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The proof on Wikipedia and the textbook both parameterize $z=x+iy$ in polar coordinates around the singularity as, $z=z_0 + re^{i\theta}$, and so I believe that $dz=e^{i\theta}dr + ire^{i\theta}d\theta$, but they make no mention of dropping the $dr$ term in the infinitesimal. Then the contour integral, of $\oint \frac{f(z_0+re^{i\theta})}{re^{i\theta}}dz$ around the singularity $z_0$ would have a $dr$ term, not mentioned in my reading. Please help me make sense of the $dr$ term missing, and also* explaining how it is justified to take $r\to 0$ before the $\theta$-integration is executed.

*edited

Gabe Fernandez
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1 Answers1

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Both $z_{0}$ and $r$ are constant, the only variable is the $\theta\in[0,2\pi]$, and this describes the path along the circle.

user284331
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  • $r$ is constant for the line integral, and so the term in the integrand of the arbitrary holomorphic function $f(z)=f(z_0+re^{i\theta})$, $r$, corresponds to the path of the integral, but the proof seems like it says $r$ is arbitrary and not corresponding to anything other than zero. – Gabe Fernandez May 04 '21 at 18:41
  • I don't follow? The radius of the circle is fixed here. – user284331 May 04 '21 at 18:43
  • But to complete the theorem $f(z)\to f(z_0)$? – Gabe Fernandez May 04 '21 at 18:49