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We cannot apply the winding number formula here since the curve pass through the point $1$. How can we evaluate the integral $\int_\gamma \frac{dz}{z-1}$, where $\gamma$ is the simple unit circle?

Can we say $\int_\gamma \frac{dz}{z-1} = \int_\sigma \frac{dz}{z-1}$, where $\sigma$ is a circle about $0$ with radius $1/2$?

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  • $\begingroup$ You can build contour which is unit circle everywhere except a small infinitesimal circular neighbourhood of 1. Then you build a half circle around the pole there. If you go outside or inside the difference will be with or without residue. Can you show they will be same then the integral will converge to this same value. (I think). $\endgroup$ – mathreadler Mar 4 at 5:46
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    $\begingroup$ It's a rather dubious integral, certainly not absolutely convergent. You could define a "principal value" as $\pi i$, if you like, since that's the mean of $0$ and $2\pi i$, the integrals you get if you deform the circle to pass $1$ on the inside and outside respectively. $\endgroup$ – Lord Shark the Unknown Mar 4 at 5:48
  • $\begingroup$ Check this link: math.stackexchange.com/questions/246701/… $\endgroup$ – Sujit Bhattacharyya Mar 4 at 5:49
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    $\begingroup$ Otherwise en.wikipedia.org/wiki/Cauchy_principal_value Cauchy principal value can maybe help you. $\endgroup$ – mathreadler Mar 4 at 5:51
  • $\begingroup$ @mathreadler Thank you for the suggestion $\endgroup$ – user398843 Mar 4 at 6:06
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$$ \mbox{Note that}\quad \overbrace{\oint_{\left\vert{z}\right\vert\ =\ \color{red}{\large 1^{+}}} {\mathrm{d}z \over z - 1}}^{\displaystyle =\ 2\pi\mathrm{i}}\ -\ \overbrace{\oint_{\left\vert{z}\right\vert\ =\ \color{red}{\large 1^{-}}} {\mathrm{d}z \over z - 1}}^{\displaystyle =\ 0}\ =\ \bbox[10px,border:1px groove navy]{2\pi\mathrm{i}} $$

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