Question

z=-1+i√3. Найдите z^9

Answer 1

$z=-1+isqrt3; z^9-?\ z=-1+sqrt{3}i=re^{iphi};\ r=sqrt{left(-1right)^2+left(sqrt{3}right)^2}=sqrt{1+3}=sqrt{4}=2;\ phi=arctanleft(frac{sqrt{3}}{-1}right)=arctanleft(-sqrt{3}right)=-arctanleft(sqrt3right)=\ =-frac{2pi}{3}+pi n;\ z=2cdot e^{frac{2pi}{3}};\ z^9=2^9cdot e^{frac{18pi}{3}+9pi}=512cdot e^{6pi}=512left(cosleft(6piright)+isinrightleft(6piright))=512cdotleft(1+0right)=\ =512.$