I think I have the following problem solved, but I'm not completely sure my reasoning is sound:
Let $n\in\mathbb{N}$ and let $C$ denote the unit circle with the counterclockwise orientation. Evaluate $$\frac{1}{2\pi i}\int_{C}\frac{z^{n-1}}{3z^{n}-1}~dz.$$
My attempt: We first recall the Argument Principle:
Theorem. Let $G$ be a domain in $\mathbb{C}$ and let $\gamma$ be a simple contour whose interior is contained in $G$. Let $f$ be a holomorphic function in $G$ without zeros on $\gamma$, then the number of zeros of $f$ in the interior of $\gamma$ (taking into account multiplicities) is $$\frac{1}{2\pi i}\int_{\gamma}\frac{f'(z)}{f(z)}~dz.$$
In our case, we note that $$\frac{1}{2\pi i}\int_{C}\frac{z^{n-1}}{3z^{n}-1}~dz=\frac{1}{3n}\cdot\frac{1}{2\pi i}\int_{C}\frac{nz^{n-1}}{z^{n}-\frac{1}{3}}~dz.$$ Now, $g(z)=z^{n}-\frac{1}{3}$ is an entire function (since $n>0$ by assumption) and $g\not\equiv 0$ on the unit circle $C$. Moreover, $g'(z)=nz^{n-1}$, so it follows by the Argument Principle that \begin{equation}\frac{1}{2\pi i}\int_{C}\frac{nz^{n-1}}{z^{n}-\frac{1}{3}}~dz=\frac{1}{2\pi i}\int_{C}\frac{g'(z)}{g(z)}~dz\tag{*}\end{equation} is none other than the number of zeros of $g(z)=z^{n}-\frac{1}{3}$ in the interior of the unit circle. Counting multiplicities, there are $n$ zeros of $g$ in the interior of $C$ (namely the $n$th roots of $\frac{1}{3}$), so we conclude that the quantity in (*) is exactly $n$. Hence, $$\frac{1}{2\pi i}\int_{C}\frac{z^{n-1}}{3z^{n}-1}~dz=\frac{n}{3n}=\frac{1}{3}.$$
My questions: Does the above work look okay? I think it's right, but it felt like too simple of an argument (I know that the Argument Principle is a pretty useful tool, but still).
Thank you in advance for any comments!