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## Homework Statement

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

## Homework 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.

## 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?