MHB Integration around circular arcs

AI Thread Summary
The discussion focuses on proving that if a function \( f \) has a simple pole at \( z=c \), the limit of the integral over a circular arc \( C_r \) centered at \( c \) approaches \( i(\theta_2 - \theta_1) \text{Res}(f;c) \) as \( r \to 0^+ \). The proof utilizes the Laurent series expansion of \( f \), confirming that the series converges uniformly on compact subsets of the annular region around the pole. By parameterizing the circular arc and applying integration, the limit is established through the properties of the residue and the behavior of the integral as \( r \) approaches zero. The conclusion validates the original claim regarding the integral's limit. This result is significant in complex analysis, particularly in the study of residues and contour integrals.
alyafey22
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Prove the following

If $$f$$ has a simple pole at $$z=c$$ and $$C_r$$ is any circular arc bounded by $$\theta_1 , \theta_2$$ and centered at $$c $$ with radius $$r$$

$$\lim_{r \to 0^+} \int_{C_r} f(z) \, dz = i ( \theta_2 - \theta_1 ) \text{Res} (f;c)$$​
 
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Hi ZaidAlyafey,

I'm assuming we are taking $$f$$ to be meromorphic on an open subset, say $$U$$, of $$\mathbb{C}$$ that contains $$c.$$ I also assume the circular arcs we are considering contain the endpoints of the arc (i.e. are compact). That said, we can prove the claim by looking at the Laurent series expansion for $$f$$.

Since $$z=c$$ is a simple pole of $$f,$$ we know that the Laurent series expansion for $$f$$ is of the form

$$f(z)=\frac{Res(f;c)}{z-c}+\sum_{n=0}^{\infty}a_{n}(z-c)^{n}\qquad (*),$$

where the above holds on an annulus of inner radius $$\rho$$ and outer radius $$R.$$ Now

$$\rho = \limsup_{n\rightarrow\infty}|a_{-n}|^{1/n}=0,$$

and since $$f$$ is meromorphic on $$U,$$ $$R>0;$$ the fact that $$\rho=0$$ is what allows us to let $$r\rightarrow 0^{+}.$$ We must note that $$(*)$$ converges uniformly on compact subsets of our annular region (Laurent series - Wikipedia, the free encyclopedia).

Since, for small enough $$r,$$ $$C_{r}$$ is a compact subset of our annular region, $$(*)$$ converges uniformly on $$C_{r}.$$ Using the uniform convergence to justify interchanging integral and sum we have

$$\int_{C_{r}}f(z)dz=Res(f;c)\int_{C_{r}}\frac{1}{z-c}dz+\sum_{n=0}^{\infty}a_{n}\int_{C_{r}}(z-c)^{n}dz.$$

Parameterizing $$C_{r}$$ via $$z=c+re^{i\theta}$$ and integrating gives

$$\int_{C_{r}}f(z)dz=i(\theta_{2}-\theta_{1})Res(f;c)+ir\sum_{n=0}^{\infty}a_{n}\int_{\theta_{1}}^{\theta_{2}}r^{n}e^{i(n+1)\theta}d \theta$$

The sum on the right exists, because the integral on the left exists. Hence, taking the limit we obtain

$$\lim_{r\rightarrow 0^{+}}\int_{C_{r}}f(z)dz=i(\theta_{2}-\theta_{1})Res(f;c),$$

as desired
 
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