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I was reading through the proof of the Cauchy Integral Formula here. I do not understand how the transition is made from equation (8) to equation (9). While taking the limit as $r\to 0$, doesn't the closed curve $\gamma_r$ also vanish? So, by then the closed curve $\gamma_r$ around $z_0$ is degenerate(a point), I think.

Can you please explain what is going on? Thank you.

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1 Answer 1

The integrate $\oint_{\gamma_r} f(z_0 + r \mathrm{e}^{i \theta}) i \mathrm{d} \theta$ really means $\int_0^{2\pi} f( z_0 + r(\theta) \mathrm{e}^{i \theta}) i \mathrm{d} \theta$, where $r(\theta)$ is bounded by $r$, i.e. $\sup_{0\le \theta < 2\pi} r(\theta) \le r$.

With this in mind, the eq. (9) will follow by continuity of $f(z)$.

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I would say that it's equation (11) that follows directly. Neither (9) nor (10) make sense really, since they refer to $r$ after having taken the limit as $r \to 0$. – Hans Lundmark Oct 6 '11 at 16:48
@Hans, I believe you are right. Once, the limit has been taken, the equations 8 and 9 do not make sense. – Herband Oct 6 '11 at 18:01
Yes, and my answer kind of suggests what they should be replaced with. – Sasha Oct 6 '11 at 18:03
@Sasha, good efforts. What is the difference between $r$ and $r(\theta)$? – Herband Oct 6 '11 at 18:03
@Sasha, since you are an R&D at Wolfram Research, you might kindly invite editors to give their views here. – Herband Oct 6 '11 at 18:06

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