# Complex number inequality, $|e^{z_1}-e^{z_2}| \leq |z_1 - z_2|$ if $Re(z_1),Re(z_2) \leq 0$

I'm trying to show the complex inequality $|e^{z_1}-e^{z_2}| \leq |z_1 - z_2|$ holds if $Re(z_1),Re(z_2) \leq 0$. It seems intuitively obvious but I haven't been able to find something that works. Would appreciate a hint.

• You should probably start with proving $|e^z - 1| \le |z|$ for $Re(z) \le 0$. – Shitikanth Aug 27 '13 at 17:33

Let $$\gamma$$ be a straight line from $$z_2$$ to $$z_1$$ (since the left half plane is convex, $$\operatorname{Re} z \le 0$$ for all $$z\in\gamma$$). Then
$$e^{z_1} - e^{z_2} = \int_\gamma e^z\,dz,$$
$$\lvert e^{z_1} - e^{z_2} \rvert= \left\lvert \int_\gamma e^z\,dz \right\rvert\le \ell(\gamma) \max_{z\in\gamma} |e^z| \le |z_1 - z_2|$$
since $$|e^z| = |e^{x+iy}| = e^{x} \le 1$$ on $$\gamma$$ because $$x\le 0$$.