Complex Analysis Inequality Involving Exponential I am trying to prove the following result:
For $0<|z|<1$, $\frac{|z|}{4}< |e^z - 1| < \frac{7|z|}{4}$.
My attempt :
First, I converted the problem to (after mult. by 4):
$$|z|< 4|e^z - 1| < 7|z|.$$
Then, I subtracted $|z|$ so that:
$$0< 4|e^z - 1|-|z| < 6|z|.$$
Note that:
$$4|e^z - 1| \leq 4\left(\sum_{i=0}^\infty \frac{|z|^k}{k!}-1\right) \\ \leq|z|(4\frac{4}{1-|z|} -1) \\ = |z|(\frac{3+|z|}{1-|z|}).$$
Which from then we could take the "Sup" over all $\exp|z|$ so that
we get the above is bounded by something like $4 + \exp|z|$.
This is where I get stuck on this. Any pointers would definitely be appreciated.
Thanks.
 A: From the Taylor series for the exponential function you get
for $0 < |z| < 1$ (with $r := |z|$):
$$
\left| \frac{e^z - 1}{z} - 1 \right | =
\left| \frac{z}{2!} +  \frac{z^2}{3!} +  \frac{z^3}{4!} + ... \right | \\
\le \frac{r}{2!} +  \frac{r^2}{3!} +  \frac{r^3}{4!} + ...\\
= \frac{r}{2} \left( 1 + \frac{r}{3} + \frac{r^2}{3\cdot 4} + ...\right) \\ \le
\frac{r}{2} \left( 1 + \frac{r}{3} + \frac{r^2}{3^2} + ...\right)
$$
Since $r <1$, the last expression is less than
$$
\frac{1}{2} \left( 1 + \frac{1}{3} + \frac{1}{3^2} + ...\right) 
= \frac{1}{2} \frac{1}{1-\frac{1}3} = \frac 34
$$
using the geometric series. It follows that
$$
1 -\frac 34 < \left| \frac{e^z - 1}{z} \right | < 1 + \frac 34
$$
which is the desired inequality.
A: We start with the second inequality.
You have more or less established that $|e^z-1|\leq|e^{|z|}-1|$ so it is sufficient to prove the second inequality for positive $z.$
Now for such $z$
$$|e^z-1|=z\sum_{n=1}^\infty\frac{z^{n-1}}{n!}=z\left(1+z\sum_{k=0}^\infty\frac{z^k}{(k+2)!}\right)$$
But $(k+2)!\geq2k!$ so the right hand side is bounded by $z(1+\frac z2e^z).$ On the other hand the number $e$ is bounded by
$$1+\frac11+\frac1{1.2}+\frac1{1.2.3}\sum_{n=0}^\infty\frac1{4^n}<3$$
So our expression is no larger than $z(1+\frac32).$
Now the first inequality.
We can express $|e^z-1|$ as
$$z(1+\frac z2(1+\frac z 3(1+\frac z 4\sum_{k=0}^\infty\frac{z^k}{(k+4)!})))$$
the absolute value of which is at least
$$|z|(1-|\frac z2(1+\frac z 3(1+\frac z 4\sum_{k=0}^\infty\frac{z^k}{(k+4)!}))|)$$
The expression after the minus sign is bounded by
$$\frac12(1+\frac13(1+\frac14e))<\frac{17}{24}$$
so that
$$|e^z-1|>|z|(1-\frac{17}{24}).$$
