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Homework Help: Complex number inequality

  1. Mar 31, 2010 #1
    1. The problem statement, all variables and given/known data

    Let [tex]z,w[/tex] be complex numbers.

    2. Relevant equations

    Prove there is a real number [tex]\alpha < 1[/tex] such that
    [tex]
    \left|\frac{z^7 + z^3 - i}{9} - \frac{w^7 + w^3 - i}{9}\right| \leq \alpha
    \left|z - w\right|
    [/tex]

    The goal is to show that [tex]\displaystyle q(z) = \frac{z^7 + z^3 - i}{9}[/tex] is a contraction mapping in a real analysis contraction mapping problem. I am stuck here maybe because of algebraic manipulation.

    3. The attempt at a solution

    Is this the same as proving the following inequality?
    [tex]
    \begin{align*}
    \frac{\left|\frac{z^7 + z^3 - i}{9} - \frac{w^7 + w^3 -
    i}{9}\right|}{\left|z - w\right|} < 1
    \end{align*}
    [/tex]

    If so then
    [tex]
    \begin{align*}
    & \frac{\left|\frac{z^7 + z^3 - i}{9} - \frac{w^7 + w^3 -
    i}{9}\right|}{\left|z - w\right|} = \left|\frac{\frac{z^7 + z^3 - i}{9} -
    \frac{w^7 + w^3 - i}{9}}{z - w}\right| = \left|\frac{\frac{(z^7 + z^3) -
    (w^7 + w^3)}{9}}{z-w}\right| = \frac{1}{9} \left|\frac{(z^7 + z^3) - (w^7 +
    w^3)}{z-w}\right| \\
    = & \frac{1}{9} \left|\frac{z^7 - w^7 + z^3 - w^3}{z-w}\right| =
    \frac{1}{9} \left| \frac{(z-w)(z^6 + z^5w + \cdots + zw^5 + w^6) +
    (z-w)(z^2 + zw + w^2)}{z-w} \right| \\
    =& \frac{1}{9} |(z^6 + z^5w + z^4w^2 + z^3w^3 + z^2w^4 + zw^5 +
    w^6) + (z^2 + zw + w^2)|
    \end{align*}
    [/tex]

    How can I proceed from here?
     
  2. jcsd
  3. Apr 2, 2010 #2

    Mark44

    Staff: Mentor

    Are there some conditions on z and w that you don't show and aren't using? It doesn't seem to me that q(z) = (z7 + z3 - i)/9 is a contraction mapping, in general. For example, q(2+0i) = (128 + 8 - i)/9 has a magnitude considerably larger than 2. If q were a contraction mapping, I would expect |q(z)| <= |z|.
     
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