Suppose $f: \mathbb C \times (-1,1) \to \mathbb C$ is a smooth function that satisfies $f(0,t)=1$ for all $t\in (-1,1)$. Assume that for any $k\in \mathbb N$, any $z \in \mathbb C$ and any $t \in (-1,1)$ there holds
$$ \frac{\partial^k f}{\partial t^k}(z,0)=z^k.$$
Does there exist constants $c,\delta>0$ such that
$$ e^{-c|z|} \leq |f(z,t)| \leq e^{c|z|},$$
for all $|t|\leq \delta$ and all $z\in \mathbb C$?



Family of functions with prescribed derivatives
A counterexample:
$$f(z,t):=e^{tz}[1+(e^{|z|^2}-1)h(t)],$$
where $h(t):=e^{-1/|t|}$ for $t\ne0$, with $h(0):=0$.
Then all the assumptions on $f$ hold, but the conclusion
$$|f(z,t)|\le e^{c|z|}\ \;\forall z\in\mathbb C \tag{1}\label{1} $$
fails to hold for any real $t\ne0$ and any real $c$.




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If now
$$f(z,t):=e^{tz}[1+e^{|z|^2}\sin(|z|^2)\,h(t)],$$
with the same $h$ as before, then all the assumptions on $f$ hold, but both the conclusion \eqref{1} and the conclusion
$$e^{-c|z|}\le|f(z,t)|\ \;\forall z\in\mathbb C \tag{2}\label{2} $$
will fail to hold for any real $t\ne0$ and any real $c$.




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In the counterexamples above, we can replace $e^{|z|^2}$ and $\sin(|z|^2)$ by $e^{z^2}$ and $\sin z$, respectively (and then consider only the real values of $z$ to see that the conclusions \eqref{1} and \eqref{2} will still fail to hold for any real $t\ne0$ and any real $c$). The bonus provided by this replacement is that then $f(z,t)$ will be analytic in $z$, which will address the [comment by G. Fougeron](https://mathoverflow.net/questions/436577/family-of-functions-with-prescribed-derivatives/436578#comment1125207_436577).


