Answer to Question #138918 in Complex Analysis for KHANYA

Question

Use de Moivre's Theorem to derive a formila for the 4th root of -8

Answer 1

Let $z= \cos \theta +i\sin\theta.$ Then $z^4= \cos 4\theta +i\sin4\theta.$ Hence $z^4=-1\Rightarrow \cos 4\theta =-1, \sin 4\theta=0.$ Hence $4\theta= 2k\pi+\pi.$ Hence $\theta= k\pi/2 +\pi/4.$

Hence $z=\cos (k\pi/2 +\pi/4)+isin(k\pi/2+\pi/4).$ Now $sin , cos$ being periodic of period $2\pi,$ $z$ has distinct values for $k=0,1,2,3.$ Hence for 4th root of -8 the solutions are $8^{1/4}[cos(k\pi/2+\pi/4)+i sin(k\pi/2+\pi/4)]$