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}`