Question

Express −1 + i in polar form. Hence show that (−1 + i )^16 is real and that 1/(−1 + i )^6
is purely imaginary, giving the value for each.

Answer 1

$let \ z=-1+i\ , \ x=-1\ and \ y=1\\ \text{we get}\\ \ r= \sqrt{x^2+y^2}=\sqrt{1+1}=\sqrt{2} \ and\\ \ \theta=\arctan \frac{y}{x}=\arctan \frac{1}{-1}=\frac{3 \pi}{4}\\ \text{Since, in the polar form, we have}\\ \ z=r(\cos \theta+i \sin \theta)\ .\\ \text{So, we get}\\ \ z=\sqrt{2}(\cos \frac{3 \pi}{4}+i \sin\frac{3 \pi}{4}).\\ \text{and}\\ \ z^{16}=(\sqrt{2})^{16}(\cos \frac{3 \pi}{4}+i \sin\frac{3 \pi}{4})^{16}\ .\\ \text{From De Moivre's Theorem, we have}\\ (\cos \theta+i \sin \theta)^{n}=\cos n \theta+i \sin n \theta\\ \text{So, we get}\\ \ z^{16}=(\sqrt{2})^{16}(\cos 12 \pi+i \sin 12 \pi)\\ \ \ \ \ \ \ \ =(\sqrt{2})^{16}(\cos 0+i \sin 0)=256\ .\\ \text{Also, we get}\\ \ \frac{1}{z^{6}}=z^{-6}=(\sqrt{2})^{-6}(\cos \frac{3 \pi}{4}+i \sin\frac{3 \pi}{4})^{-6}\ .\\ \ z^{-6}=(\sqrt{2})^{-6}(\cos \frac{-9 \pi}{2}+i \sin \frac{-9 \pi}{2})\\ \ \ \ \ \ \ \ \ =(\sqrt{2})^{-6}(\cos \frac{- \pi}{2}+i \sin \frac{- \pi}{2})=-\frac{1}{8}i\ .\\ \text{So, we can conclude that}\\ 1. \ (-1+i)^{16}\text{ is real and equal} \ 256\ .\\ 2.\ (-1+i)^{-6} \text{ is purely imaginary and equal} \ -\frac{1}{8}i\ \\$