This question is from ponnusamy and silvermann Complex variables with applications ( Subsection: Argument principle) and I was unable to solve it.
Show that $z^4+ 4z-1$ has 1 root in the disk |z|<1/3 and remaining three roots in annulus 1/3 <|z|<2.
Attempt: I have proved the existence of 1 root in the disk |z|<1/3 using rouche's theorem and polynomial 1-4z but I am unable to choose 2nd function to use rouche's theorem whose roots lie in 1/3<|z|<2.
Kindly tell me which polynomial to choose.
Best Answer
Let $\varepsilon(z)=4z$ and let $f(z)=z^4-1$. Then $z^4+4z-1=f(z)+\varepsilon(z)$. Furthermore, if $|z|=2$, then\begin{align}\bigl|\varepsilon(z)\bigr|&=|4z|\\&=8\\&<15\\&\leqslant|z^4-1|\\&=\bigl|f(z)\bigr|.\end{align}So, your function has as many zeros on $D(0,2)$ as $f$, which has $4$ zeros there.