Cube roots of the complex numbers 1+i?

Question

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??

Answer 1

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 $2^{1/6}$.

Answer 2

Hint: if you let $\theta\in[0,2\pi)$, then every complex number can be uniquely represented by $re^{i\theta}$. Now the cube roots are just this number raised to the $1/3$ power.

Answer 3

Since $1+i = \sqrt{2} e^{i\pi/4}$, we have

$$(1+i)^{1/3} = 2^{1/6} e^{i\pi/12}.$$

Alternatively, $1+i$ could be written $\sqrt{2} e^{9i\pi/4}$ or $\sqrt{2} e^{17i\pi/4}$, providing two more cube roots. WolframAlpha provides a nice illustration of this: