Question

sendo z=1+i, então z²², sera?

Answer 1

$z=1+i$

$z^2=(1+i)^2$
$z^2=1^2+2\cdot1\cdot i+i^2$
$z^2=1+2i-1$
$z^2=2i$

Note que $z^{22}=(z^2)^{11}$

Desse modo:

$z^{22}=(2i)^{11}=2^{11}\cdot i^{11}$

Observe que $i^{11}=(i^2)^{5}\cdot i^{1}=(-1)^{5}\cdot i=(-1)\cdot i=-i$

Logo:

$z^{22}=2^{11}\cdot(-i)$
$z^{22}=2048\cdot(-i)$
$\boxed{z^{22}=-2048i}$