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Let $f$ be a function with a simple pole at $z_0$. Let $C_{\epsilon}$ be an arc from the point $z_0$ in the angle $\alpha$. It means, that if we take a circle of radius $\epsilon$ around $z_0$, then $C_{\epsilon}$ is an arc on this circle with the angle of $\alpha$.

I want to prove that $$\lim_{\epsilon \to 0} \int_{C_{\epsilon}} f(z)\,dz = i\alpha \operatorname{res}_{z_0} f$$

It seems obvious to use the residue theorem. However I can't think of an appropriate contour. I can the the "pizza slice" around $z_0$ because then I won't be able to use the theorem. All other contour I could think of seem really complicated.

Help would be appreciated.

Moreover, it is asked to answer what happens when $z_0$ is not a simple pole. I don't really see how anything will differ in this case.

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    $\begingroup$ Obvious parametrization together with an appropriate convergence theorem will do the job. $\endgroup$ Commented Jun 11, 2019 at 23:06
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    $\begingroup$ The residue theorem is for closed contours. $dz = i \epsilon e^{i t} dt, f(z) = \frac{B}{z-z_0}+O(1)$ $\endgroup$
    – reuns
    Commented Jun 11, 2019 at 23:07
  • $\begingroup$ To expand on what reuns said, you need to do this one explicitly. This does not follow from the residue theorem - this is, in some sense, a generalization of the residue theorem (for this specific case) $\endgroup$ Commented Jun 11, 2019 at 23:10

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Let $r=\operatorname{res}_{z=z_0}f(z)$. Since $z_0$ is a simple pole,then, near $z_0$, $f(z)$ can be written as$$\frac r{z-z_0}+\sum_{n=0}^\infty a_n(z-z_0)^n.$$Let $\varphi(z)=\sum_{n=0}^\infty a_n(z-z_0)^n$ and let $\eta$ be a primitive of $\varphi$. Then, if $C_\varepsilon(t)=z_0+\varepsilon e^{it}$ ($t\in[a,b]$, with $b-a=\alpha$), we have\begin{align}\int_{C_\varepsilon}\frac r{z-z_0}\,\mathrm dz&=r\int_a^b\frac{\varepsilon ie^{it}}{z_0+\varepsilon e^{it}-z_0}\,\mathrm dt\\&=ir\int_a^b1\,\mathrm dt\\&=ir(b-a)\\&=i\alpha r.\end{align}So,\begin{align}\int_{C_\varepsilon}f(z)\,\mathrm dz&=\int_{C_\varepsilon}\frac r{z-z_0}\,\mathrm dz+\int_{C_\varepsilon}\varphi(z)\,\mathrm dz\\&=i\alpha r+\eta(b_\varepsilon)-\eta(a_\varepsilon),\end{align}where $a_\varepsilon$ and $b_\varepsilon$ are the extreme points of the arc $C_\varepsilon$. And, clearly, $\lim_{\varepsilon\to0}\eta(b_\varepsilon)-\eta(a_\varepsilon)=0$.

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  • $\begingroup$ Could you explain why the first integral after the $=$ sign is equal to $i\alpha r$ ? $\endgroup$
    – Gabi G
    Commented Jun 12, 2019 at 8:27
  • $\begingroup$ I've edited my answer. Is it clear now? $\endgroup$ Commented Jun 12, 2019 at 8:54
  • $\begingroup$ Yes now all is clear, thank. $\endgroup$
    – Gabi G
    Commented Jun 12, 2019 at 8:59
  • $\begingroup$ I'm glad I could help. $\endgroup$ Commented Jun 12, 2019 at 9:00
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Answer for the last part: let $z_0=0$ and $f(z)=\frac 1 {z^{2}}$. For any $\alpha \in (0,2\pi)$ we have $\int_{C_{\epsilon}} f(z)dz=\int_0^{\alpha} \frac {\epsilon i e^{it}} {\epsilon ^{2}e^{2it}} dt$. Clearly the limit of this does not exist.

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I don't think you can get this from the residue theorem. If $r$ is the residue then $$f(z)=\frac r{z-z_0}+g(z),$$where $g$ is bounded near $z_0$.

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