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If $z$ is in a closed disc $\bar{D}(0;1)$, how do we prove that $(3-e)|z|\leq |e^z-1| \leq (e-1)|z|$ ?

I could attempt: $(3-\sum \frac{1}{n!})|z| \leq |\sum \frac{z^n}{n!} -1| \leq (\sum \frac{1}{n!}-1)|z|$, but I'm stuck...

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    $\begingroup$ I think you should write out the terms first. $\endgroup$ – IAmNoOne Feb 12 '15 at 2:38
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RHS

\begin{align} \left | \left (1 + z + \frac{z^2}{2!} + \dots \right ) -1\right | &= \left | z \left( 1 + \frac{z}{2} + \frac{z^2}{3!} + \dots \right ) \right | \\ &\leq |z|\left | \sum_{n = 0}^{\infty}\frac{z^n}{(n+1)!} \right | \\ &\color{green}{\leq |z|\left | \sum_{n = 0}^{\infty}\frac{1}{(n+1)!} \right | }\\ &= |z|\left | e^z -1 \right | \end{align} LHS

I believe we get $|z| \leq |1 - e^z|$ via minimum principle using $g(z) = \frac{|1 - e^z|}{|z|}$ which has a removable singularity at $z = 0$. Multiplying both sides by $3 - e$ completes the LHS. (gotta eat, will check later.)

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  • $\begingroup$ thanks !...filling spaces... $\endgroup$ – rebc Feb 12 '15 at 2:56
  • $\begingroup$ I'm probably missing something incredibly obvious but $2-(e^z-1)=3-e^z$. How did you get equality in the third line? $\endgroup$ – Tim Raczkowski Feb 12 '15 at 2:59
  • $\begingroup$ @TimRaczkowski, I made a misprint, because I was supposed to prove for $e$, not $e^z$. $\endgroup$ – IAmNoOne Feb 12 '15 at 3:12
  • $\begingroup$ OK. I'm still not understanding how you derive line 3 from line 2. $\endgroup$ – Tim Raczkowski Feb 12 '15 at 3:19
  • $\begingroup$ No, it's a mistake. Let me try to fix. $\endgroup$ – IAmNoOne Feb 12 '15 at 3:31

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