Complex Analysis: Proving Bounds for |e^z-1|

[the1ceman]

Hi.
I need to show that for all $$|z|\leq1$$:

$$(3-e)|z|\leq|e^{z}-1|\leq(e-1)|z|$$
Now
$$|e^{z}-1|=\left|\sum^{\infty}_{n=1}\frac{z^{n}}{n!}\right|\leq\sum^{\infty}_{n=1}\left|\frac{z^{n}}{n!}\right|=|z|\sum^{\infty}_{n=1}\frac{\left|z\right|^{n-1}}{n!}\leq|z|\sum^{\infty}_{n=1}\frac{1}{n!}=\left(\sum^{\infty}_{n=0}\frac{1}{n!}-1\right)|z|=(e-1)|z|$$

So the upper bound is done. I have no idea how to do the lower bound. I have tried different techniques but just can't seem to get it.
Please someone help!

 

[Jitse Niesen]

Start with
$$|e^z-1| = |z + (e^z-z-1)| \ge \ldots$$

 

[the1ceman]

AHHh

$$|e^z-1| = |z + (e^z-z-1)|=|z-(1+z-e^z)|\ge|z|-\left|-\sum^{\infty}_{n=2}\frac{z^n}{n!}\right|= |z|-|z|\left|\sum^{\infty}_{n=2}\frac{z^{n-1}}{n!}\right|\ge\left(1-\sum^{\infty}_{n=2}\frac{|z|^{n-1}}{n!}\right)|z|$$
$$\ge\left(1-\sum^{\infty}_{n=2}\frac{1}{n!}\right)|z|=(1-(e-2))|z|=(3-e)|z|$$

Brilliant how did you come up with the initial inequality:

$$|e^z-1| = |z + (e^z-z-1)|$$

 

[Jitse Niesen]

I'm not sure I can explain where I got the idea from. The first step was to guess that you have to start by using the inequality $|a+b| \ge |a| - |b|$, because that's one of the few inequalities I know that give a lower bound of the absolute vale of something. Then the trick is to choose a and b properly. To get something useful, $|b|$ had better be small. And 1+z are the first two terms of the Taylor series for e^z, so $|e^z-z-1|$ is small (if z is close to zero). Furthermore, $b=e^z-z-1$ implies that a=z, so you get a term |z| which looked promising.