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Could you please have a look on my solution of the following exercise? The text said that no advanced material like residue theorem and so on are allowed; only elementary methods

Let $f$ be holomoprhic function on $\{z \in \mathbb{C} \mid \lvert z - z_0 \rvert < r\}$ and let $f(z_0) = 0$, but $f^\prime(z_0) \ne 0$. Show that for $\epsilon>0$ small enough holds:

$$\int_{\lvert z - z_0 \rvert = \epsilon} \frac{dz}{f(z)} \ dz = \frac{2 \pi i }{f^\prime(z_0)}$$

My Attempt:

\begin{align} \int_{\lvert z - z_0 \rvert = \epsilon} \frac{dz}{f(z)} \ dz &= \int_0^{2\pi} \frac{1}{f(z_0 + \epsilon e^{it})} \cdot \epsilon i e^{it} \ dt \\ &=i \cdot \int_0^{2\pi} \frac{\epsilon e^{it}}{f(z_0 + \epsilon e^{it})-f(z_0)} \ dt = i \cdot \int_0^{2\pi} \frac{z_0+\epsilon e^{it}-z_0}{f(z_0 + \epsilon e^{it})-f(z_0)} \ dt\\ &= i \cdot \int_0^{2\pi} \frac{1}{f^\prime(z_0)} \ dt = \frac{2\pi i}{f^\prime(z_0)} \end{align}

For the last line I used that $\epsilon$ can be chosen arbitrarily small. Is this correct?

EDIT: Following Gabriele Cassese's comment I make a small adjustment to my proof. For a sequence $\epsilon_n \rightarrow 0$ we define the following sequence:

$$ f_n(z) := \int_{\lvert z - z_0 \rvert = \epsilon_n} \frac{dz}{f(z)}$$

Then it holds by the fact that the integral does not depend on $\epsilon$ that:

\begin{align} \lim_{n \rightarrow \infty} \ f_n(z) &= \lim_{n\to \infty}\int_{\lvert z - z_0 \rvert = \epsilon_n} \frac{dz}{f(z)} \ dz = \int_0^{2\pi} \lim_{n\to \infty}\frac{1}{f(z_0 + \epsilon_n e^{it})} \cdot \epsilon_n i e^{it} \ dt \\ &=i \cdot \int_0^{2\pi} \lim_{n \rightarrow \infty} \frac{\epsilon_n e^{it}}{f(z_0 + \epsilon_n e^{it})-f(z_0)} \ dt\\ &= i \cdot \int_0^{2\pi} \lim_{n \rightarrow \infty} \frac{z_0+\epsilon_n e^{it}-z_0}{f(z_0 + \epsilon_n e^{it})-f(z_0)} \ dt\\ &= i \cdot \int_0^{2\pi} \frac{1}{f^\prime(z_0)} \ dt = \frac{2\pi i}{f^\prime(z_0)} \end{align}

Is this correct now?

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3nondatur
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    Your last passage needs to be motivated: $\lim_{\varepsilon\to 0}\int=\int\lim_{\varepsilon \to 0}$ is not true in general but only under special conditions –  Nov 29 '19 at 16:39
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    Defining $f_n(z)$ makes little sense, in it does not depend on $z$ (and there is no need to make a sequence of $\varepsilon_n$, you can take directly $\varepsilon\to 0$). You can, however, define the function $h(\varepsilon)=\int_{|z-z_0|=\varepsilon}\frac{dz}{f(z)}$, and then argue that, since the integral does not depend on $\varepsilon$, $h$ is actually a constant, and determine it with $h(\varepsilon)=\lim_{\varepsilon \to 0}h(\varepsilon)$ –  Nov 30 '19 at 13:17
  • Ok I think I got it now. Thanks for your help. – 3nondatur Nov 30 '19 at 16:52

1 Answers1

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As I stated in the comments, your idea is right, but your last passage, as it is, is not formally correct. There are two problems:

  1. Taking the limit under the integral is a process that is not always possible (e.g $\lim_{n}\int{\frac{1}{n}}\chi_{[0,\frac1n]}dx=1\neq 0=\int \lim_n\frac{\chi_{[0,\frac1n]}}{n})$. The problem is avoided if one note that $f$ is a continuously differentiable function (which is true since every holomorphic function is analytic). Then, as shown here, this function is uniformly differentiable on a compact set, and the exchange of limit and integral can be performed, thanks to the uniforme convergence criterion.

  2. You are not sure that the integral is not dependent on the value of $\varepsilon$. It is enough, however, to state the homotopic invariance of the integral on the pointed neighbourhood of $z_0$with radius small enough, thanks to Cauchy theorem.

  • Thanks for your answer. I think that problem 1 can be solved by using the fact that the integral does not depend on $\epsilon$ though, so there is not even a need to consider uniform convergence. I edited my proof. Maybe you could have a look at it. – 3nondatur Nov 30 '19 at 12:09
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    @3nondatur Your adjustment is on the right way. It would still benefit of a few correction, as I said in the comments to your answer, but the idea is correct and avoids the objections I made to your first proof. –  Nov 30 '19 at 13:24