# If $f(z)$ has order of growth $\rho$, then $f(z)/z^{\ell}$ has order of growth $\leqslant\rho$

Let $f$ be an entire function. We say that $f$ has order of growth $\leqslant\rho$ if there exists constants $A, B>0$ such that for all $z\in\mathbb{C}$, we have $$\large{|f(z)|\leqslant Ae^{B|z|^{\rho}}}$$ And we define the order of growth of $f$ to be $\inf \rho$ as where the infimum ranges over all $\rho$ such that $f$ has order of growth $\leqslant\rho$.

I am looking at the proof of Chapter 5, Proposition 2.1 (page 138) in Stein and Shakarchi's Complex Analysis. The following observation has been made:

If $f(z)$ is an entire function with order of growth $\rho$, and $z=0$ is zero of $f$ with multiplicity $\ell$, then $F(z)=f(z)/z^{\ell}$ has order of growth $\leqslant\rho$.

Could someone explain why this is true?

My attempt: If we try to estimate naively, we get $$\large{|F(z)|=\left|\frac{f(z)}{{z^{\ell}}}\right|\leqslant \frac{Ae^{B|z|^{\rho}}}{|z|^{\ell}}}$$ But the right hand side will blow-up as $|z|\to 0$.

• As $|z|\rightarrow 0$, you need to use the fact that $z=0$ is a zero of $f$ with multiplicity $\ell$ to show that the ratio is bounded. – mjqxxxx Nov 10 '13 at 1:52
• @mjqxxxx: Right. So I can see that $f(z)/z^{\ell}$ is bounded near $0$ (because $\ell$ is the multiplicity). But how does it let us extract information about $\rho$? Perhaps I need a further hint :) – Prism Nov 10 '13 at 2:03

Up to Wiki, the order of an entire function is expressed in terms of its coefficients by $$\rho= \limsup_{n \to \infty} \frac {n\log n}{-\log |a_n|} .$$ The $\ell$-shift of $a_n$ does not change $\rho.$