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Let $P(z)=\sum_{k=0}^{n}{a_kz^k}$ be polynomial that is injective in the open unit disc. Show that $|a_n|\le |a_1|/n$. I know that if $P$ is injective function than $P$ is conformal map and therefore $P'(z)\not=0$ for each $z$ such that $|z|<1$. I've tried using proof by contradiction, assuming $|a_n|>|a_1|/n$ and showing that there is $z_0$ such that $|z_0|<1$ and $P'(z_0)$=0 using Rouché's theorem but that only works when $P(z)=a_nz^n+a_1z+a_0$. I would like a hint.

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Let $z_1,\ldots,z_{n-1}$ be the zeros of $P'(z)$, (each repeated according to its multiplicity.) We have $$ P'(z)=na_n(z-z_1)(z-z_2)\cdots(z-z_n) $$ Since $P$ is injective then $P'$ does not vanish inside the unit disk and consequently $|z_k|\geq1$ for $1\leq k\leq n-1$. Hence $$ |a_1|=|P'(0)|=n|a_n|\cdot|z_1|\cdot|z_2|\cdots |z_{n-1}|\geq n|a_n|, $$ and the desired inequality follows.$\qquad\square$

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