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I need to take: $$\int \frac{1}{z-2r}\ dz$$

over the contour $\alpha(t) = re^{it}, t\in [0, 2\pi]$

I know I can take the antiderivative if the contour is contained in the domain of the antiderivative. So, the antiderivative is $\log z-2r$ and since this is a circle shifted $2r$ to the left, I think that the contour is contained in the domain of the antiderivative, because its domain is the entire complex plane minus $\{x+iy; x< 2, y=0\}$ So the integral becomes $0$ because it's a closed path.

Am I right?

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1 Answer 1

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Which analytic branch of $\log(z-2r)$ will you take ???

Let $f(z):=\frac{1}{z-2r}$ and $D:=\{z \in \mathbb C:|z|<\frac{3}{2}r\}$.

Then $f$ is analytic in $D$. Monsieur Cauchy says:

$$\int_{\alpha}f(z)dz=0$$

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