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Let $a \neq b\in \mathbb{C} $ and $U := \mathbb{C} -[a,b] $

Let $\Gamma$ be a cycle in $U$. The following equality is true?

$$\int_{\Gamma} \frac{1}{(z-a)(z-b)}dz=0$$

I saw it some notes of a friend. I know it's true for every cycle $\Gamma$ that is homologous to zero in $U$. But what about the others cycles?

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  • $\begingroup$ What do you know about the winding numbers of $\Gamma$ about $a$ and $b$? $\endgroup$
    – Daniel Fischer
    Commented Sep 29, 2014 at 21:48
  • $\begingroup$ In the notes there is noting about it. But the winding numbers of Γ about a and b maybe there are equal ? $\endgroup$
    – Porufes
    Commented Sep 29, 2014 at 21:56

1 Answer 1

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Expanding the comment:

$$\frac{1}{(z-a)(z-b)} = \frac{1}{a-b}\left(\frac{1}{z-a}-\frac{1}{z-b}\right),$$

hence

$$\int_\Gamma \frac{1}{(z-a)(z-b)}\,dz = \frac{1}{a-b}\left(\int_\Gamma\frac{dz}{z-a} - \int_\Gamma \frac{dz}{z-b}\right) = \frac{2\pi i}{a-b} \left(n(\Gamma,a) - n(\Gamma,b)\right).$$

Since $a$ and $b$ belong to the same component of the complement of $U$, we have $n(\Gamma,a) = n(\Gamma,b)$ for every cycle $\Gamma$ in $U$.

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  • $\begingroup$ of course partial fraction, thanks alot! $\endgroup$
    – Porufes
    Commented Sep 29, 2014 at 22:04

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