Question

Calculați NUMERE COMPLEXE

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Answer 1

$\it (1+i)^2=1+2i-1=2i\\ \\ (1-i)^2=1-2i-1=-2i\\ \\ Expresia\ \ devine:\\ \\ \Big(\dfrac{\ \ 2i}{-2i}\Big)^n+\Big(\dfrac{-2i}{\ \ 2i}\Big)^n =(-1)^n+(-1)^n=\begin{cases}\it -1-1=-2,\ dac\breve a\ n=impar\\ \\ \it 1+1=2,\ dac\breve a\ n=par\end{cases}$