TITLE: If $f(z)=z+a_2z^2+...+a_nz^n$ is injective on the unit disk, then $|a_2|≤\frac12(n-1)$
QUESTION [7 upvotes]: Suppose $f(z)=z+a_2z^2+...+a_nz^n$ is injective in $D:=\{z:|z|<1\}$.  Show that $|a_2|≤(n-1)/2$ and $|a_n|≤1/n$.

Partial Solution
Assume without loss of generality that $a_n\ne0$. If $f$ is injective, then $f'(z)≠0$ in $D$. Now, $$f'(z)=1+2a_2z+...+na_nz^{n-1}$$ By the fundamental theorem of algebra, $$f'(z)=na_n(z-r_1)...(z-r_{n-1})$$ where $r_1,...,r_{n-1}$ are the roots of $f'(z)$, and $$f'(0)=1=na_n(-1)^nr_1r_2...r_{n-1}$$ Note that $|r_i|\geq1$ for each $i=1,...,n-1$, hence $$|r_1r_2...r_{n-1}|\geq1$$ and, finally, $$|a_n|=\frac1{n\,|r_1r_2...r_{n-1}|}\leq\frac{1}{n}$$
What I cannot figure out, though, is how to prove that $|a_2|≤(n-1)/2$.  I have tried similar arguments, but cannot seem to get it.  Any hints would be greatly appreciated.

REPLY [5 votes]: Figured it out.  Just to answer my own question:  $$f'(z)=1+2a_2z+\ ...\ +na_nz^{n-1}=na_n(z-r_1)...(z-r_{n-1})$$
$$\implies2a_2=na_n(r_2r_3...r_{n-1}(-1)^{n-2}+r_1r_3...r_{n-1}(-1)^{n-2}+\ ...\ +r_1r_2...r_{n-2}(-1)^{n-2})$$
$$\begin{align} 
\implies|2a_2| & = n|a_n||r_2r_3...r_{n-1}(-1)^{n-2}+r_1r_3...r_{n-1}(-1)^{n-2}+\ ...\ +r_1r_2...r_{n-2}(-1)^{n-2}|\\
& \leq n\bigl(\frac{1}{n}\bigr)\bigl(|r_2r_3...r_{n-1}(-1)^{n-2}|+|r_1r_3...r_{n-1}(-1)^{n-2}|+\ ...\ +|r_1r_2...r_{n-2}(-1)^{n-2}|\bigr)\\
\end{align}$$
$$\implies |2a_2|\leq1+1+\ ...\ +1=n-1$$
$$\implies |a_2|\leq\frac{n-1}{2}$$