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Will trying to integrate $\int \frac{1}{z-z_0}$ over $|z-z_0|=r$

I have tried to look for an antiderivative and was asked to point out where it is not defined.

So the antiderivative is $ln(z-z_0)$

  1. If I look at the log function over the $\mathbb{R}$ the function is not defined in $(-\infty,0]$ so in my case the function is not defined at $(-\infty+z_0,z_0]$?

  2. If I look at the log function over the $\mathbb{C}$ log is defined as $ln(z-z_0)=ln|z-z_0|+iArg(z-z_0)$ where is it not defined?

  3. in general how does the complex log function look?

gbox
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1 Answers1

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That integral is equal to $2\pi i$, because\begin{align*}\int_{|z-z_0|=r}\frac1{z-z_0}\,dz&=\int_0^{2\pi}\frac{(z_0+e^{it})'}{z_+e^{it}-z_0}\,dt\\&=\int_0^{2\pi}\frac{ie^{it}}{e^{it}}\,dt\\&=\int_0^{2\pi}i\,dt\\&=2\pi i.\end{align*}So, you can deduce that the function that you were trying to integrate has no primitive, because if it had a primitive $F$, then$$\int_{|z-z_0|=r}\frac1{z-z_0}\,dz=F\bigl(e^{2\pi i}\bigr)-F(e^0)=0.$$

There is no differentiable (or even continuous) $\log$ function from $\mathbb{C}\setminus\{0\}$ into $\mathbb C$. Because if there was, its derivative would be $\frac1z$ and, by the same argument as above, this function has no primitive.

José Carlos Santos
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