If $0<|z|<1$, show that $\frac{1}{4}|z|<|1-e^z|<\frac{7}{4}|z|$ My question:
If $0<|z|<1$, show that $\frac{1}{4}|z|<|1-e^z|<\frac{7}{4}|z|$ ($z$ is complex)
what I have tried:
I tried to expand the middle term in its Taylor series but I can't get the appropriate bound.

The Maclaurin expansion of $e^z - 1$ is 
$$z + \frac{z^2}{2!} + \cdots + \frac{z^N}{N!} +\cdots \tag{*}$$
Since $0 < |z| < 1$ and $k! \ge 2\cdot 3^{k-2}$ for all $k \ge 2$, then by the triangle inequality, (*) is strictly bounded above in modulus by 
$$\left[1 + \frac{1}{2} + \frac{1}{2}\left(\frac{1}{3}\right) + \frac{1}{2}\left(\frac{1}{3}\right)^2+\cdots\right]|z| = \left(1 + \frac{1/2}{1-1/3}\right)|z| = \frac{7}{4}|z|.$$
By a similar argument, (*) is strictly bounded below in modulus by 
$$\left[1 - \left(\frac{1}{2} + \frac{1}{2}\left(\frac{1}{3}\right) + \cdots\right)\right]|z| = \left(1 - \frac{1/2}{1 - 1/3}\right)|z| = \frac{1}{4}|z|.$$
Thus,
$$\frac{1}{4}|z| < |e^z - 1| < \frac{7}{4}|z|\qquad (0 < |z| < 1).$$
Since $|e^z - 1| = |1 - e^z|$, the above inequality is the same as your inequality.
