Analysis: fixed point, contraction mapping

[complexnumber]

Let $$p,q : \mathbb{C} \to \mathbb{C}$$ be
defined by
$$\begin{align*}<br /> p(z) =& z^7 + z^3 - 9z - i, \\<br /> q(z) =& \frac{z^7 + z^3 - i}{9}<br /> \end{align*}$$

1. Prove that $$p$$ has a zero at $$z_0$$ if and only if $$z_0$$ is a
fixed point for $$q$$.

If $$z_0$$ is a fixed point for $$q$$ then
$$\begin{align*}<br /> q(z_0) = \frac{z_0^7 + z_0^3 - i}{9} =& z_0 \\<br /> z_0^7 + z_0^3 - i =& 9 z_0 \\<br /> z_0^7 + z_0^3 - 9 z_0 - i =& p(z_0) = 0<br /> \end{align*}$$
Hence $$z_0$$ is a zero for $$p$$.

If $$p$$ has a zero at $$z_0$$ then
$$\begin{align*}<br /> p(z_0) = z_0^7 + z_0^3 - 9 z_0 - i =& 0 \\<br /> z_0^7 + z_0^3 - i =& 9 z_0 \\<br /> \frac{z_0^7 + z_0^3 - i}{9} =& q(z_0) = z_0<br /> \end{align*}$$
Hence $$z_0$$ is a fixed point for $$q$$.

2. Hence or otherwise show that $$p$$ has exactly one zero in
the closed unit disk $$D = \{ z \in \mathbb{C} : |z| \leq 1 \}$$.

How can I solve this?

3. Where are the other zeros?

Does this follow part 2?

 

[Tinyboss]

Show that q is a contraction map from D into D, i.e. that $$|q(w)-q(z)|<\alpha|w-z|$$ with $$\alpha<1$$. Then use a theorem about contraction maps.

 

[complexnumber]

Show that q is a contraction map from D into D, i.e. that $$|q(w)-q(z)|<\alpha|w-z|$$ with $$\alpha<1$$. Then use a theorem about contraction maps.

What metric should I use for complex number space? The question did not mention any metric function at all.

 

[complexnumber]

In $$D$$ the derivative $$q'$$ is not less than 1.

$$\begin{align*}<br /> |q'(z)| = <br /> |\frac{7z^6 + 3z^2}{9}| \leq \frac{1}{9}<br /> (|7z^6| + |3z^2|) \leq \frac{1}{9} (7 + 3) = \frac{10}{9}.<br /> \end{align*}$$

Is $$q : D \to D$$ is a contraction mapping in D?