Prove that $(-1)^{a/b}= \cos(180(b-a)/b)+i \sin(180(b-a)/b)$

Question

I wonder why $(-1)^{a/b}= \cos(180(b-a)/b)+i \sin(180(b-a)/b)$

I got a feeling that this has something to do with complex plane.

Answer 1

Note that $$ -1 = e^{i\pi } \implies (-1) ^{a/b}= e^{i\pi(a/b)} $$

$$= \cos \pi a/b +i\sin \pi a/b$$

$$= - \cos ( \pi - \pi a/b) +i \sin (\pi -\pi a/b)$$

$$= -\cos \pi (1 - a/b) +i \sin \pi(1-a/b)$$

$$= -\cos \pi (b - a)/b +i \sin \pi(b-a)/b$$