# How to count the total zeros of a complex polynomial outside a closed curve?

### 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 $$N$$?

• 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.
– Bma
Jun 1 at 6:25
• 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.
– Bma
Jun 1 at 6:48
• 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.
– Bma
Jun 1 at 6:56

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 $$w=0$$).

• 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 $f(z)$? May 26 at 1:32
• 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. May 26 at 2:21
• 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., $M=\frac{1}{2\pi i}(w^nf(w))'/(w^nf(w))dw$? May 26 at 2:23
• @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 $n$. May 26 at 2:28
• @Guoqing I'm not sure I see how you can integrate $f()$ without knowing $n$ — after all, it's in the very definition of $f$! May 26 at 5:33

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.

• 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? May 26 at 6:59
• Our ability to chose a "sufficiently large circle" depends on the a priori knowledge of an upper estimate for the moduli of zeros. May 26 at 11:19
• @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 $\mathbb{CP^1}$). May 26 at 14:09