If $f(z)=z+a_2z^2+…+a_nz^n$ is injective on the unit disk, then $|a_2|≤frac12(n-1)$
7
1
$begingroup$
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 $a_nne0$ . 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, $$|a_n|=frac1{n,|r_1r_2...r_{n-1}|}leqfrac{1}{n}$$ 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.
...
