Suppose that an entire function f(z) satisfies \left|f(z)\right|\leq k\left|z\right|^n for sufficiently large \left|z\right|, where n\in\mathbb{Z^+} and k>0 is constant. Show that f is a polynomial of degree at most n.

**Answer**

Since f is entire, it is equal to a power series centered at zero with radius of convergence \infty, which must match its Taylor series there.

f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}z^n

Since |f(z)|\leq k|z|^m, Cauchy’s estimate gives

|f^{(n)}(0)|\leq \frac{n!k|z|^m}{R^n}

for all |z|=R. For n>m, letting R\rightarrow\infty, we see that |f^{(n)}|=0. It follows that f is a polynomial of degree \leq m.

**Attribution***Source : Link , Question Author : ron , Answer Author : Community*