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Say i have the following integral:

$$\int_\gamma \frac{1}{z}dz$$

where $\gamma$ is half of the unit circle lying in the region defined by $Re(z)<0$. It's pretty clear that the integral crosses the branch cut of the the antiderivative $log(z)$. My solution to avoid this problem is to define $\gamma_1$ & $\gamma_2$, where $\gamma_1$ is the path on the same unit circle as $\gamma$, but starting at $i$ and ending at $e^{i(\pi-\epsilon)}$ and $\gamma_2$ would be similar to $\gamma_1$, but starts at $e^{i(-\pi+\epsilon)}$ and ends at $-i$ .Now to compute the value of the integral I would like to say that: $$\int_\gamma \frac{1}{z}dz=\lim_{\epsilon\rightarrow 0}\left(\int_{\gamma_1} \frac{1}{z}dz+\int_{\gamma_2} \frac{1}{z}dz\right)$$ and then proceed to evaluate the integrals using the fundamental theorem of calculus and take the limit. Is this approach valid?

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  • $\begingroup$ It is nowhere near that complicated. Parameterize the curve as $z = e^{i t}$ for $t \in [\pi/2, 3\pi/2]$ depending on your orientation. $\endgroup$ – user296602 Feb 6 '18 at 21:39
  • $\begingroup$ The exercise explicitly asks me to use the fundamental theorem of calculus and to use the argument convention $(-\pi;\pi]$ so i can’t consider a “different” logarithm with the branch cut elsewhere, say along positive real axis $\endgroup$ – Abrb Feb 6 '18 at 21:43

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