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Please note: this question was asked before, but the solution provided does not work as far as I know; see How to find the number of roots using Rouche theorem?

We have $f(z) = z^{10} + 10z + 9$ and have to use Rouché's theorem to estimate the number of zeroes in $D(0,1)$. The theorem requires that we find a $g(z)$ for which

$|f(z)-g(z)|<|f(z)|+|g(z)|$ for all $z \in \partial D(0,1)$.

Just like in the aforementioned question, I wanted to divide by a factor $(z+1)$, and use Rouché's theorem on $h(z) = f(z)/(z+1)$ and $g(z) = 9$. However, if $z = -1$, then $|h(z)-g(z)| = 9$ and $|h(z)|+|g(z)| = 9$, so we do not have a strict inequality.

The reason for this is that $-1$ is a zero of multiplicity $2$ in $f(z)$. I then thought that dividing by $(z+1)$ again would help me, but now I am left with having to prove that

$|z^8 - 2z^7+3z^6-4z^5+5z^4-6z^3+7z^2-8z| < |z^8 - 2z^7+3z^6-4z^5+5z^4-6z^3+7z^2-8z+9| + |9|$ for all $z \in \partial D(0,1)$.

Maple and Wolfram both agree this is valid for all $z \in \partial D(0,1)$, but how do I prove this? Or is there a better function $g(z)$ that I can use?

Thanks

Community
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1 Answers1

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This needs a little consideration.

We show $f_1(z)=\frac{f(z)}{z+1}=z^9-z^8+z^7+\dots-z^2+z+9$ has no root in the disk $D(0,r)$ for any $0<r<1$, hence your assertion follows.

Since $|f_1-9|\le r^9+r^8+\dots+r<9$ on $\partial D(0,r) $, you can use the theorem.

Ma Ming
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