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I am asked to calculate the integral $$\int_C \frac{1}{z-a}dz$$ where $C$ is the circle centered at the origin with radius $r$ and $|a|\neq r$

I parametrized the circle and got $$\int_0^{2\pi}\frac{ire^{it}}{re^{it}-a}dt=\text{log}(re^{2\pi i}-a)-\text{log}(r-a)=\text{log}(r-a)-\text{log}(r-a)=0$$

Because of how the complex logarithm is defined, I am pretty sure that the first equaility is wrong, it it is, what is the correct solution?

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If $r<|a|$ the integral is indeed $0$, but if $r>|a|$, then using Cauchy integral formula (or more generally, Residue theorem), you get $2i\pi$.


You can also do what you did, but :

Notice that if $a\in \mathbb R$, $a>0$, then :

  • If $r>a$, then $$\ln(re^{2i\pi}-a)=\ln(e^{2i\pi}(r-a))=2i\pi+\ln(r-a),$$

  • If $r<a$, then $$\ln(re^{2i\theta }-a)-\ln(r-a)=\ln(e^{i\pi}(a-r))-\ln(e^{i\pi}(a-r))=0.$$

You can generalize this to the case $a\in \mathbb C$, and distinguish the case $r>|a|$ or $r<|a|$.

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By Cauchy's Integral Formula, we get $2\pi i$, if $|a|\lt r$. Otherwise we get $0$, by Cauchy's theorem.

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