Question

How do you find the 4th root of `2 (cos (pi/4) + isin(pi/4))`?

Answer 1

The principal `4`th root is:

`root(4)(2(cos(pi/4)+i sin(pi/4)))=root(4)(2)(cos(pi/16) + i sin(pi/16))`

There are `3` other `4`th roots.

Explanation:

In general, if `n` is a positive integer and `theta in (-pi, pi]` then the principal `n`th root is given by:

`root(n)(r(cos theta + i sin theta)) = root(n)(r)(cos (theta/n) + i sin (theta/n))`

So in our example:

`root(4)(2(cos(pi/4)+i sin(pi/4)))=root(4)(2)(cos(pi/16) + i sin(pi/16))`

Note that the primitive Complex `4`th root of `1` is `i`, so the other `4`th roots are:

`i root(4)(2)(cos(pi/16) + i sin(pi/16)) = root(4)(2)(cos((5pi)/16)+i sin((5pi)/16))`

`i^2 root(4)(2)(cos(pi/16) + i sin(pi/16)) = root(4)(2)(cos(-(7pi)/16)+i sin(-(7pi)/16))`

`i^3 root(4)(2)(cos(pi/16) + i sin(pi/16)) = root(4)(2)(cos(-(3pi)/16)+i sin(-(3pi)/16))`