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I'm reading Conway's complex analysis book and I'm trying to prove this theorem left to the reader on page 124:

I tried to use integration by parts without success. I need some hint how prove this theorem.

user42912
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1 Answers1

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You can proceed similarly as in the proof of the "ordinary" argument principle. Write $$ f(z) = \frac{\prod_{j=1}^n (z-z_j)}{\prod_{k=1}^m (z-p_k)} h(z) $$ where $h$ is holomorphic and $\ne 0$ in $G$. Now take the logarithmic derivative $$ \frac{f'(z)}{f(z)} = \sum_{j=1}^n \frac{1}{z-z_j} - \sum_{k=1}^m \frac{1}{z-p_k} + \frac{h'(z)}{h(z)} $$ and multiply with $g(z)$ $$ g(z) \frac{f'(z)}{f(z)} = \sum_{j=1}^n \frac{g(z)}{z-z_j} - \sum_{k=1}^m \frac{g(z)}{z-p_k} + g(z) \frac{h'(z)}{h(z)} \\ = \sum_{j=1}^n \frac{g(z_j)}{z-z_j} - \sum_{k=1}^m \frac{g(p_k)}{z-p_k} + \left( g(z) \frac{h'(z)}{h(z)} + \sum_{j=1}^n \frac{g(z)-g(z_j)}{z-z_j} - \sum_{k=1}^m \frac{g(z)-g(p_k)}{z-p_k} \right) \, . $$ The term in parentheses has only removable singularities in $G$, therefore $$ \tag{*} g(z) \frac{f'(z)}{f(z)} = \sum_{j=1}^n \frac{g(z_j)}{z-z_j} - \sum_{k=1}^m \frac{g(p_k)}{z-p_k} + \phi(z) $$ with an holomorphic function $\phi$ in $G$.

The assertion now follows since for $a =z_1, \ldots, z_n, p_1, \ldots, p_m$ $$ \frac{1}{2 \pi i} \int_\gamma \frac{g(a)}{z-a}\, dz = g(a) \operatorname{n}(\gamma; a) $$ and $$ \frac{1}{2 \pi i} \int_\gamma \phi(z) \, dz = 0 $$ due to the Cauchy integral theorem.

Remark: Another method to obtain $(*)$ would be to observe that both $$ g(z) \frac{f'(z)}{f(z)} $$ and $$ \sum_{j=1}^n \frac{g(z_j)}{z-z_j} - \sum_{k=1}^m \frac{g(p_k)}{z-p_k} $$ are holomorphic in $A$ with the only exception of (at most) simple poles at the zeros and poles of $f$, and that the principle parts at those points are the same. Therefore the difference has only removable singularities.

Martin R
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    Thank you very much for your answer. I didn't understand just one thing. Why is $\phi$ an holomorphic function? the term in parentheses may have removable singularities, right? – user42912 May 30 '16 at 03:55
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    @user42912: A function has a removable singularity at a point $a$ if it can be defined in the point $a$ in such a way that the continuation is holomorphic in $a$ as well. In this case, Riemann's theorem tells us that the term in parentheses can be continued holomorphically to the entire domain $A$ (because it is bounded in the neighborhood of each $z_j$ and $p_k$). – Martin R May 30 '16 at 05:15
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    Yes, i understand that if $a$ is a removable singularity, then you can redefine $\phi(a)$ such that $\phi$ is holomorphic. The problem is when you redefine $\phi (a)$ you change the function represented by the parentheses terms at this point and we don't have an equality anymore. I hope you have understood my doubt. Thank you again. – user42912 May 30 '16 at 08:15
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    @user42912: You still have the equality in $G$ minus the $z_j$ and $p_k$. And that is sufficient to compute the integral because $\gamma$ avoids these points. – Martin R May 30 '16 at 08:18
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    Yes, of course. You answer is perfectly clear to me now. Thank you again! – user42912 May 30 '16 at 08:21
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    May I ask why you have to write $g(z)=(g(z)-g(z_i))+g(z_i)$ ? Since we already have the analyticity of $g$ and also the cauchy integral formula – mnmn1993 Sep 27 '17 at 13:50
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    @mnmn1993: The above proof (if I remember correctly) uses only the Cauchy integral theorem and the definition of the winding number. If you assume the Cauchy integral formula as known then you can omit one step. – Martin R Sep 27 '17 at 14:20
  • Anyone knows why is needed the hypothesis that $G$ is a region? In the original statement of the Principle argument $G$ can be any open set of $\mathbb{C}$, right? Thanks in advance – Alejandro Tolcachier Dec 31 '19 at 10:10
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    @AleTolcachier: It is not needed, $G$ can be an arbitrary open set. It is assumed that $\gamma$ is homologous to zero in $G$ ($\gamma \approx 0$), so that $ \int_\gamma \phi(z) , dz = 0$. – If $G$ is a simply-connected domain then $\gamma \approx 0$ is automatically satisfied. – Martin R Dec 31 '19 at 12:26
  • @MartinR in the initial logarithmic derivative, the numerators are all 1. how is this possible? whenever i take the logarithmic derivative, i end up with a constant. when you're multiplying the zeros and poles, don't you need to include the multiplicities and orders as exponents? – ekorel Dec 01 '23 at 21:36
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    @ekorel: The numerators are equal to one because multiple zeros or poles are repeated in $z_1, \ldots, z_n$ and $p_1, \ldots, p_m$ according to their multiplicities. Example: If $f$ has a triple zero at $z=a$ then three of the $z_j$ are equal to $a$, so the total contribution to the logarithmic derivative is $3 \cdot \frac{1}{z-a}$. – Martin R Dec 01 '23 at 21:48
  • @MartinR so are the terms (z-a)^n being split into (z-a)(z-a)...(z-a) in your initial f(z) before you take the derivative? that way you end up only with powers of 1, and 1 as a coefficient? – ekorel Dec 01 '23 at 22:03
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    @ekorel: Yes, exactly. – Martin R Dec 01 '23 at 22:16
  • @MartinR just so you have an idea of where my confusion is, in the book that I'm reading through, the logarithmic derivative is written out like this, so I'm just confused as to where all of those constants in the numerator went, because when I go to do the Cauchy integral formula on the sums of the integrals (g(z)/(z-z^k)), they're still there.

    edit: oh, alright then. i had no clue that was possible, it felt like cheating! thank you for answering.

     – ekorel Dec 01 '23 at 22:18