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.
Best Answer
Notice
$$f(z) = z^{10} + 10z + 9 = (z+1)(z^9-z^8+z^7-z^6+z^5-z^4+z^3-z^2+z+9)$$
and for $|z| < 1$, the non-constant part of the $2^{nd}$ factor is bounded above by:
$$\begin{align} & |z^9-z^8+z^7-z^6+z^5-z^4+z^3-z^2+z|\\ \le & |z^9|+|z^8|+|z^7|+|z^6|+|z^5|+|z^4|+|z^3|+|z^2|+|z|\\ < & 9|z| < 9 \end{align} $$ This means the $2^{nd}$ factor cannot vanish. As a result, $f(z)$ doesn't have any root "inside" $D(0,1)$.