Question

Care este partea reala a numarului x=(1+i) la puterea 6

Answer 1

$x=(1+i)^6 \\ \\ x = (1+i)^{3\cdot 2} \\ \\ x = \Big((1+i)^3\Big)^2\\ \\ x = (1^3+3\cdot 1^2\cdot i+3\cdot 1\cdot i^2+i^3)^2 \\ \\ x = (1+3i-3-i)^2 \\ \\ x = (2i-2)^2 \\ \\ x = (2i)^2-2\cdot 2i\cdot 2+2^2 \\ \\ x = -4-6i+4 \\ \\ x = 0-6i \\ \\ \Rightarrow \boxed{ Re (x) = 0}$