# show $\lim \limits_{\epsilon \to 0} \int_{C_{\epsilon}} f(z)dz = i\alpha\:\text{res}_{z_0} f$ when $z_0$is a simple pole and $C_{\epsilon}$ is an arc

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.

• Obvious parametrization together with an appropriate convergence theorem will do the job. Commented Jun 11, 2019 at 23:06
• The residue theorem is for closed contours. $dz = i \epsilon e^{i t} dt, f(z) = \frac{B}{z-z_0}+O(1)$ Commented Jun 11, 2019 at 23:07
• 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) Commented Jun 11, 2019 at 23:10

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, $$\lim_{\varepsilon\to0}\eta(b_\varepsilon)-\eta(a_\varepsilon)=0$$.
• Could you explain why the first integral after the $=$ sign is equal to $i\alpha r$ ? Commented Jun 12, 2019 at 8:27
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.
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 $$z_0$$.