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My text gives a much more complicated proof of this result, which makes me wonder if the argument I have in my head for this has something wrong with it. Does this work, or have I made a bad assumption somewhere along the line?

Let $U \subseteq \mathbb{C}$ be open, $\overline{D}(z_0, r) \subset U$, and $f : U \rightarrow \mathbb{C}$ be holomorphic with a zero of order $n$ at $z_0$ and no other zeroes in $U$. Taking a power series expansion at $z_0$, $f(z) = a_n(z-z_0)^n + o((z-z_0)^{n+1}).$ Differentiating, $f'(z) = n a_n(z-z_0)^{n-1} + o((z-z_0)^n),$ so we have $\lim_{z \rightarrow z_0} \frac{(z-z_0) f'(z)}{f(z)} = \lim_{z \rightarrow z_0} \frac{n a_n(z-z_0)^n + o((z-z_0)^{n+1})}{a_n(z-z_0)^n + o((z-z_0)^{n+1})} = n.$ Define the function $g : U \rightarrow \mathbb{C}$ by $g(z) = \begin{cases} (z-z_0) f'(z)/f(z) & \text{if } z \neq z_0 \\ n & \text{otherwise} \end{cases}$ By the Cauchy integral formula, $n = g(z_0) = \frac{1}{2 \pi i} \oint_{\partial\overline{D}(z_0, r)} \frac{g(z)}{z - z_0} dz = \frac{1}{2 \pi i} \oint_{\partial\overline{D}(z_0, r)} \frac{(z-z_0)f'(z)}{(z-z_0)f(z)} dz= \frac{1}{2 \pi i} \oint_{\partial\overline{D}(z_0, r)} \frac{f'(z)}{f(z)} dz$ as desired.

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    Could you please add the factor $2\pi i$ that is missing in the title and on the last line? :)2010-11-09

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If $f$ has a zero of order $n$ at $z_0$, then there is an function $h$ holomorphic near $z_0$ and such that $f(z)=(z-z_0)^nh(z)$ near $z_0$ and $h(z_0)\neq0$. Computing, $\frac{f'}{f}=\frac{n}{z-z_0}+\frac{h'(z)}{h(z)}.$ If you use the Cauchy formula now over a sufficiently small circle around $z_0$ you get (a corrected version of) what you want.

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    @user3296: I don't understand what you mean by "going to the length of producing the function $f/f'$ out of thin air" when the function $f'/f$ appears right there in the integral you want to compute!2010-12-13
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According to my professor the approach I've laid out here is fine. It does appear, though, that the more complicated technique that involves splitting (fg)'/fg into f'/f + g'/g has some uses, so I can see why they'd bring that up.

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    Looking at the theorem, I came up with the proof described above in my head. I then saw a proof of the form Mariano describes in the book, which uses a trick that wouldn't normally occur to me. When I see something like this, it generally leads me to assume that the "obvious" way of doing things must have been wrong in order to merit a more complicated proof, but I couldn't find anything wrong with the argument outlined above. If the answer is that$f'/f$is a natural thing that analysts automatically think about, that may explain things better.2011-11-17