Let $a,b\in \{\operatorname{Re}(z) <0\}$ then $|e^a - e^b | < |a-b|$.

Question

Let $a,b\in \{ \operatorname{Re}(z) <0\}$. Then I want to show $|e^a - e^b | < |a-b|$.

Here is my trial :

Let $a = a_1 + i a_2$ and $b=b_1 + ib_2$, then \begin{align} &e^{a} - e^b = e^{a_1} ( \cos(a_2) + i \sin(a_2) ) - e^{b_1} (\cos (b_2) + i \sin(b_2)) \end{align} and I tried to use $|x|-|y| \leq |x-y| \leq |x|+|y|$, but this does not give desired result.

Answer 1

It follows from the mean value theorem and from the fact that $\exp'=\exp$ that$$\left\lvert\frac{e^a-e^b}{a-b}\right\rvert\leqslant\max_{t\in[0,1]}\left\lvert e^{(1-t)a+tb}\right\rvert.$$But, for each $t\in[0,1]$,$$\left\lvert e^{(1-t)a+tb}\right\rvert=e^{\operatorname{Re}\left((1-t)a+tb\right)}<1,$$since $\operatorname{Re}\bigl((1-t)a+tb\bigr)=(1-t)\operatorname{Re}(a)+t\operatorname{Re}(b)<0$.

Answer 2

If $\gamma$ is the line segment from $a$ to $b$ then $\int_{\gamma} e^{z}dz=e^{b}-e^{a}$. This gives $|e^{b}-e^{a}| < |b-a|$ because $Re z<0$ for all $z$ on $\gamma$ and so $|e^{z}|=e^{Re z} <1$.