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I'm trying to use Rouché's theorem to find the number of roots of the function $f(z)=z^5+3z^4+9z^3+10$ on $\vert z\vert<2$. I know that the answer is $3$, but I have been unsuccessful in proving it. I've tried nearly every combination of the components of $f$ which also have $3$ roots, applying the theorem up to $3$ times, but to no avail. I've even resorted to trying unrelated polynomials, but there's a lot of those! A hint would surely benefit my sanity.

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Taking the geometric sequence $1,3,9$ in the leading coefficients into account, multiply with $(z-3)$ to get $$ g(z)=(z-3)f(z)=z^6-27z^3+10z-30 $$ which has the same number of roots inside the disk $|z|<2$. Now $$ g(2w)=64w^6-216w^3+20w-30 $$ on $|w|=1$ has clearly the third order term as the dominating one.

Lutz Lehmann
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