Amin12 wrote: Let $P(z)=a_d z^d+\dots+ a_1z+a_0$ be a polynomial with complex coefficients. The $reverse$ of $P$ is defined by $$P^*(z)=\overline{a_0}z^d+\overline{a_1}z^{d-1}+\dots+\overline{a_d}$$(a) Prove that $$P^*(z)=z^d \overline{ P\left( \frac{1}{\overline{z}} \right) } $$(b) Let $m$ be a positive integer and let $q(z)$ be a nonconstant polynomial with complex coefficients. Suppose that all roots of $q(z)$ lie inside or on the unit circle. Prove that all roots of the polynomial $$Q(z)=z^m q(z)+ q^*(z)$$lie on the unit circle. (a) this is obvious: $\overline{ P\left( \frac{1}{\overline{z}} \right) } = \bar{a_0}+\bar{a_1} \frac{1}{z}+...+ \bar {a_d} \frac {1}{z^d} \implies P^*(z)=z^d \overline{ P\left( \frac{1}{\overline{z}} \right) }$ (b)(Note: we should have $q(0) \not = 0$.) write $q(x)=(x-z_1)...(x-z_n)$ where $|z_i| \le 1$. from (a) we have $q^{*}(x)=x^n \overline{ q\left( \frac{1}{\overline{x}} \right) }=x^n(\frac{1}{x}-\bar{z_1})...(\frac{1}{x}-\bar{z_n})$ $\implies q^*(x)=(1-x\bar{z_1})...(1-x\bar{z_n})$. if $Q(z)=0$ for some complex $z$; then we have:$$z^{m}q(z)=-q^*(z) \iff z^m(z-z_1)...(z-z_n)=-(1-z\bar{z_1})...(1-z\bar{z_n}) \implies |z^m(z-z_1)...(z-z_n)|=|(1-z\bar{z_1})...(1-z\bar{z_n})|(*)$$we can't have $z=0$; otherwise the $LHS$ of $(*)$ is zero, while the $RHS$ is equal to 1. if $z=z_i$ from $(*)$ we get $1-z \bar{z_j}=0$, therefore $1=z_i \bar{z_j}$ which means $|z_i|=|z_j|=1$ so $|z|=|z_i|=1$. So assume $z \not= z_i ,1\le i \le n$. We have 2 cases: case 1. $|z| \geq 1$ We will prove the following inequality : $$|z-z_i| \geq |1- \bar {z_i}z|(**)$$proof. it is sufficient to prove $(z-z_i)(\bar{z}-\bar{z_i}) \geq (1- \bar{z_i} z)(1- \bar{z} z_i) \iff |z|^2+|z_i|^2\geq 1+|z|^2|z_i|^2 \iff (|z|^2-1)(1-|z_i|^2)\geq 0$ which is true. from $(*),(**)$ we have: $|z^m|= \frac {1-z\bar{z_1}}{z-z_1}...\frac{1-z\bar{z_n}}{z-z_n}\le 1.$So we have $|z| \le 1$, but we knew $|z|\geq 1$, Thus $|z|=1$ case 2. $|z| \le 1$ prove the following inequality, every thing is similar to case 1:$$|z-z_i| \le |1- \bar {z_i}z|$$