- #1

complexnumber

- 62

- 0

defined by

[tex]

\begin{align*}

p(z) =& z^7 + z^3 - 9z - i, \\

q(z) =& \frac{z^7 + z^3 - i}{9}

\end{align*}

[/tex]

1. Prove that [tex]p[/tex] has a zero at [tex]z_0[/tex] if and only if [tex]z_0[/tex] is a

fixed point for [tex]q[/tex].

If [tex]z_0[/tex] is a fixed point for [tex]q[/tex] then

[tex]

\begin{align*}

q(z_0) = \frac{z_0^7 + z_0^3 - i}{9} =& z_0 \\

z_0^7 + z_0^3 - i =& 9 z_0 \\

z_0^7 + z_0^3 - 9 z_0 - i =& p(z_0) = 0

\end{align*}

[/tex]

Hence [tex]z_0[/tex] is a zero for [tex]p[/tex].

If [tex]p[/tex] has a zero at [tex]z_0[/tex] then

[tex]

\begin{align*}

p(z_0) = z_0^7 + z_0^3 - 9 z_0 - i =& 0 \\

z_0^7 + z_0^3 - i =& 9 z_0 \\

\frac{z_0^7 + z_0^3 - i}{9} =& q(z_0) = z_0

\end{align*}

[/tex]

Hence [tex]z_0[/tex] is a fixed point for [tex]q[/tex].

2. Hence or otherwise show that [tex]p[/tex] has exactly one zero in

the closed unit disk [tex]D = \{ z \in \mathbb{C} : |z| \leq 1 \}[/tex].

How can I solve this?

3. Where are the other zeros?

Does this follow part 2?