I had a question saying "find the sum of the principal arguments for the cubic roots of $z=e^{7 \pi i/2}$. I calculated the roots then added their principal arguments, in the answers it was solved like this:
The sum of the principal arguments of cubic roots of $z$ equals the principal argument of $z$.
Is this correct? I tried a bunch of numbers and it seems to be true for any $n$-th root except for square roots, any explanation would be helpful!
Assuming a principal value being in $(-\pi,\pi]$, then it is true. If you're taking the cubic roots of $w \in \mathbb{C}$, they are: $$|w|^{1/3} e^{i \theta} \text{ where }\theta = \text{Arg}(w)/3, \text{Arg}(w)/3 + 2\pi/3, \text{Arg}(w)/3 - 2\pi/3.$$
Summing them up you have $\text{Arg}(w)$. In fact, this will hold when you're taking the $n$th roots of any complex number, provided that $n$ is odd. (One way to see that it doesn't hold for even $n$, even if you allow for non-principal values: finding $n$th roots is the same as solving the equation $z^n - w = 0$, but the product of roots when $n$ is even is $-w$, so the sum of the arguments of the $n$th roots must differ from the argument of $w$ by $\pi$.)
However, it isn't true when you define your principal argument to lie in $[0,2\pi)$. For example, take the cubic roots of $1$. Their principal arguments add up to $2\pi$ which does not equal to the principal argument of $1$.