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This is an exercise from Stein-Shakarchi's Complex Analysis: evaluate integration $$\int_{|z|=r} \frac{1}{(z-a)(z-b)}dz, \,\,\,\, |a|<r<|b|. $$

The problem I am facing is the following. It is sufficient to find $\int_{|z|=r} \frac{1}{z-a}dz$ and $\int_{|z|=r} \frac{1}{z-b}dz$ (and use partial fraction methd).

This exercise is in first chapter, where the author introduces the integration of $f$ over a parametrized smooth curve $\gamma$. However, I didn't find any theorem in first chapter applicble to evaluate this integration. I tried to evaluate it through parametrization $\gamma(t)=re^{it}$ for $0\leq t\leq 2\pi$. Then $$\int_{|z|=r} \frac{1}{z-a}dz=\int_0^{2\pi} \frac{rie^{it}}{re^{it}-a}dt$$. But I couldn't solve this last integration. Can you help me?

I have seen that this can be solved using some Cauchy's integration formua; BUT, this is taken in second chapter of the book, whereas this exercise is in first chapter.

Groups
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  • Wait, which integral you need help with...? – qqo Jan 20 '15 at 07:08
  • I tried to evaluate integration in title; I thought it is sufficient to find $\int_{|z|=r} \frac{1}{z-a}$. I then thought that if we solve this second integration, we can solve similarly $\int_{|z|=r} \frac{1}{z-b}$. By partial fraction method, these two integration will give answer of integration in the title. – Groups Jan 20 '15 at 07:11
  • You are supposed to let $z = a + re^{it}$ – qqo Jan 20 '15 at 07:23
  • This will not be circle around origin; in question(=exercise), $\gamma$ is circle around origin of radius $r$. – Groups Jan 20 '15 at 07:26
  • Then can't the contour be deformed to the circle centered at $a$? – qqo Jan 20 '15 at 07:34
  • I know this theory a little bit, involving topology. My problem is, I want to solve this with the material in first chapter of the book, which includes only parametrization of the curve. (Otherwise, this can be easily computed by Cauchy integral formula, but it is in second chapter.) – Groups Jan 20 '15 at 07:40
  • This has been asked at least twice here. – Git Gud Jan 20 '15 at 10:40
  • See this. If you can't use Cauchy's Integral Formula (or the idea behind it), you need logarithms. You need to use something... – Git Gud Jan 20 '15 at 10:43
  • @GitGud "you need logarithms" Well, no. – Did Jan 09 '18 at 22:26
  • @Did Completely unrelated matter. I just finished reading The Black Swan: The Impact of the Highly Improbable and as I read the final pages it hit me that the author reminds me of you. The subject should also interest you. – Git Gud Feb 01 '18 at 08:32

4 Answers4

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why don't you expand $$\frac{1}{z - a} = \frac{1}{z} \frac{1}{1 - a/z} = \frac{1}{z}\{1 + a/z + a^2/z^2 + \cdots \}$$ and $$\frac{1}{z - b} = -\frac{1}{b} \frac{1}{1 - z/b} = -\frac{1}{b}\{1 + z/b + z^2/b^2 + \cdots \}$$ and use the fact $\int_{|z| = r} z^n dz = 0$ for $n \neq - 1$ and $\int_{|z| = r} z^{-1} dz = 2\pi i.$

abel
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    I don't know how the author expects the solution as long as only content of chapter 1 is considered. I think, in your solution, we have to use also the fact that if a power series converge uniformly on some domain, then we can interchange integration and infinite sum. – Groups Jan 20 '15 at 08:30
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$I=\int_\gamma \frac{1}{z-a}dz=\int_0^{2\pi}\frac{rie^{it}}{re^{it}-a}dt=\int_0^{2\pi}\frac{ri}{r-ae^{-it}}dt$

Now,$r>|a|\implies |a/r|<1$

So,$I=i\int_0^{2\pi}(1+\frac{a}{r}e^{-it}+(\frac{a}{r})^2 e^{-2it}+...)dt=i(2\pi+0+0+...)=2\pi i$ (Integrating term by term as the convergence is uniform)

Similarly,$\int_\gamma \frac{1}{z-b}dz=0$

So,we are done.

Kishalay Sarkar
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If you are just interested in evaluating the last integral then note that

$$ \int \frac{f'(t)}{f(t)}dt = \ln(f(t))+C $$

Mhenni Benghorbal
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    And yet sometimes we never see the most obvious step. – qqo Jan 20 '15 at 08:24
  • @Nameless: Do not worry, it happens a lot. We are humans. – Mhenni Benghorbal Jan 20 '15 at 08:28
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    Nameless: I understand this; but the author has not defined this logarithmic function; we should be careful about it, since there is no unique way to define logarithmic function. I didn't think how can I proceed with this. Also, the function $\frac{1}{z-a}$ has no primitive in the disc $|z|=r$ since $a$ is inside this disc. I cant directly apply second fundamental theorem of calculus. – Groups Jan 20 '15 at 08:43
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    But... but... this is completely wrong! (Three upvoters saw fit to vote on a post about complex analysis without having the slightest idea of the subject.) – Did Jan 09 '18 at 22:25
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Assigned

$$f(z) ≔ \frac{1}{(z-a)(z-b)},\space\space\space\space\space\space\space z\in\mathbb{C}\setminus\{ a,b\}$$

and given the parametrization

$$γ(t) ≔ re^{it},\space\space\space\space\space\space\space0≤t≤2π,$$

observe that $a$ and $b$ are single poles of $f$, holomorphic over its domain, with $a$ lying within the circle of boundary $γ$, oriented in counterclockwise direction. Hence

$$∮_γ f = 2πi \space \mathrm{res}_a ⁡f = 2πi \space \lim_{z→a}⁡(z-a)f(z) = \frac {2π}{a-b}i.$$

Moli
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