Need help in proving that 3 roots of $z^4-6z+3=0$ lies in $|z|<2$

Question

This question is from Ponnusamy and Silvermann Complex Variables with Applications (subsection: Argument principle) page 346.

Show that there exists $1$ root in $|z|<1$ and $3$ roots in $|z|<2$ of the equation $z^4 -6z +3=0$.

I have proved that there exists a root in $|z|<1$ but I am unable to prove the assertion for $|z|<2$. The problem arises as I have to look at a third degree equation and I am unable to find a suitable candidate .

So, I am looking for your help!

Answer 1

Let $\varepsilon(z)=-6z+3$ and let $f(z)=z^4$. Then $z^4-6z+3=f(z)+\varepsilon(z)$. Furthermore, if $|z|=2$, then\begin{align}\bigl|\varepsilon(z)\bigr|&=|-6z+3|\\&\leqslant15\\&<16\\&=|z^4|\\&=\bigl|f(z)\bigr|.\end{align}So, your function has as many zeros on $D(0,2)$ as $f$, which has $4$ zeros there (if we count them with their multiplicities).