Explicație pas cu pas:
[tex](1+i)^{2n}+(1+i)^{2n+1}=[/tex]
[tex]=(\sqrt{2}e^{i\frac{\pi}{4}})^{2n}+(\sqrt{2}e^{-i\frac{\pi}{4}})^{2n+1}=[/tex]
[tex]=2^n[(e^{i\frac{\pi}{2}})^n+(1-i)(e^{-i\frac{\pi}{2}})^n]=[/tex]
[tex]=2^n[i^n+(1-i)(-i)^n]=[/tex]
[tex]=\begin{cases}2^n(2-i)&\text{dacă }n\equiv 0(\text{mod } 4)\\-2^n&\text{dacă }n\equiv 1(\text{mod }4)\\2^n(-2+i)&\text{dacă }n\equiv 2(\text{mod } 4)\\2^n&\text{dacă }n\equiv 3(\text{mod } 4)\end{cases}.[/tex]
