complexnumber
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Homework Statement
Let z,w be complex numbers.
Homework Equations
Prove there is a real number \alpha < 1 such that
<br /> \left|\frac{z^7 + z^3 - i}{9} - \frac{w^7 + w^3 - i}{9}\right| \leq \alpha<br /> \left|z - w\right|<br />
The goal is to show that \displaystyle q(z) = \frac{z^7 + z^3 - i}{9} is a contraction mapping in a real analysis contraction mapping problem. I am stuck here maybe because of algebraic manipulation.
The Attempt at a Solution
Is this the same as proving the following inequality?
<br /> \begin{align*}<br /> \frac{\left|\frac{z^7 + z^3 - i}{9} - \frac{w^7 + w^3 -<br /> i}{9}\right|}{\left|z - w\right|} < 1<br /> \end{align*}<br />
If so then
<br /> \begin{align*}<br /> & \frac{\left|\frac{z^7 + z^3 - i}{9} - \frac{w^7 + w^3 -<br /> i}{9}\right|}{\left|z - w\right|} = \left|\frac{\frac{z^7 + z^3 - i}{9} -<br /> \frac{w^7 + w^3 - i}{9}}{z - w}\right| = \left|\frac{\frac{(z^7 + z^3) -<br /> (w^7 + w^3)}{9}}{z-w}\right| = \frac{1}{9} \left|\frac{(z^7 + z^3) - (w^7 +<br /> w^3)}{z-w}\right| \\<br /> = & \frac{1}{9} \left|\frac{z^7 - w^7 + z^3 - w^3}{z-w}\right| =<br /> \frac{1}{9} \left| \frac{(z-w)(z^6 + z^5w + \cdots + zw^5 + w^6) +<br /> (z-w)(z^2 + zw + w^2)}{z-w} \right| \\<br /> =& \frac{1}{9} |(z^6 + z^5w + z^4w^2 + z^3w^3 + z^2w^4 + zw^5 +<br /> w^6) + (z^2 + zw + w^2)|<br /> \end{align*}<br />
How can I proceed from here?