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Complex analysis proof

  1. Jun 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Given |z|<1 and n a positive integer prove that
    \left|\frac{1-z^n}{1-z}\right|\le n

    3. The attempt at a solution
    I try to find the maximum of the function by differentiation

    \frac{d}{dz}\frac{1-z^n}{1-z}=\frac{-nz^{n-1}*(1-z)+(1-z^n)}{(1-z)^2}=0\Rightarrow (1-z^n)=nz^{n-1}*(1-z)

    I then plug this in

    \left|\frac{nz^{n-1}*(1-z)}{1-z}\right|=n\left|z^{n-1}\right|\le n

    I guess this works. Does someone have another way to prove it?
  2. jcsd
  3. Jun 15, 2009 #2


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    Homework Helper

    Divide the expression inside absolute values and see what you have - use the bound on |z|.
  4. Jun 15, 2009 #3
    Ah that is clever

    \left|\frac{1-z^n}{1-z}\right|=\left|\sum_{k=0}^{n-1}z^k\right|\le \sum_{k=0}^{n-1}|z|^k\le n
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