Application of Rouche's theorem in complex analysis Let $p(z)=\sum_{j=0}^np_jz^j\in\mathbb{C}[z]$. Define
\begin{equation}
\tilde{p}(z)=z^n\overline{p(1/\overline{z})}=\sum_{j=0}^n\overline{p}_{n-j}z^j.
\end{equation}
Assume $p$ and $\tilde{p}$ have no common factors and $|p_0|>|p_n|$. Then $p$ and $\overline{p_0}p-p_n\tilde{p}$ have the same number of zeros in $\mathbb{D}.$
My guess is that since the statement is related to the number of zeros, I think we could use the Rouche's theorem. Then can we first show $p$ and $\overline{p_0}p$ has same number of zeros and then show $\overline{p_0}p$ and $\overline{p_0}p-p_n\tilde{p}$ has same number of zeros? I also notice that $\deg(\overline{p_0}p-p_n\tilde{p})<n$ if that helps.
 A: The result follows by an easy application of Rouche. First wlog assume $p_n \ne 0$ as otherwise there is nothing to prove since $|p_0|>0$ so $\bar p_0p$ has same zeroes as $p$.
The hypothesis implies that there is no root of $p$ on the unit circle and numbering them $a_1,..a_k, b_1,..b_m, |a_l|<1, |b_l|>1, k+m=n$, we also get that $|a_1..a_kb_1..b_m|>1$ (that is $|p_0/p_n|$ by the usual considerations), so in particular $m \ge 1$ and at least a root is outside the unit disc.
Clearly $p(z)=p_n(z-a_1)..(z-a_k)(z-b_1)..(z-b_m)$) , while the roots of $\tilde{p}$ are $1/\bar a_l, 1/\bar b_l$ so $\tilde{p}(z)=\bar p_0(z-1/\bar a_1)..(z-1/\bar a_k)(z-1/\bar b_1)..(z-1/\bar b_m)$
But now notice that on the unit circle $|\frac{z-a_l}{1-\bar a_lz}|=1$ so $|\frac{z-a_l}{z-1/\bar a_l}|=|\bar a_l|=|a_l|$ and similarly for $b_l$ we have $|\frac{z-1/\bar b_l}{1- z/b_l}|=1$ so $|\frac{z-b_l}{z-1/\bar b_l}|=|b_l|$
But this means that on the unit circle $|\frac{\bar p_0p(z)}{p_n\tilde{p}(z)}|=|\frac{\bar p_0p_n}{p_n \bar p_0}||\frac{(z-a_1)..(z-a_k)(z-b_1)..(z-b_m)}{(z-1/\bar a_l)..(z-1/\bar a_k)(z-1/\bar b_1)..(z-1/\bar b_m)}|=|a_1..a_kb_1..b_m|>1$, hence indeed $|p_n\tilde{p}(z)|<|\bar p_0p(z)|$ and Rouche applies, so we are done!
