<s>[INST] Number of zeros of a polynomial in the unit disk Suppose $m$ and $n$ are two nonnegative integers. What is the number of zeros  of the polynomial $(1+z)^{m+n}-z^n$ in the unit ball $|z|<1$?
Some calculations for small values of $m$ and $n$ suggests the following formula:
$$n-1+2\left\lfloor\frac{m-n+5}{6} \right\rfloor$$
Does this formula work for all values of $m$ and $n$?
 [/INST] Here is my solution of this problem. So we have next equation for the zeros:
$$
(1+z)^{n+m}=z^n
$$
We can modify it like that:
$$
\Bigl(1+\frac{1}{z}\Bigr)^n \Bigl(1+z\Bigr)^m = 1
$$
Next we can mark the first factor as $re^{i\varphi}$, so the equation splits into two ones:
$$
\Bigl(1+z\Bigr)^m = re^{i\varphi} \\
\Bigl(1+\frac{1}{z}\Bigr)^n = \frac{1}{r}e^{-i\varphi}
$$
Let's consider the first equation. We have to extract the root:
$$
1 + z = r^{\frac{1}{m}} e^{i\frac{\varphi + 2 \pi k}{m}}
$$
Next we take away $1$ and write down the module of $z$ using the condition $|z|<1$:
$$
|z|^2 = |r^{\frac{1}{m}} e^{i\frac{\varphi + 2 \pi k}{m}} - 1|^2 = r^{\frac{2}{m}} - 2 r^{\frac{1}{m}} \cos \frac{\varphi + 2 \pi k}{m} + 1 < 1
$$
Finally, we divide by $r^{\frac{1}{m}}$, so:
$$
r^{\frac{1}{m}} < 2 \cos \frac{\varphi + 2 \pi k}{m}
$$
At the same time we know that $r^{\frac{1}{m}}<1$, so we have to understand, which inequation is stronger. Let's mark $z=\rho e^{i\psi}$ and write down the following inequation:
$$
\Bigl|1 + \frac{1}{z}\Bigr|^2 = \rho^{-2} + 2\rho^{-1} \cos \psi + 1 > 1
$$
Next,
$$
\rho \cos \psi + 1 > \frac{1}{2}
$$
It's clear that the left part equals $\Re \sqrt[m]{re^{i\varphi}}$, so:
$$
r^{\frac{1}{m}} \cos \frac{\varphi + 2 \pi k}{m} > \frac{1}{2}
$$
Because of $r^{\frac{1}{m}} < 1$ we have:
$$
\cos \frac{\varphi + 2 \pi k}{m} > \frac{1}{2}
$$
Therefore $r^{\frac{1}{m}} < 1$ is stronger and we have $\cos \frac{\varphi + 2 \pi k}{m} > \frac{1}{2}$. Next, we perform the same procedure for the second equation, and finally obtain another inequation:
$$
\cos \frac{-\varphi + 2 \pi k'}{n} < \frac{1}{2}
$$
Let's transform these inequations:
$$
-\frac{m}{6} < \frac{\varphi}{2\pi} + k < \frac{m}{6} \\
\frac{n}{6} < -\frac{\varphi}{2\pi} + k' < \frac{5n}{6}
$$
We have to take away $\varphi$, so:
$$
-\frac{m}{6} - \frac{\varphi}{2\pi} < k < \frac{m}{6} - \frac{\varphi}{2\pi}
$$
And:
$$
-\frac{m}{6} + \frac{n}{6} - k' < k < \frac{m}{6} + \frac{5n}{6} - k'
$$
So, finally:
$$
\frac{n-m}{6} < \ell < \frac{5n+m}{6}
$$
where $\ell = k + k'$. The number of $\ell$'s satisfying this inequation defines the number of zeros in the unit disk.
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