How many zero points does the polynomial $z^8-5z^3+z-2$ have within the open unit circle?
Hello,
consider $$ \gamma\colon [0,2\pi]\to\mathbb{C}, \varphi\longmapsto\exp(i\varphi) $$ and define the following functions $$ f(z):=-5z^3,~~~~~~~~~~g(z):=z^8+z-2. $$
This two functions are holomorphic and for all $z\in\mbox{rg}(\gamma)$ it is $$ \lvert f(z)\rvert=5,~~~~~~~~~~\lvert g(z)\rvert\leq 4, $$ so it is $\lvert g\rvert < \lvert f\rvert$ on $\mbox{rg}(\gamma)$.
Concerning to the theorem of Rouché the number of zero points of $f+g$ is the same as that of $f$ and this number is given by $$ \frac{1}{2\pi i}\int_{\gamma}\frac{f'(z)}{f(z)}\, dz. $$ It is $$ \frac{f'(z)}{f(z)}=\frac{3}{z},~~~~~~~~~~\frac{1}{2\pi i}\int_{\gamma}\frac{3}{z}\, dz=3 $$ So the given polynomial does have 3 zero points within the open unit circle.
Would be great to get a confirmation whether my result is correct.
With kindly regards!
math12