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Set up

Suppose $\gamma$ a simple closed curve, oriented in a counterclockwise direction. $f(z)$ is a complex polynomial $$ f(z)=a_nz^{n}+a_{n-1}z^{n-1}+\cdots+a_0. $$ We already know that the integral $$ N=\frac{1}{2\pi i}\oint_{\gamma}{\frac{f'(z)}{f(z)}dz} $$ which we called the winding number, gives the total zeros $N$ of $f(z)$ inside the closed curve $\gamma$. Now I want to know the total zeros $M$ outside $\gamma$ and this can be done exactly by the fundamental theorem of algebra, which leads to $$ M=\mathrm{total\ zeros\ of}\ f(z)\ \mathrm{in\ whole\ plane}\ -N. $$ My question is: is there an "integral way" instead of the "algebraic way" to count the number of zeros $M$ outside $\gamma$ like what we did for $N$?

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  • $\begingroup$ It seems like an equivalent problem is to ask whether there's any effective integral formula which can find the degree of $f$, given no information about $f$ other than the fact that it is a polynomial. Asking for an effective formula rules out doing things like using the argument principle in order to find the order of the pole at infinity, since that requires finding a contour which does not enclose any of the zeros, and we do not know how to do this a priori. It also rules out things like taking derivatives of $f$ until we reach one that vanishes identically. $\endgroup$
    – Bma
    Commented Jun 1, 2022 at 6:25
  • $\begingroup$ My intuition is that it is impossible to do something like this in general. If we wish to apply the argument principle, the requirements of the problem as I stated above would seem to essentially limit us to looking at the integrals $\int\limits_{\gamma_j} S_k(f(T_k(z))) dz$, where the $\gamma_j$ $(0 \le j < p)$ are a fixed multiset of closed contours and $S_k$, $T_k$ $(0 \le k < q)$ are a fixed multiset of rational functions. I think it should not be hard to show we can always find polynomials of distinct degrees which will give the same values for all these integrals. $\endgroup$
    – Bma
    Commented Jun 1, 2022 at 6:48
  • $\begingroup$ What I said above may be a little rough around the edges-- for instance we should probably also allow the integrands to depend on $f$ and any fixed number of its derivatives, since the argument principle involves the logarithmic derivative. $\endgroup$
    – Bma
    Commented Jun 1, 2022 at 6:56

2 Answers 2

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Yes: change variables to $w = 1/z$. Then $f(z) = f(1/w)$ has an a pole of multiplicity $n$ at $w=0$, and a zero at $1/z$ for each zero $z$ of $f$. The zeros $z$ not enclosed by $\gamma$ are precisely those for which $1/z$ is enclosed by the image of $\gamma$ in the $w$-plane. But the path switches orientation when we go from $z$ to $w$, so the integral is $-N$, which recovers $M = n - N$ (since that $-N$ is $M$ minus the multiplicity of the pole at $w=0$).

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    $\begingroup$ Thanks, your answer is very clear and helpful! I want to ask another question: Suppose $f(z)$ has a zero at $z_h$ inside $\gamma$ with a highest order (multiplicity) $h$. Is it possible to calculate $h$ by using some integral? In other words, is $h$ a topological aspect of $f(z)$? $\endgroup$
    – Guoqing
    Commented May 26, 2022 at 1:32
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    $\begingroup$ This answer assumes knowledge of $n$. But if you know $n$, there is a simpler answer: the number of zeros outside the curve is $n$ minus the number of zeros inside the curve. $\endgroup$
    – Alexandre Eremenko
    Commented May 26, 2022 at 2:21
  • $\begingroup$ By the way, this integral $\frac{1}{2\pi i}\oint_{\gamma}f'(w)/f(w)dw$ gives $-N$, does this mean that I need to drop the denominator $w^n$ if I just want to get $M$, i.e., $M=\frac{1}{2\pi i}(w^nf(w))'/(w^nf(w))dw$? $\endgroup$
    – Guoqing
    Commented May 26, 2022 at 2:23
  • $\begingroup$ @AlexandreEremenko Yes! This is exactly what I mean. I'm wondering that is there a way to directly calculate $M$ using a integral along $\gamma$ without any knowledge of $n$. $\endgroup$
    – Guoqing
    Commented May 26, 2022 at 2:28
  • $\begingroup$ @Guoqing I'm not sure I see how you can integrate $f()$ without knowing $n$ — after all, it's in the very definition of $f$! $\endgroup$
    – Steven Stadnicki
    Commented May 26, 2022 at 5:33
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This is just an extended comment to answer your concerns. Maybe it's helpful to treat $f$ as a black-box function, where all we know is that it's meromorphic on $\mathbb CP^1$ (of course, this means it's actually rational, but let's suppress that for now). In particular, we happen to know there's a pole at $\infty$. We can calculate the order by taking a sufficiently small circle around it (ie, a sufficiently large circle) that doesn't contain any other zeroes or poles of $f$. Then we can calculate the mentioned integral, call it $-n$. By symmetry, $n$ will equal the number of zeroes $-$ poles on the other side of the circle, ie all the others besides $\infty$. If we know $f$ is entire on $\mathbb C$ this tells us that the total number of zeroes of $f$ is equal to the order of the pole at $\infty$, and we can calculate this quantity directly with an integral on a sufficiently large circle, which is what you wanted.

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  • $\begingroup$ I'm confused about "we can calculate this quantity directly with an integral on a sufficiently large circle". The integral in question along a very large circle will give the total zeros $n$ instead of $M$, I guess? Could you give me more details about calculating $M$ by a single integral that you mentioned in your answer? $\endgroup$
    – Guoqing
    Commented May 26, 2022 at 6:59
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    $\begingroup$ Our ability to chose a "sufficiently large circle" depends on the a priori knowledge of an upper estimate for the moduli of zeros. $\endgroup$
    – Alexandre Eremenko
    Commented May 26, 2022 at 11:19
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    $\begingroup$ @AlexandreEremenko: It should be easy for the OP to get an effective bound $|P(z)| \geq C|z|^n$ for some constant $C = C(a_0, \dots, a_n)$ and sufficiently large $z$ (or to simply prove that $P$ has only finitely many zeros on $\mathbb{C}$ or $\mathbb{CP^1}$). $\endgroup$
    – anomaly
    Commented May 26, 2022 at 14:09

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