How do you find the fourth roots of #625i#?

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Answer 1 · 2016-09-07T03:41:55

Using #cis theta = cos theta + i sin theta#, the roots are

#5cis(pi/8), 5 cis(5/8pi), 5 cis(9/8pi) and 5 cis(13/8pi)#
.
=#5(+-sqrt(2+sqrt 2)/2+-isqrt(2-sqrt 2)/2)#

Using # cis theta# for cos theta +i sin theta#,

#cis((2kpi+pi/2)i)=cis(pi/2i)=i#, for k=0, +_1, +-2, +-3, ...

So, #(625i)^(1/4)=625^(1/4)( cis((2kpi+pi/2)i)=cis(pi/2i))^(1/4)#

#=5( cis((2kpi+pi/2)/4)i), k =0, 1, 2, 3#, omitting other k values that

produce repetitions. So, the root are

#=5cis(pi/8), 5 cis(5/8pi), 5 cis(9/8pi) and 5 cis(13/8pi)#.

Using #cos (pi/8)=sqrt((1+cos(pi/4))/2)=sqrt(2+sqrt 2)/2# and,

likewise,

#sin(pi/8)= sqrt(2-sqrt 2)/2#, the answer can be presented as

#5(+-sqrt(2+sqrt 2)/2+-isqrt(2-sqrt 2)/2)#