Question

Prove that (- 1 + i)^7 = -8(1 + i)​

Answer 1

Convert
`-1+i` to polar form.

`z = -1 + i \implies \begin{cases}|z| = √((-1)^2 + 1^2) = \sqrt2 \\\\ \arg(z) = \pi + \tan^(-1)\left(\frac1{-1}\right) = \frac{3\pi}4 \end{cases}`

By de Moivre's theorem,

`\left(-1+i\right)^7 = \left(\sqrt2 \, e^{i\,\frac{3\pi}4}\right)^7 \\\\ ~~~~~~~~ = \left(\sqrt2\right)^7 e^{i\,\frac{21\pi}4} \\\\ ~~~~~~~~ = 8\sqrt2 \, e^{-i\,\frac{3\pi}4} \\\\ ~~~~~~~~ = 8\sqrt2 \left(\cos\left(\frac{3\pi}4\right) - i \sin\left(\frac{3\pi}4\right)\right) \\\\ ~~~~~~~~ = 8\sqrt2 \left(-\frac1{\sqrt2} - \frac1{\sqrt2}\,i\right) \\\\ ~~~~~~~~ = -8 (1 + i)`

QED