Question

Question #dd5ca

Answer 1

Start with the Euler's Formula: `1 = cos (2n) pi + i sin (2n) pi = e^(i 2n pi)`

So `1^(1/3) = (e^(i2n pi))^(1/3) = e^((i 2n pi)/3)`

`n = 0: = cis 0 = 1` This is the principal (real) root.

`n = 1: = cis (2pi)/3 =-1/2 +sqrt3/2 i`

`n = 2: = cis (4pi)/3 =-1/2 - sqrt3/2 i`

Thereafter the roots repeat. In the complex plane, each number has n distinct nth roots.