# Finding a upper bound for a complex and algebraic function.

I am stuck in understanding a proof, where I don't really understand one step, which i specify below: (suposse that $$f$$ is holomorphic, thus analytic)

(I have no information about the holomorphism domain of $$f$$).

Result. If $$\phi(x,y)$$ is an algebraic function in both variables, i.e. there are polynomials $$P_k(x,y)$$ such that $$\begin{equation*} \sum_{k=0}^{N}P_k(x,y)[\phi(x,y)]^k = 0 \end{equation*}$$ and also let $$f$$ be a holomorphic function such that $$\begin{equation*} |f(z)|^2 = \phi(x,y) \end{equation*}$$ then $$|f(z)|^2 \leq C|z|^m$$, for some $$m$$.

EDIT. Based on @MartinR comment, $$f(z)=z+1$$ would be a counterexample for the inequality above, which makes it invalid. Instead, he suggested (and very well) that: $$\begin{equation*} |f(z)|^2 \leq C(1+|z|^m), \hspace{.2cm} \text{ for some m} \end{equation*}$$ which would also verify the demonstration I am talking about in the beginning.

This is a step that appears in a proof, and I tried to write it formally as a proposition/theorem above. I would like to know where this comes from, and even if it's valid, since it's not making much sense to me. Mainly coming here to understand the result and the proof itself, not building it by myself.

Thanks for all the help in advance.

• Replacing $\phi$ by $\phi f$ for some polynomial $f$ you can assume that $p$ monic in $\phi$. If $\phi(a,b)$ is too large then $p(\phi(a,b),a,b)=0$ can't hold. May 22, 2021 at 22:20
• By the way, holomorphic is equivalent to being analytic, but these are not equivalent to being entire. Entire functions are those which are holomorphic, or equivalently, analytic, in a neighbourhood of every point of $\mathbb{C}$. May 23, 2021 at 7:04
• @GradStudent So a function that is holomorphic in $\mathbb{C}$ is entire or that's not the case? Thanks for your help!
– xyz
May 23, 2021 at 10:03
• Entire $\Leftrightarrow$ holomorphic in $\mathbb{C}$ May 23, 2021 at 10:25
• Btw, you probably mean something like $|f(z)|^2 \le C(1+|z|)^m$, otherwise $f(z) = z+1$ is a counterexample. May 26, 2021 at 15:20