\frac{\pi }{2})+i\sin (\alpha -\frac{\pi }{2})=e^{i(\alpha -\frac{\alpha }{2})}$