# Proof: complex exponential inequality [duplicate]

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?

## marked as duplicate by Martin R, user21820, Daniel Fischer complex-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 17 '18 at 9:08

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 $\left|e^z\right|=e^{\operatorname{Re}(z)}\leqslant1$.
• Regarding the last inequality, I would better say $\vert e^z \vert = e^{{Re}(z)} \le 1$. – mathcounterexamples.net Jun 14 '18 at 12:53
Another way is to consider a segment connecting $z_1$ and $z_2$ that has $Re(z) \leq 0$. Then $|e^{z_2} - e^{z_1}| = |\int_{z_1}^{z_2} e^z dz| \leq \int_{z_1}^{z_2} |e^z| dz \leq \int_{z_1}^{z_2} 1 dz = |z_2 - z_1|$.