Fix $\sigma >0$. I'm looking for a function $g:\mathbb R^2\to\mathbb R_{\geq 0}$ that can be the modulus of an analytic function (so $g=|q|$ for some analytic $q$) and grows like the exponential function $e^{\sigma t}$ "in every radial direction". More precisely, I want $g$ to have the property that for any $\theta\in [0,2\pi)$,
$$\sigma=\inf\left\{\rho:g(r,\theta)=O(e^{\rho r})\text{ as }r\to\infty\right\}$$
where the argument of $g$ is in polar coordinates. Is there a "canonical" or "simplest" $g$ with this property?
Context:
Suppose $f$ is an entire function that is not a polynomial. We say that $f$ is of exponential type $\sigma\in[0,\infty]$ if
$$\sigma=\inf\left\{A>0:|f(z)|=O\left(e^{A|z|}\right)\text{ as }|z|\to\infty\right\}$$
Suppose the type of $f$ is $\sigma\in[0,\infty)$. I want to see if there's a decomposition of $f$ that writes it as a product of two entire functions, one of type $\sigma$ and the other of type $0$. My first idea was to write $f(z)=e^{\sigma z}h(z)$ and try to prove that $h(z)$ has type $0$, but it occurred to me that $e^{\sigma z}$ "grows at a $\sigma$ rate" only as $\Re (z)\to\infty$, so this won't work. This thinking leads to the explicit counterexample $f(z)=e^{-\sigma z}$, which would necessitate $h(z)$ to be $e^{-2\sigma z}$ and hence not have type $0$.
Thus, it seems like a decomposition $f(z)=q(z)\cdot r(z)$ ($q$ has type $\sigma$ and $r$ had type $0$) must have the property that $q$ "grows with a $\sigma$ rate" in every direction on the complex plane. Taking norms and writing $z=x+yi$ yields $|f(x+yi)|=|q(x+yi)|\cdot|r(x+yi)|$, so it seems like we need
$$g(x,y):=|q(x+yi)|$$
to grow like $e^{\sigma t}$ in a radially symmetric fashion. My original question was an attempt to work backwards, that is, find a $g$ with the desired property and use $g(x,y)=|q(x+yi)|$ to try and figure out a closed-form for $q$.
How about $$g(z) = |e^{\rho_1 z}+e^{-\rho_2 z}+e^{i \rho_3 z}+e^{-i \rho_4 z}|$$, $\rho_i >0$ is a positive constant.
Also Check Hadamard factorization theorem. https://en.wikipedia.org/wiki/Hadamard_factorization_theorem#:~:text=In%20mathematics%2C%20and%20particularly%20in,is%20named%20for%20Jacques%20Hadamard.