Question

Find the 8th roots of 3i-3 Urgent...please help!!!!

Answer 1

$z=-3+3i=3(-1+i)=3\left(\sqrt2e^{i3\pi/4}\right)=3\sqrt2e^{i3\pi/4}$

Let $z=w^8$. Then the eight roots of $z$ are

$w=\left(3\sqrt2e^{i3\pi/4}\right)^{1/8}$
$w=3^{1/8}2^{1/16}}e^{i(3\pi/4+2\pi k)/8}$

where $k=0,1,\ldots,7$. So the eighth roots are

$w=\begin{cases}3^{1/8}2^{1/16}e^{i3\pi/32}&\text{for }k=0\\\\3^{1/8}2^{1/16}e^{i11\pi/32}&\text{for }k=1\\\\3^{1/8}2^{1/16}e^{i19\pi/32}&\text{for }k=2\\\\3^{1/8}2^{1/16}e^{i27\pi/32}&\text{for }k=3\\\\3^{1/8}2^{1/16}e^{i35\pi/32}&\text{for }k=4\\\\3^{1/8}2^{1/16}e^{i43\pi/32}&\text{for }k=5\\\\3^{1/8}2^{1/16}e^{i51\pi/32}&\text{for }k=6\\\\3^{1/8}2^{1/16}e^{i59\pi/32}&\text{for }k=7\end{cases}$