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I need to show that for all [tex]|z|\leq1 [/tex]:

[tex] (3-e)|z|\leq|e^{z}-1|\leq(e-1)|z|[/tex]

Now

[tex]|e^{z}-1|=\left|\sum^{\infty}_{n=1}\frac{z^{n}}{n!}\right|\leq\sum^{\infty}_{n=1}\left|\frac{z^{n}}{n!}\right|=|z|\sum^{\infty}_{n=1}\frac{\left|z\right|^{n-1}}{n!}\leq|z|\sum^{\infty}_{n=1}\frac{1}{n!}=\left(\sum^{\infty}_{n=0}\frac{1}{n!}-1\right)|z|=(e-1)|z|[/tex]

So the upper bound is done. I have no idea how to do the lower bound. I have tried different techniques but just can't seem to get it.

Please someone help!

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# Complex Analysis

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