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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.

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

$$ e^{z_1} - e^{z_2} = \int_\gamma e^z\,dz, $$

so

$$\lvert e^{z_1} - e^{z_2} \rvert= \left\lvert \int_\gamma e^z\,dz \right\rvert\le \ell(\gamma) \max_{z\in\gamma} |e^z| \le |z_1 - z_2| $$

since $|e^z| = |e^{x+iy}| = e^{x} \le 1$ on $\gamma$ because $x\le 0$.

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