1. \(1+i\) przedstawiamy w postaci trygonometrycznej \(|1+i|=\sqrt{2}\).
Stąd \(1+i=\sqrt{2}(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})\). Wyznaczamy argument z układu \(\left\{\begin{array}{l}\cos(\varphi)=\frac{\sqrt{2}}{2}\\ \sin(\varphi)=\frac{\sqrt{2}}{2}\end{array}\right.\)
Stąd \(\varphi=\frac{\pi}{4}\). Czyli \(1+i=\sqrt{2}(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))\)
Stosujemy wzór de Moivre'a (\(z^n=|z|^n(\cos(n\varphi)+i\sin(n\varphi))\)) i otrzymujemy \((1+i)^n=(\sqrt{2})^n(\cos(\frac{n\pi}{4})+i\sin(\frac{n\pi}{4}))\). Pamiętając, że cosinus i sinus są okresowe otrzymujemy \((1+i)^n=\left\{\begin{array}{ll}
(\sqrt{2})^n(\cos(\frac{(2+k8)\pi}{4})+i\sin(\frac{(2+k8)\pi}{4}))=(\sqrt{2})^n(0+i) & ,\textrm{dla}n=2+8k\\
(\sqrt{2})^n(\cos(\frac{(3+k8)\pi}{4})+i\sin(\frac{(3+k8)\pi}{4}))=(\sqrt{2})^n(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}) & ,\textrm{dla}n=3+8k
(\sqrt{2})^n(\cos(\frac{(4+k8)\pi}{4})+i\sin(\frac{(4+k8)\pi}{4}))=(\sqrt{2})^n(-1+i0) & ,\textrm{dla}n=4+8k
(\sqrt{2})^n(\cos(\frac{(5+k8)\pi}{4})+i\sin(\frac{(5+k8)\pi}{4}))=(\sqrt{2})^n(-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}) & ,\textrm{dla}n=5+8k
(\sqrt{2})^n(\cos(\frac{(6+k8)\pi}{4})+i\sin(\frac{(6+k8)\pi}{4}))=(\sqrt{2})^n(0-i) & ,\textrm{dla}n=6+8k
(\sqrt{2})^n(\cos(\frac{(7+k8)\pi}{4})+i\sin(\frac{(7+k8)\pi}{4}))=(\sqrt{2})^n(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}) & ,\textrm{dla}n=7+8k
(\sqrt{2})^n(\cos(\frac{(8+k8)\pi}{4})+i\sin(\frac{(8+k8)\pi}{4}))=(\sqrt{2})^n(1+i0) & ,\textrm{dla}n=8+8k
(\sqrt{2})^n(\cos(\frac{(9+k8)\pi}{4})+i\sin(\frac{(9+k8)\pi}{4}))=(\sqrt{2})^n(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}) & ,\textrm{dla}n=9+8k
\end{array}\right.\)