The problem is as follows:
$P(z) = z^n + a_1z^{n-1} + \cdots + a_n$ satisfies $|P(z)|\leq M$ when $|z|\leq 1$, prove:
1)If $R > 1$, then $|P(z)|\leq M R^n$ when $|z|\leq R$.
2)All zeros of $P(z)$ lie in $|z|<1+M$.
I have proved 1), and didn't figure out how it may be used to prove 2)...
I think on $|z|=1+M$ we might have $|P(z)-z|<|z^n|$, if so, Rouche's theorem will do the work. But I can't prove the inequality. I also tried induction on n, that didn't work either.
Any help will be appreciated!
For $|z| \le M+1$, $ |P(z)| \le M(M+1)^n $
Clearly, for $|z|=M+1$ , $|z^n|= (M+1)^n$
Now, let $Q(z) = P(z) - z^n = a_1z^{n-1} + a_2z^{n-2} + ...a_n$
Clearly, for $|z| \le1$, $|Q(z)| - |z^n| \le |Q(z) +z^n| = |P(z)| \le M$
$\implies |Q(z)| \le M+1 $, for $|z| \le 1$
Using the first part, you can show, $ |Q(z)| < (M+1)^n$ for $|z|<M+1$
So, on the boundary of circle $|z| = M+1$ , $Q(z) < |z^n|$
So, in the interior of circle, $z^n$ and $z^n + Q(z) = P(z)$ will have same no. of zeroes.