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.

  • $\begingroup$ You should probably start with proving $|e^z - 1| \le |z|$ for $Re(z) \le 0$. $\endgroup$ – 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 $x\le 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.