TITLE: Finding number of zero of $z^{10}+10z+9$ in the unit disc.
QUESTION [7 upvotes]: I am trying to find the number of zero of the polynomial $f(z)=z^{10}+10z+9$ in the disc $D(0,1)$.
So far I used Rouché's theorem with $g(z)=z^{10}$ to find that there are 10 zeroes in $D(0,2)$. However, as $-1$ is a zero of $f$ and is in $D(0,1)$, I think Rouché cannot be used directly.
Could anyone give me a advice on how to proceed ? Thanks.

REPLY [2 votes]: A possible approach is to use the logarithmic indicator principle. Since, by approximating the integrand function:
$$\frac{1}{2\pi i}\oint_{|z|=\frac{10}{9}}\frac{f'(z)}{f(z)}\,dz = 2,$$
it follows that there are just two roots (accounted with multiplicity) of $f(z)$ inside the disk $|z|\leq\frac{10}{9}$. Since $z=-1$ is a double root, there are no roots of $f(z)$ in the open unit disk.