Finding the nth root of complex number. Example z=(1-i)^1/5 I am confused after seeing some examples of solving problems like these online and in reference books. When finding the nth root we typically need to get the value of the radius and the angle. 
My confusion lies in with the angle part. I thought that we needed to use the principal argument in order to find all the roots later in z=rcis(2nπ+θ) where θ is NOT the regular argument. In my example the principal argument is -(π/4) and not (7π/4). 
Similarly, for finding the three cube roots of (−2−2√3i) the principal argument -(2π/3) should be used is what I thought but in many worked examples online and in books they're simply using just the argument for finding the roots. So θ=4π/3 (from π+tan^-1(-2√3/-2)=4π/3) is used and not -(5π)/6 which is the principal argument.
So what is happening, which θ  or argument should be used? 
(I have attached a screenshot of such a worked problem).
 A: What you're in the process of discovering is that there isn't a canonical, absolutely correct way of defining an $n$th root function on the complex numbers.
There are $n$th roots, of course: the fundamental theorem of algebra ensures that. The problem is that there are always (except for $0$) $n$ $n$th roots. And unlike the real numbers (where we can deal with the problem of having $2$ square roots by insisting that the positive square root is better, thereby ending up with a continuous function) there's no way to consistently choose a complex $n$th root in a nice way.
Fortunately, the questions you are looking at aren't saying to find a particular $n$th root. They want all of the $n$th roots.
Depending on which definition of argument you use, you'll get a different $n$th root in the first step. But that doesn't matter, because after that you find all the other $n$th roots by going around in $\frac{2 \pi}{n}$ steps. Even though you and others started that process in different places, you'd get the same set of $n$ $n$th roots.
(edited) There are $n$ $n$th roots of $1$, which are $1$, $e^{\frac{2 i \pi}{n}}$, $e^{\frac{4 i \pi}{n}}$, ... , $e^{\frac{(2n-2) i \pi}{n}}$: $n$ points going around the unit circle in the plane with equal angles. Once you've found any $n$th root $w$ of a complex number $z$, the entire list $n$th roots are $w$, $w e^{\frac{2 i \pi}{n}}$, $w e^{\frac{4 i \pi}{n}}$, ... , $w e^{\frac{(2n-2) i \pi}{n}}$: $n$ points going around the $|z|^{1/n}$ circle in the plane. Start at any one of those points, and you'll reach all the other points by multiplying by an $n$th root of one.
