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Complex Analysis

  1. Jun 21, 2008 #1
    Hi.
    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!
     
  2. jcsd
  3. Jun 21, 2008 #2
    Start with
    [tex] |e^z-1| = |z + (e^z-z-1)| \ge \ldots [/tex]
     
  4. Jun 22, 2008 #3
    AHHh

    [tex]|e^z-1| = |z + (e^z-z-1)|=|z-(1+z-e^z)|\ge|z|-\left|-\sum^{\infty}_{n=2}\frac{z^n}{n!}\right|= |z|-|z|\left|\sum^{\infty}_{n=2}\frac{z^{n-1}}{n!}\right|\ge\left(1-\sum^{\infty}_{n=2}\frac{|z|^{n-1}}{n!}\right)|z|[/tex]
    [tex]\ge\left(1-\sum^{\infty}_{n=2}\frac{1}{n!}\right)|z|=(1-(e-2))|z|=(3-e)|z|[/tex]

    Brilliant how did you come up with the initial inequality:

    [tex]|e^z-1| = |z + (e^z-z-1)|[/tex]
     
  5. Jun 22, 2008 #4
    I'm not sure I can explain where I got the idea from. The first step was to guess that you have to start by using the inequality [itex] |a+b| \ge |a| - |b| [/itex], because that's one of the few inequalities I know that give a lower bound of the absolute vale of something. Then the trick is to choose a and b properly. To get something useful, [itex] |b| [/itex] had better be small. And 1+z are the first two terms of the Taylor series for e^z, so [itex] |e^z-z-1| [/itex] is small (if z is close to zero). Furthermore, [itex] b=e^z-z-1[/itex] implies that a=z, so you get a term |z| which looked promising.
     
    Last edited: Jun 22, 2008
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