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Induction: Absolute value of complex sequence

  1. Oct 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Define sequence [tex]\left[ {z_n} \right]} _{n=1}^{\infty}[/tex] of points in the complex plane with [tex]z_1 = 1[/tex] and [tex]z_{n+1} = (4+3i)z_n - 1[/tex]. Show that [tex]|z_n| \ge 4^{n-1}[/tex] for all [tex]n \ge 1[/tex].


    2. Relevant equations



    3. The attempt at a solution

    The base case of induction is true, since [tex]|z_1| \ge 4^{1-1} \Leftrightarrow 1 \ge 1[/tex].

    [tex]z_1 = 1[/tex]
    [tex]z_2 = 3+3i[/tex]
    [tex]z_3 = 2+21i[/tex]
    [tex]z_4 = -56+90i[/tex]
    [tex]z_5 = -2557-717i[/tex]
    [tex]z_6 = -8078-10539i[/tex]
    ...

    I figured I can find a find some [tex]f(n)[/tex], I'll perhaps be able to show the inductive step. However, I'm getting nowhere on that approach. Perhaps I am not intuitive enough, but the points seem to be all over the complex plane. Thoughts? Guidance? Ideas?!
     
  2. jcsd
  3. Oct 24, 2009 #2

    Dick

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    The absolute value |4+3i|=5. Think about that.
     
  4. Oct 27, 2009 #3
    I slept on it, yet no results were yielded. That usually means the problem is impossible, or retardedly simple. I'm pretty sure I know which one of those it is...

    |4+3i|=5 would seem very useful if [tex]z_{n+1} = (4+3i)z_n[/tex], but unfortunately, [tex]z_{n+1} = (4+3i)z_n - 1[/tex]. I don't know what to do with it, since I can't figure out how to treat [tex]|(4+3i)z_n - 1|[/tex].
     
  5. Oct 27, 2009 #4

    Dick

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    It's not retardedly simple, but it's not impossible either. There's a thing called the inverse triangle inequality. |a-b|>=||a|-|b||. Can you find a way to use that as well?
     
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