Question

How do you find the 5th root of `1-i`?

Answer 1

The answer is `=2^(1/10)(cos(-pi/20+(2kpi)/5)+isin(-pi/20+(2kpi)/5))`

Explanation:

Let `z=1-i`
We need to write `z` in trigonometric form
`∣z∣=sqrt(1+1)=sqrt2`
So the trigonometric form is
`z=sqrt2(1/sqrt2-i/sqrt2)`
then there is an angle `theta` such that `cos theta=1/sqrt2` and `sintheta=-1/sqrt2`
So `theta` lies in the 4th quadrant and is `-pi/4`
So the trigonometric form is
`z=sqrt2(cos-pi/4+isin-pi/4)`
we need to find `w` such that `w^5=z`
Then `w=z^(1/5)`
so we have to apply Demoivre's theorem
if `z=costheta+isintheta`
then `z^n=cosntheta+isinntheta`
So here we have
`z^(1/5)=(sqrt2)^(1/5)[cos(-pi/20+(2kpi)/5)+i*sin(-pi/20+(2kpi)/5)]`
where `k=0,1,2,3,4`