TITLE: Cube roots of the complex numbers 1+i?
QUESTION [5 upvotes]: I cant get any good reference in my books regarding cube of complex numbers. Please help me find cube roots of the Complex number i+1??

REPLY [7 votes]: The way for finding the nth roots of a complex number, is to express it in the form of $r.cis(\theta)$.  The function $cis()=cos()+i.sin()$.  
For example, if one has $x+iy$, one can find $\theta=arctan(y/x)$, with appropriate quadrant adjustment, and $r^2=x^2+y^2$.  
The value of the nth roots, are then $r^{1/n} cis(2\pi*m/n+\theta/n)$, for $m = 0, \ldots, n-1$.
In your case 1+i gives $r=\sqrt{2}$, and $\theta=45°$.
The angles of the cube root are at $15°$, $135°$, and $255°$, with a radius of $2^{1/6}$.