Arg(z) from $z^n$

Question

$z \in \mathbb{C}$. If the principal argument of $z^n$ is in the quadrant $q$, what is the complete set of values for $Arg(z)$?

For example if $n = 3$ and $q = 2$, how could I find all values of $Arg(z)$?

Answer 1

For any $z \in \mathbb{C},\,\, z = r e^{i\theta}$, to find the args of its $n^{th}$ roots we need to divide $\theta + 2k\pi$ by $n$, where $k$ is any integer that puts the result, $\dfrac{\theta + 2k\pi}{n}$, in the interval $(-\pi,\pi]$.

These values of $k$ are: $\qquad\dfrac{-n}{2} \leq k \leq \dfrac{n-1}{2}$.

Our number $z$ is in the second quadrant, so $\theta \in (\dfrac{\pi}{2}, \pi)$, and our $n$ resulting intervals are:

$\qquad\left(\dfrac{\frac{\pi}{2} +2k\pi}{n},\,\, \dfrac{\pi + 2k\pi}{n} \right)$

or, slightly simpler:

$\qquad\left(\dfrac{4k+1}{2n} \pi,\,\, \dfrac{2k+1}{n} \pi \right)\qquad$ for $\,\,\dfrac{-n}{2} \leq k \leq \dfrac{n-1}{2}$.

The set of possible values for the arg is the union of these $n$ intervals.

Example: $n=3$:

$\dfrac{-n}{2} \leq k \leq \dfrac{n-1}{2}$ gives us $k = -1, 0, 1$.

Our interval for each of these:

$k = -1$: $\qquad\left(\dfrac{4(-1)+1}{6} \pi,\,\, \dfrac{2(-1)+1}{3} \pi \right) = \left(\dfrac{-\pi}{2} ,\,\, \dfrac{-\pi}{3} \right)$

$k = 0$: $\qquad\left(\dfrac{\pi}{6} ,\,\, \dfrac{\pi}{3} \right)$

$k = 1$: $\qquad\left(\dfrac{4(1)+1}{6} \pi,\,\, \dfrac{2(1)+1}{3} \pi \right) = \left(\dfrac{5\pi}{6} ,\,\, \pi \right)$

So the set of possible values for the arg is the union of these three intervals.