Question

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

Answer 1

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)`

Explanation:

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)`