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I am trying to prove this inequality

$$|e^z-1|\leq e^{|z|}-1\leq |z|e^{|z|}$$

I've tried calculating the difference in their power series

$$\sum_{k=0}^\infty\frac{|z|^k}{k!}-1-\left|\,\sum_{k=0}^\infty\frac{z^k}{k!}-1\,\right|$$

but did not get anywhere.

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    $\begingroup$ You know that $\frac{a^0}{0!} = 1$ presumably. And you probably know the triangle inequality. That suffices. $\endgroup$ Oct 19, 2015 at 22:39

1 Answer 1

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By Taylor series $$ |e^z-1|=\left|\sum_{n=1}^{\infty}\frac{z^n}{n!}\right|\leqslant\sum_{n=1}^{\infty}\frac{|z|^n}{n!}=e^{|z|}-1=|z|\sum_{n=1}^{\infty}\frac{|z|^{n-1}}{n!}<|z|e^{|z|} $$

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