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

1 Answer
Narad T. · Zor Shekhtman
Oct 31, 2016

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

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