# Proof that $\frac{1}{2\pi i}\oint f'(z)/f(z) \, dz = n$

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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In your last equation, the second member should probably be $g(z_0)$ instead of $f(z_0)$. To use the Cauchy formula to get from the second member to the third, you need to know that $g$ is holomorphic, not only that it is continuous. Moreover, with your hypotheses you don't know that $g$ is holomorphic in all of $D(z_0,r)$. –  Mariano Suárez-Alvarez Nov 9 '10 at 4:04
(By the way, to get a \\ to work in TeX you need to type \\\\) –  Mariano Suárez-Alvarez Nov 9 '10 at 4:30
Why isn't $g$ holomorphic on $D(z_0, r)$? It's clearly holomorphic everywhere but $z = z_0$, and the limit as $z \rightarrow z_0$ couldn't exist if there was a pole or essential singularity there... –  Daniel McLaury Nov 9 '10 at 4:33
Well, now you've proved it is holomorphic at $z_0$, and added an hypothesis that implies that it is holomorphic everywhere in $D(z_0,r)$. –  Mariano Suárez-Alvarez Nov 9 '10 at 4:45
So with the corrected hypothesis about zeroes inside U (which is an assumption in my text as well), this approach is valid? –  Daniel McLaury Nov 9 '10 at 5:00
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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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This is essentially the solution given in my text. What I want to know is why we have to go to the length of producing the function f'/f out of thin air when the argument outlined above is more constructive. –  Daniel McLaury Nov 10 '10 at 20:47
In any case, it's useful to know about logarithmic derivatives. –  lhf Dec 12 '10 at 22:37
@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! –  Mariano Suárez-Alvarez Dec 13 '10 at 12:20
I still don't understand why you would think of the proof given by Mariano above as being "more complicated". Of course the argument you give - which essentially observes that the function f'/f has a simple pole at $z_0$, with residue $n$ - is correct. However, it seems far more enlightening to consider that $f'/f$ is the derivative of $\log f$ (which of course is only locally defined), and hence computing the integral comes down to seeing how often the image of the curve winds around zero. –  mathstribble Nov 16 '11 at 12:58
The fact that $(fg)'/fg=f'/f + g'/g$ should not be seen as something complicated that comes from thin air, but it just comes from the law on the logarithm of a product. Proving it is also just one line, so I would say that Mariano's answer, with all details given, would still be shorter than the one you propose. Of course it is always good to find several different ways of understanding a result, and I commend you for that. –  mathstribble Nov 16 '11 at 13:00