Question

Give `Z` complex, what is `Z^(1/n)` ?

Answer 1

See below.

Explanation:

Any complex number `Z` can be represented as

`Z = r e^(i (phi + 2k pi))`.

For instance, if `Z = x + i y` then

`r = sqrt(x^2+y^2)` and `phi= arctan(y/x)`.

Here we are using de Moivre's identity which reads

`e^(i phi) = cos(phi)+i sin(phi)`

so

`Z^(1/n)=r^(1/n) e^(i (phi + 2k pi)/n)` or equivalently

`Z^(1/n)=r^(1/n)(cos((phi + 2k pi)/n)+isin((phi + 2k pi)/n))`

for `k = 0,1,cdots,n-1`