Question

1) se considera numarul complex z egal cu 1-i. Aratati ca z la puterea a 2+2i egal 0........astept raspunsul..... multumesc anticipat

Answer 1

$z = 1-i \\ \\ z^{2+2i} = (1-i)^{2+2i} = (1-i)^2\cdot (1-i)^{2i} = (1-2i-1)\cdot\big[(1-i)^2\big]^i= \\ =-2i\cdot(1-2i-1)^i = -2i\cdot(-2i)^i = -2i\cdot (-2^i)\cdot i^i= \\ = -6.147417534038984 +7.4008126711400495i \\ \\ (in niciun caz nu este 0)$

Answer 2

z²+2i=(1-i)²+2i=1-2i+i²+2i=1+i²=1-1=0
as simple as that!