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The sharp upper bound is relatively easy to find: $$|e^z - 1| = \left|\sum_{n = 1}^\infty \frac{z^n}{n!} \right| \leq \sum_{n = 1}^\infty \frac{|z|^n}{n!} = e^{|z|} - 1 = e - 1$$ and it is attained at $z = 1$.

I am wondering if there is a simple way to obtain a positive lower bound.

I am suspecting a sharp lower bound is $(1 - 1/e)$ but I cannot prove it. I was told that $(3 - e)$ is a positive lower bound, but I could not prove it neither.

Remark: One can do some horrible single variable calculus by writing $z = e^{it}$, where $t \in [0, 2\pi]$ but I would like to know if there is another (possibly much simpler) way to find a nontrivial lower bound.

Edit: It seems that I got many answer like the following but they are deleted by the author very soon. I think it is a good idea to show a wrong attempt so I put it here:

Putting $z = e^{it}$, then

\begin{align*}|e^z - 1|^2 & = |e^{2 \cos t} - 2e^{\cos t} \cos (\sin t)) + 1| \\ & \geq |e^{\cos t} - 1|^2 \\ & \geq (1 - 1/e)^2 \end{align*}

The last estimation is WRONG when $t = \pi/2$.

(End of edit)

E.E.
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  • Studying $e^{2\cos t} +1-2e^{\cos t}\cos(\sin t)$ should be enough in order to obtain the desired lower bound. Hint: this function is even in $t$ so you can assume $t\in [0,\pi]$, and you can put $x=\cos t$ in this range. – F.Battistoni Jun 17 '21 at 14:02

1 Answers1

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Just got it from my friend how to get (3 - e) as a lower bound:

\begin{align*} |e^z - 1| & = \left| \sum_{n = 1}^\infty \frac{z^n}{n!} \right| \\ & \geq |z| - \left| \sum_{n = 2}^\infty \frac{z^n}{n!} \right| \\ & = 1 - \left| \sum_{n = 2}^\infty \frac{z^n}{n!} \right| \\ & \geq 1 - \sum_{n = 2}^\infty \frac{|z|^n}{n!} \\ & = 1 - \sum_{n = 2}^\infty \frac{1}{n!} \\ & = 3 - e \end{align*}

However, the sharp bound remains open.

E.E.
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