# Complex Contour Integration along a Circular Arc and Residues

Recently I stumbled upon this answer in which the author derives a very nice identity for when one does not integrate along a whole circle but rather only a circular arc. Their proof is concise but easily understandable to me. How this works for the corollary, however, is not apparent to me, even after thinking about it for a while.

Allow me to "set the scene":
Lemma:
Let $$f\left(z\right)$$ be holomorphic near $$z=z_0$$. Fix $$\vartheta_0 \in \left] 0, 2\pi \right[$$. Let $$\gamma_\varepsilon$$ denote a postively oriented circular arc subtending an angle of $$\vartheta_0$$ on a circle of radius $$\epsilon$$, centred at $$z_0$$.
Then $$\lim\limits_{\varepsilon \to 0}\left(\displaystyle{\int\limits_{\gamma_\varepsilon}}\frac{f\left(\zeta\right)}{\zeta-z_0} d\zeta\right) = i\vartheta_0 f\left( z_0\right).$$
Proof:
By the substitution $$\zeta = z_0 + \varepsilon e^{i\vartheta}$$, we have:

\begin{align*} \left\vert \int\limits_{\gamma_\varepsilon} \frac{f(\zeta)}{\zeta-z_0}d\zeta - i\vartheta_0 f(z_0)\right\vert &= \left\vert i \int\limits_{\vartheta_1}^{\vartheta_1+\vartheta_0} f(z_0 + \varepsilon e^{i\vartheta})d\vartheta - i\vartheta_0 f(z_0)\right\vert \\ &\leq \int\limits_{\vartheta_1}^{\vartheta_1+\vartheta_0} \left\vert f(z_0 + \varepsilon e^{i\vartheta}) - f(z_0) \right\vert d\vartheta, \end{align*} which clearly goes to zero when $$\varepsilon \to 0$$.
q.e.d So far, so great, this all seems pretty easy to follow and understand to me.\

They then claim, that with the same notation and the same "machinery" for the proof, the following holds if $$f\left(z\right)$$ possesses a simple pole at $$z_0$$: $$\lim\limits_{\varepsilon \to 0}\left(\int\limits_{\gamma_\varepsilon}f\left(\zeta\right) d\zeta\right) = i \vartheta_0 \text{Res}_{z_0}\left(f\left(z\right)\right) ,$$

where $$\text{Res}_{z_0}\left(f\left(z\right)\right)$$ denotes the residue of $$f\left(z\right)$$ at $$z_0$$.
And this is where I no longer follow. The result seems plausible to me and I have tried to set up a similar proof but I think I am missing the right expression for the residue to make this work or using the wrong substitution.

\begin{align*} \left\vert \int\limits_{\gamma_\varepsilon} f\left(\zeta\right) d\zeta - i\vartheta_0 \text{Res}_{z_0} \left( f \left( z \right) \right) \right\vert &= \left\vert i\int\limits_{\vartheta_1}^{\vartheta_1+\vartheta_0} f(z_0 + \varepsilon e^{i\vartheta}) \varepsilon e^{i\vartheta} d\vartheta - i\vartheta_0\text{Res}_{z_0}\left(f\left(z\right)\right)\right\vert\\ &\leq \int\limits_{\vartheta_1}^{\vartheta_1+\vartheta_0} \left\vert \varepsilon e^{i\vartheta} f\left(z_0 + \varepsilon e^{i\vartheta}\right) - \lim\limits_{z\to z_0}\left(\left( z-z_0\right)f\left( z\right)\right)\right\vert d\vartheta, \end{align*} where I have inserted the definition of the residue of a function with a simple pole, does, in my opinion, not prove that this relation should hold. So I would be very thankful for any advice on how to see this proof through.

• Use the lemma with $g(z) = (z - z_0)f(z)$ which is holomorphic because $g$ has a simple pole at $z_0$ and $\mathrm{Res}_{z_0}(f) = g(z_0)$. Jun 16, 2023 at 13:29
If $$f$$ has a simple pole at $$z_0$$ with residue $$R$$ then $$f(z) = \frac{R}{z-z_0} + g(z)$$ where $$g$$ is holomorphic in a neighborhood of $$z_0$$. Then (for sufficiently small $$\epsilon$$) $$\int_{\gamma_\epsilon} f(\zeta) \, d\zeta = \int_{\gamma_\epsilon} \frac{R}{\zeta - z_0} \, d\zeta + \int_{\gamma_\epsilon} g(\zeta) \, d\zeta \, .$$ The first integral on the right-hand side is equal to $$i \theta_0 R$$, and the second integral converges to zero for $$\epsilon \to 0$$ (since the integrand is bounded and the length of the curve converges to zero).