# Residues on a boundary

1. Oct 18, 2010

### hunt_mat

Suppose I want to integrate $$f(z)(z-a)^{-1}$$ where $$|a|=1$$ over the circle $$|a|=1$$, why is it that:
$$f(a)=\frac{1}{\pi i}\int_{|z|=1}\frac{f(z)}{z-a}dz$$
instead of:
$$f(a)=\frac{1}{2\pi i}\int_{|z|=1}\frac{f(z)}{z-a}dz$$

2. Oct 19, 2010

### jackmell

Just take $a=e^{\pi i/4}[/tex] and consider the unit circle contour with an indentation around the pole so that the pole is inside the contour C. Then: $$\int_C \frac{f(z)}{z-a}dz=2\pi i f(a)$$ via the Residue Theorem. Now suppose I just want the integral over the the part of the contour not including the indentation? What is the integral over just the indentation as the radius of the indentation goes to zero? Isn't that just [itex]\pi i f(a)$? So that the other piece is $\pi i f(a)$.

Last edited: Oct 19, 2010
3. Oct 19, 2010

### hunt_mat

I don't get you. First you deform the circle so that the pole in inside the contour, which the integral then yields $$2\pi if(a)$$ as you say. So do you deform again so the pole is outside the contour? Are you integrating over a closed contour here?

4. Oct 19, 2010

### jackmell

Here's the contour:

$$\int_C \frac{f(z)}{z-a}dz=\int_{red}+\int_{Blue}=2\pi i f(a)$$

But we know for a simple pole, as the radius of the indentation goes to zero, the value of the integral is just $\theta i$ times the residue where theta is the arc length of the contour. So I could write:

$$\lim_{\rho\to 0} \int_{Blue} \frac{f(z)}{z-a}dz=\pi i f(a)$$

Then the integral over the red must be $\pi i f(a)$

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Last edited: Oct 19, 2010
5. Oct 19, 2010

### hunt_mat

I understand you more now. I didn't know the thing about the simple pole and the indentation. I will have to think how I can prove that.

Thank you.

6. Oct 19, 2010

### jackmell

See "Basic Complex Analysis" by Marsden and Hoffman:

Let f(z) be analytic with a simple pole at z0 and c be an arc of a circle of radius r and angle a centered at z0. Then:

$$\lim_{r\to 0} \int_C f dz=ai Res(f;z0)$$

7. Oct 19, 2010

### hunt_mat

I think I see the proof now. I use the substitution $$w=a+\rho e^{i\theta}$$ in my integral and integrate over the interval $$0\leqslant\theta\leqslant\alpha$$. then as $$\rho\rightarrow 0$$ I am integrating a constant f(a), which gives the answer.

Right?

8. Oct 19, 2010

### jackmell

If f(z) is analytic except for a simple pole at z=a, then we can write f(z) as:

$$f(z)=\frac{b}{z-a}+g(z)$$

where g(z) is analytic at a, Then:

$$\int_{Blue} f(z)dz=\int_{Blue}\frac{b}{z-a}dz+\int_{Blue} g(z)dz$$

Note that since g(z) is analytic at a, then it's bounded say [itex]|g(z)|<M[/tex] there. Now try and evaluate:

$$\lim_{r\to 0}\left\{\int_{Blue}\frac{b}{z-a}dz+\int_{Blue} g(z)dz\right\}$$

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