# Complex exponentials inequality series proof disc

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...

• I think you should write out the terms first. – IAmNoOne Feb 12 '15 at 2:38

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.)
• I'm probably missing something incredibly obvious but $2-(e^z-1)=3-e^z$. How did you get equality in the third line? – Tim Raczkowski Feb 12 '15 at 2:59
• @TimRaczkowski, I made a misprint, because I was supposed to prove for $e$, not $e^z$. – IAmNoOne Feb 12 '15 at 3:12