# Show that for all $z \in \overline{D}(0;1)$, $(3-e)|z| \leq |e^z - 1|\leq |z|(e-1)$

Show that for all $z \in \overline{D}(0;1)$, $(3-e)|z| \leq |e^z - 1|\leq |z|(e-1)$

I think I'm supposed to use the following chain of inequalities $$|e^z -1|\leq e^{|z|}-1 \leq |z|e^{|z|}$$

But every time I try to solve it I get stuck because some term doesn't vanish or I get something similar but no the same I'm asked to prove. Any help or hint will be appreciate and thanks in advance!

• For the upper bound, use the fact that the maximum modulus of the function $\frac{e^z-1}z$ (which is analytic on $\bar D(0;1)$, right?) must be achieved on the boundary, that is, on the unit circle. Indeed, this analytic function doesn't vanish in the disk, so the minimum modulus principle also applies; that leads to the superior lower bound $(1-\frac1e)|z| \le |e^z-1|$. – Greg Martin Jul 20 '16 at 17:38
• @GregMartin Thanks for the comment! I didn't cover minimum modulus principle when I found this exercise, but it's a great way to solve it. – user313212 Jul 21 '16 at 10:43
• And just to remember why the minimum modulus principle holds and requires the function $f$ to be nonvanishing: just apply the maximum modulus principle to $1/f$! – Greg Martin Jul 21 '16 at 17:31

$$e^z - 1 = z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots$$
By the triangle inequality and the condition $\lvert z\rvert \le 1$, we have\begin{align}\lvert e^z - 1\rvert &\le \lvert z \rvert + \frac{\lvert z\rvert^2}{2!} + \frac{\lvert z\rvert^3}{3!} + \cdots\\ &\le \lvert z\rvert + \frac{\lvert z\rvert}{2!} + \frac{\lvert z\rvert}{3!} + \cdots = \lvert z\rvert\left(1 + \frac{1}{2!} + \frac{1}{3!} + \cdots\right) = \lvert z\rvert(e - 1)\end{align} On the other hand, by the reverse triangle inequality,\begin{align}\lvert e^z - 1\rvert &\ge \lvert z\rvert - \left\lvert\frac{z^2}{2!} + \frac{z^3}{3!} + \cdots\right\rvert \ge \lvert z\rvert - \frac{\lvert z\rvert^2}{2!} - \frac{\lvert z\rvert^3}{3!} -\cdots \\ &\ge \lvert z\rvert - \frac{\lvert z\rvert}{2!} - \frac{\lvert z\rvert}{3!} - \cdots = \lvert z\rvert\left[1 - \left(\frac{1}{2!} + \frac{1}{3!} + \cdots\right)\right] = \lvert z\rvert(1 - (e - 2)) = \lvert z\rvert(3 - e)\end{align}Hence, $(3 - e)\lvert z\rvert \le \lvert e^z - 1\rvert \le (e - 1)\lvert z\rvert$ on $\overline{D}(0;1)$.
• And by the way, I think it should be $(e-2)$ instead of $(2-e)$ on the last step, or then $(1+(2-e))$ instead of $(1-(2-e))$. – user313212 Jul 21 '16 at 10:49