How to show that $|\exp(z)-1|\le2|z|$ for $|z|\le 1$


now it remains to prove that the sum, the most right is $\le|z|$

Since $\sum\limits_{k=2}^\infty\frac{|z|^k}{k!}\le\sum\limits_{k=0}^\infty\left(\frac{|z|}{2}\right)^k-\frac{|z|}{2}-1+\frac{|z|^2}{4}$ (geometric series)

the RHS is $\displaystyle \frac{1}{1-\frac{|z|}{2}}-\frac{|z|}{2}-1+\frac{|z|^2}{4}=\frac{\frac{|z|}{2}}{\frac{2}{|z|}-1}+\frac{|z|^2}{4}\le|z|$

am I right ?

  • 2
    $\begingroup$ Simpler: use $\lvert z\rvert^k \leqslant \lvert z\rvert$ for $k \geqslant 1$ and $\lvert z\rvert \leqslant 1$. That gives you a bound of $(e-1)\lvert z\rvert$. (But yes, your method also works.) $\endgroup$ – Daniel Fischer Aug 18 '15 at 14:28
  • $\begingroup$ looks fine to me $\endgroup$ – DanielWainfleet Aug 18 '15 at 17:38

Yes, your proof is correct.

Here is an alternative one: $$e^z-1 = \int_0^z e^{\xi} \, d\xi = \int_0^1 e^{tz} \, dt$$ implies, by the triangle inequality, $$|e^z-1| \leq |z| \int_0^1 \underbrace{|e^{tz}|}_{e^{t Re z} \leq e^{t}} \, dt \leq |z| \cdot |e-1| \leq 2|z|$$


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