Question

If
`z=-1-√(3)i` and
`z^(6)=a+bi`, then
`a` = _ and
`b` = _

Answer 1

Rewriting
`z` in polar form makes this trivial.

`z=|z|e^{i\mathrm{arg}(z)}`

We have

`|z|=√((-1)^2+(-\sqrt3)^2)=2`

`\mathrm{arg}(z)=\tan^(-1)(-1,-\sqrt3)=-\frac{2\pi}3`

(not to be confused with the standard inverse tangent function
`\tan^(-1)x`. Here
`\tan^(-1)(x,y)` is the inverse tangent function that takes into account position in the coordinate plane; look up "atan2" for more information)

So we have

`z=-1-\sqrt3\,i=2e^(-2\pi/3\,i)`

Then

`z^6=2^6\left(e^(-2\pi/3\,i)\right)^6=64e^(-4\pi\,i)=64`

so that
`a=64` and
`b=0`.