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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