Arguing that $\lim_{r \to 0}\int_{C_{r}}f(z) \, dz = i \pi \operatorname{Res}[f(z), z_{0}]$, where $C_{r}$ is a semicircle of radius $r$

Question

I recently made the following observation:

Assume that the function $f(z)$ can be expressed near $z=z_{0}$ in the form

$$ \sum_{k=-n}^{0} a_{2k-1} (z-z_{0})^{2k-1} + g(z) \, ,$$

where the function $g(z)$ is analytic at $z_{0}$.

If $C_{r}$ is a counterclockwise-oriented semicircle of radius $r$ centered at $z_{0}$, then

$$\lim_{r \to 0} \int_{C_{r}} f(z) \, dz = i \pi \, \text{Res}[f(z),z_{0}].$$

Is my observation correct?

Attempt at a proof:

$$ \begin{align} \int_{C_{r}} f(z) \, dz &= \int_{\alpha}^{\alpha + \pi}f(z_{0}+re^{it}) \ i r e^{it} \, dt \\ &= \int_{\alpha}^{\alpha + \pi} \left(\sum_{k=-n}^{0} a_{2k-1} (re^{it})^{2k-1} + g(z_{0}+re^{it}) \right) i r e^{it} \, dt \\ &= i \sum_{k=-n}^{-1}a_{2k-1} r^{2k} \underbrace{\int_{\alpha}^{\alpha + \pi} e^{2ikt} \, dt}_{0} + i a_{-1} \int_{\alpha}^{\alpha + \pi} \, dt + i r \int_{\alpha}^{\alpha + \pi} g(z_{0}+re^{it}) \, e^{it} \, dt \\ &= i \pi a_{-1} + i r \int_{\alpha}^{\alpha + \pi} g(z_{0}+re^{it}) \, e^{it} \, dt \end{align}$$

And since $|g(z)|$ is bounded near $z_{0}$,

$$ \lim_{r \to 0} \int_{C_{r}} f(z) \, dz = i \pi a_{-1} + 0 = i \pi \, \text{Res}[f(z),z_{0}]$$

Answer 1

As Daniel Fischer said, the computation is correct (now that the typos were fixed). Here is one more proof.

Assume $z_0=0$ for simplicity. Also ignore the $g$ part because its contribution is small (bounded by $\pi r \sup|g|$, which tends to $0$). Let $C$ and $C'$ be complementary half-circles of circle $|z|=r$. The change of variables $z=-\zeta$ yields $$\int_{C'} f(z)\,dz = \int_C -f(-\zeta) \,d\zeta$$ Notice that $f$ is an odd function: $-f(-\zeta)=f(\zeta)$. Hence
$$ \int_{C'} f(z)\,dz = \int_{C} f(z)\,dz $$ The sum of these integrals is $\int_{|z|=r} f(z)\,dz = 2\pi i \operatorname{Res}(f,0)$. The conclusion follows.