How to show that this polynomial has all it's roots in in the disk $|z|<3$

Question

This question was asked in my analysis assignment and I am unable to solve it.So, I am looking for help here.

Show that the polynomial $z^3 -z^2 +4z+5=0 $ has all its roots in the disk |z|<3.

I thought of using Rouché's theorem but I am not able to choose a $g(z)$ which would work. So, I would be unable to provide anything as attempt as this is the starting point of the answer if question has to be solved using Rouché's theorem

Can you please help me with finding $g(z)$?

I hope Rouché's theorem would work in this question.

Answer 1

Take $f(z)=z^3$ and $\varepsilon(z)=-z^2+4z+5$. Then, $z^3-z^2+4z+5=f(z)+\varepsilon(z)$ and, if $|z|=3$,\begin{align}\bigl|\varepsilon(z)\bigr|&=|-z^2+4z+5|\\&\leqslant3^2+4\times3+5\\&=26\\&<27\\&=|z^3|\\&=\bigl|f(z)\bigr|.\end{align}So, $z^3-z^2+4z+5$ has as many zeros on the disk $D(0,3)$ as $f$, which has $3$ zeros there (if we count them with their multiplicities).

Answer 2

No theorem is needed for this. If $|z| \geq 3$ we get $|z^{3}-z^{2}| \geq |z|^{2} (|z|-1)\geq 9(|z|-1)$. But $|4z+5| \leq 5+4|z| <9(|z|-1) \leq |z^{3}-z^{2}|$. Hence, the equation cannot hold if $|z| \geq 3$.

Answer 3

This is a similar approach to Kavi's answer.

Assume by contradiction that one root $z$ satisfies $|z| \geq 3$.

Then $$ 1= \frac{1}{z}-\frac{4}{z^2}-\frac{5}{z^3}= \left| \frac{1}{z}-\frac{4}{z^2}-\frac{5}{z^3}\right| \\ \leq \left| \frac{1}{z}\right| + \left| \frac{4}{z^2}\right| + \left| \frac{5}{z^3}\right| \leq \frac{1}{3}+\frac{4}{9}+\frac{5}{27}=\frac{9+12+5}{27}=\frac{26}{27} $$ This is a contradiction.