Solve complex equation $z^3=1-i$ I have attempted to solve it by converting to polar form first using $z=r\cdot e^{(i\cdot\theta)}$ and $i=e^{i\cdot(pi/2 \cdot ( 2\cdot n \cdot pi))}$. 

Then I was taught to compare like terms independently for the magnitude r and angle $\theta$. 

The issue I'm having is not knowing how to handle the "$1 -$ " term in front of i when I am matching terms. The similar example problem we solved in class was $z^4=i$ was much easier to solve since $i$ was by itself so comparing $r$ and the $\theta$ terms was simple. Any advice is much appreciated!
Find the norm and the angle to write out the polar form of $1-i$.

$$1-i =\sqrt{2} \exp((-\frac{\pi}{4} + 2n \pi)i) $$

Now, the problem reduces to $$z^3=\sqrt{2} \exp((-\frac{\pi}{4} + 2n \pi)i)$$
You are correct in your first step of converting $z^3=1-i$ to polar form. 

$r= \sqrt{1^2+(-1)^2} = \sqrt{2}$, and $\theta= \arctan(\frac{-1}{1})=\frac{7}{4}\pi$

Thus, $z^3 = \sqrt{2} \cdot e^{i(\frac{7}{4}\pi)}$.

But remember that you can use de Moivre's Theorem in that form while taking the cube root of $(1-i)$ is just the cube root of the radius times and the exponent (for simplicity since I don't know LaTeX) is just:

$i[(\frac{7}{4} + 2n)\frac{\pi}{3}]=i[\frac{7}{12}\pi + 2n\frac{\pi}{3}]$, where $n=0,1,2$.

Thus, for this problem, we get three answers:

1) $z = 2^{\frac{1}{6}} \cdot e^{i(\frac{7}{12}\pi)}$;

2) $z = 2^{\frac{1}{6}} \cdot e^{i(\frac{7}{12}\pi + \frac{2}{3}\pi)} = 2^{\frac{1}{6}} \cdot e^{i(\frac{5}{4}\pi)}$;

3) $z = 2^{\frac{1}{6}} \cdot e^{i(\frac{7}{12}\pi + \frac{4}{3}\pi)} = 2^{\frac{1}{6}} \cdot e^{i(\frac{23}{12}\pi)}$.