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":
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 $$\lim\limits_{\varepsilon \to 0}\left(\displaystyle{\int\limits_{\gamma_\varepsilon}}\frac{f\left(\zeta\right)}{\zeta-z_0} d\zeta\right) = i\vartheta_0 f\left( z_0\right).$$
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.

Thanks in advance and cheers!

  • 1
    $\begingroup$ Use the lemma with $g(z) = (z - z_0)f(z)$ which is holomorphic because $g$ has a simple pole at $z_0$ and $\mathrm{Res}_{z_0}(f) = g(z_0)$. $\endgroup$
    – Cactus
    Jun 16, 2023 at 13:29
  • $\begingroup$ Oh my God, this is embarassingly obvious. Thank You, Cactus! But doesn't f(z) have the simple pole the way you are setting it up? $\endgroup$
    – Moguntius
    Jun 16, 2023 at 13:33

1 Answer 1


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).


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