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This question already has an answer here:

Let $z_1,z_2$ be two complex numbers with $\operatorname{Re}(z_1)\leq0$ and $\operatorname{Re}(z_2)\leq0$. I want to prove: $$\big|e^{z_2}-e^{z_1}\big|\leq\big|z_2-z_1\big|$$

I began by using the reverse triangle inequality: $\big|e^{z_2}-e^{z_1}\big|\geq\bigg|\big|e^{z_2}\big|-\big|e^{z_1}\big|\bigg|$

So, it must be shown that: $$\frac{\bigg|\big|e^{z_2}\big|-\big|e^{z_1}\big|\bigg|}{\big|z_2-z_1\big|}=\bigg|\frac{e^{\operatorname{Re}(z_2)}-e^{\operatorname{Re}(z_1)}}{z_2-z_1}\bigg|\leq1$$

Why is this true?

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marked as duplicate by Martin R, user21820, Daniel Fischer complex-analysis Jun 17 '18 at 9:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Note that, by the mean value theorem, there is some $z$ in the line segment joining $z_1$ to $z_2$ such that$$\left|\frac{e^{z_2}-e^{z_1}}{z_2-z_1}\right|\leqslant\left|e^z\right|.$$But $\operatorname{Re}(z_1),\operatorname{Re}(z_2)\leqslant0\implies\operatorname{Re}(z)\leqslant0$ and therefore $\left|e^z\right|=e^{\operatorname{Re}(z)}\leqslant1$.

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  • $\begingroup$ Regarding the last inequality, I would better say $\vert e^z \vert = e^{{Re}(z)} \le 1$. $\endgroup$ – mathcounterexamples.net Jun 14 '18 at 12:53
  • $\begingroup$ @mathcounterexamples.net I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Jun 14 '18 at 12:54
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Another way is to consider a segment connecting $z_1$ and $z_2$ that has $Re(z) \leq 0$. Then $|e^{z_2} - e^{z_1}| = |\int_{z_1}^{z_2} e^z dz| \leq \int_{z_1}^{z_2} |e^z| dz \leq \int_{z_1}^{z_2} 1 dz = |z_2 - z_1|$.

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