Question

Sendo z = 2-2i, resolva: $z^{8}$

Answer 1

$z^{8}=z^{2\cdot4}=(z^{2})^{4}$

Vamos calcular $z^{2}$:

$z=2-2i\\\\z^{2}=(2-2i)^{2}\\\\z^{2}=2^{2}-2\cdot2\cdot2i+(2i)^{2}\\\\z^{2}=4-8i+2^{2}i^{2}\\\\z^{2}=4-8i+4(-1)\\\\z^{2}=4-8i-4\\\\z^{2}=-8i$

Então:

$z^{8}=(z^{2})^{4}\\\\z^{8}=(-8i)^{4}\\\\z^{8}=(-8)^{4}i^{4}\\\\z^{8}=8^{4}i^{4}\\\\z^{8}=(2^{3})^{4}i^{2\cdot2}\\\\z^{8}=2^{3\cdot4}(i^{2})^{2}\\\\z^{8}=2^{12}(-1)^{2}\\\\z^{8}=2^{12}\cdot1\\\\\boxed{\boxed{z^{8}=2^{12}=4096}}$