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Complex analysis inequality proof

  1. Sep 18, 2011 #1
    Prove for all Z E C

    |ez-1| [itex]\leq[/itex] e|z| - 1 [itex]\leq[/itex] |z|e|z|

    I think this has to be proven using the triangle inequality but not sure how.

    Please help. :)

    thanks
     
  2. jcsd
  3. Sep 18, 2011 #2

    Dick

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    Use the triangle inequality along with the power series expansion e^z=1+z+z^2/2!+...
     
  4. Sep 19, 2011 #3
    great thanks...

    quick question:

    does |ez| = e|z| ????
     
  5. Sep 19, 2011 #4

    Dick

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    You tell me. Use the series expansion to compare them.
     
  6. Sep 19, 2011 #5
    thanks!
     
  7. Sep 19, 2011 #6

    Dick

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    Glad you got it. Just out of curiousity, what did you conclude about the truth of |e^z|=e^|z|?
     
  8. Sep 19, 2011 #7
    Once i expanded them i realized they looked exactly like the triangle inequality where the modulus of the summation of terms was less than or equal to the modulus of each term summed.

    i didnt try to conclude |e^z| = e^|z|

    from what i was reading i think they are equal when z is real and the inequality holds when z has some imaginary part. but not too sure...
     
  9. Sep 19, 2011 #8

    Dick

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    Good. That's about it. Except that they aren't necessarily equal when z is real and negative either, yes? It's only clearly true if z is real and postive.
     
  10. Sep 19, 2011 #9
    yeah good point.

    thanks for all your help!
     
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