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I'm self-studying complex analysis, and in my book there are starred exercises on complex integration I'm interested in understanding.

Lemma 1 of the text states

If the piecewise differentiable closed curve $\gamma$ does not pass through the point $a$, then the value of the integral $$ \int_\gamma\frac{dz}{z-a} $$ is a multiple of $2\pi i$

in preparation for defining the winding number.

One exercise says, give an alternate proof of Lemma 1 by dividing $\gamma$ into a finite number of subarcs such that there exists a single-valued branch of $\text{arg}(z-a)$ on each subarc. Pay particular attention to the compactness argument needed to prove the existence of such a subdivision.

I thought about it a bit, and don't really know how to approach it. Is there a proof or possibly a sketch I could attempt to work through in the meantime? Thank you.

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    Here's an intuitive description to help get you started. You can find solutions online using some google searches. Since $\gamma$ is compact (every open cover has a finite subcover) you can find the appropriate subarcs $\gamma_1, \ldots ,\gamma_n$. Your function "integrates" to $\ln|z-a|+arg(z-a)$ and since $\gamma$ is closed the $\ln|z-a|$ values of the $\gamma_i$ on the endpoints of the arcs all sum to $0$. All that is left is the sum of the arguments, which is some multiple of $2\pi i$. This is just for intuition, not a proof.2012-02-14

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