The order of growth of $e^z-1$

Question

Let $f$ be an entire function. If there exist a positive number $\rho$ and constants $A,B>0$ such that $$ |f(z)|\leq Ae^{B|z|^{\rho}}\ \text{for all}\ z\in{\Bbb C} $$ then we say that $f$ has an order of growth $\leq\rho$. We define the order of growth of $f$ as $$ \rho_f=\inf\rho $$ where the infimum is over all $\rho>0$ such that $f$ has an order of growth $\leq\rho$.

Here is my question:

Using the definition above, how can I find the order of $f(z)=e^z-1$?

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It's not hard to show that $$ |e^z-1|\leq e^{|z|} $$ and thus $f$ has an order of growth $\leq 1$. I guess the order should be $1$. Then for any $\varepsilon>0$, and $A,B>0$, I need a $z\in{\Bbb C}$ such that $$ |e^z-1|> Ae^{B|z|^{1-\varepsilon}}. $$ How can I go on?

Answer 1

As you said it has order of growth $\leq 1$.

Now if it had order of growth $\leq \rho<1$, we would have $$ |e^z-1|\leq Ae^{B|z|^\rho} \quad\forall z\in\mathbb{C}\qquad\Rightarrow\qquad |e^x-1|\leq Ae^{Bx^\rho} \quad\forall x>0. $$ Hence $$ e^x\leq |e^x-1|+1\leq(A+1)e^{Bx^\rho}\quad\Rightarrow\quad e^{x-Bx^\rho}\leq A+1$$ $$ \Rightarrow\quad x\left(1 -Bx^{\rho-1}\right)= x-Bx^\rho\leq\log(A+1)\quad\forall x>0. $$ Now the lhs clearly tends to $+\infty$ as $x$ tends to $+\infty$. Contradiction.

Answer 2

For any $\rho<1$, and $A,B>0$, one needs a $z\in{\Bbb C}$ such that $$ |e^z-1|> Ae^{B|z|^{\rho}}. $$

This is actually asking if the inequality has a solution. If we focus on $z=x>0$, then the inequality is $$ e^x-1> Ae^{Bx^{\rho}} $$ which is equivalent to $$ e^{x-Bx^\rho}-e^{-Bx^\rho}>A\tag{*} $$ But $$ \lim_{x\to+\infty}(e^{x-Bx^\rho}-e^{-Bx^\rho})=\lim_{x\to+\infty} (e^{x^\rho(x^{1-\rho}-B)}-e^{-Bx^\rho})=+\infty. $$ Thus ($*$) must have a solution.