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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.

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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...

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    $\begingroup$ To show whether there are other zeroes than $-1$ on the closed disc, I proceeded as follows: Following your hint I divided $f$ by $(z+1)^2$ to obtain $h(z)=z^8-2z^7+3z^6-4z^5+5z^4-6z^3+7z^2-8z+9$. Setting $g(z)=3z^6+5z^4+7z^2+9$ and using Rouché, we have for $\vert z\vert=1$ that $\vert h(z)-g(z)\vert \leq\vert z\vert^8+2\vert z\vert^7+4\vert z\vert^5+6\vert z\vert^3+8\vert z\vert=1+2+4+6+8=21<\vert g(z)\vert=\vert 3z^6+5z^4+7z^2+9 \vert =24$. As $g$ has no root on the unit disc (to be proven), neither does $h$. Is it correct ? $\endgroup$ – Luc M Dec 23 '14 at 12:19
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    $\begingroup$ As you point out, you would still need to show that your $g$ has no root on the unit disc. I'm also worried about your statement $|g(z)|=24$. It's true that $g(1)=g(-1)=24$, but what about, say, $|g(i)|$ ? $\endgroup$ – Max Dec 24 '14 at 6:57
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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" $D(0,1)$.

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    $\begingroup$ It seems inconsiderate to post a full solution after Max posted a very nice hint to let the OP think. $\endgroup$ – Pedro Tamaroff♦ Dec 23 '14 at 11:18
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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.

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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 $\{z:\lvert z\rvert>1\}$.

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