Complex Analysis: Showing f is a Polynomial of Degree n

[AcC]

Homework Statement

Let f be analytic through C, suppose that |f(z)|<=M|z|^n for a real constant M and positive integer n. Show that f is a polynomial function of degree less than n.

 

[praharmitra]

If you can show that $f^{(k)}(0) = 0$ for $k\geq n$ that would establish that f(z) is a polynomial of degree less than n. Try using the Cauchy integral formula

 

[AcC]

If you can show that $f^{(k)}(0) = 0$ for $k\geq n$ that would establish that f(z) is a polynomial of degree less than n. Try using the Cauchy integral formula

Thanks a lot for hint..:)

I have solved this as below;

since f is analytic on C so f is differentiable on C, if we use Cauchy's Estimate then
we find
|f$$^{(k)}$$(z)|<=M.|z|^n.k!/|z|^k for k>=n

if take lim as z→0
then we find f$$^{(k)}$$(0)=0 for k>=n

is it true!