Finding number of zero of $z^{10}+10z+9$ in the unit disc. 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.
 A: I'm not sure I understand how you concluded there are 10 zeros of $f$ in $D(0,2)$. It also seems to me that you can directly use Rouche's theorem on $D(0,1)$ if you are willing to slightly modify $f$.  
Hint: Since you know $f$ has a zero at $-1$, perhaps you could divide $f$ by something...
A: 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.
A: 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" $D(0,1)$.
A: Hint. Observe first that $z^{10}+10z+9=0$ possesses a double zero at $z=-1$. 
Target: Show that our equation has exactly 8 zeroes in $\{z:\lvert z\rvert>1\}$.
