Brouwer's fixed point theorem: root of polynomial in complex plane

I wish to use Brouwer's fixed point theorem to prove that there is a solution in the closed disc $|z|\leq 1$ to the equation $z^4 - z^3 + 8z^2 + 11z + 1=0$.

Please could you help me solve this. I believe I need to find a continuous function $f:D \to D$, where $D = \{z: |z|\leq 1\}$, such that $f(z)=z$, and then rearrange this to the equation above to show there is a fixed point, but I cannot think of what function $f$ to use.

Thanks

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I you want to get the form $f(z) = z$ you have to bring $z$ on the other side of your equation. But that alone won’t be enough. The easiest thing to do then is to divide your equation by a number $n$ in the beginning. Those two steps will give us:

$$f_n(z) := \frac1n(z^4-z^3+8z^2+(11-n)z+1) \stackrel{!}{=} z$$

To apply Brouwers fixed point theorem we need to show $f_n$ maps $D$ into itself. So taking the triangle inequality we get

$$|f_n(z)| ≤ \frac1n(|z^4|+|z^3|+8|z^2|+|11-n|⋅|z|+1)$$

and for $|z|≤1$ this gives

$$|f_n(z)| ≤ \frac{11+|11-n|}{n} = 1,\quad\text{as long as }n≥11$$

So $f_{11}$ is the continuous function you are looking for, mapping $D$ into $D$, thus having a fixed point in it, which is the solution to your polynomial equation.

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Rewrite the equation as $$z=\frac{-z^4+z^3-8\,z^2-1}{11}=f(z).$$ Then if $|z|\le1$ $$|f(z)|\le\frac{1+1+8+1}{11}=1.$$

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