3
$\begingroup$

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.

1 Answers 1

3

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 $f(z)$.

  • 0
    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 $r \to 0$.2011-10-06
  • 0
    @Hans, I believe you are right. Once, the limit has been taken, the equations 8 and 9 do not make sense.2011-10-06
  • 0
    Yes, and my answer kind of suggests what they should be replaced with.2011-10-06
  • 0
    @Sasha, good efforts. What is the difference between $r$ and $r(\theta)$?2011-10-06
  • 1
    @Sasha, since you are an R&D at Wolfram Research, you might kindly invite editors to give their views here.2011-10-06
  • 0
    $r$ is a constant radius, and $r(\theta)$ is the function that parametrically defines the contour $\gamma_r$ in polar coordinates.2011-10-06