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Define $f(z)=z^4-4z^3+8z-2$. Find how many zeros (including multiplicity) the function has in $\{z\in\mathbb{C}:|z|<3\}$.

I tried using Rouché's-theorem on $\{z\in\mathbb{C}:|z|<3\}$. The dominant factor is $z$, so $$|f(z)|\le|z+8z|=g(z)$$g(z) has only one root so $f(z)$ also has one root. for $1<|z|<3$$$|f(z)|<|z^4+8z|=h(z)$$ and $h(z)$ has three roots in this range (the ones for the equation $z^3=-8$) so I get it has four roots in this domain but when I solve it using MATLAB, I get that in $1<|z|<3$ there are two roots and not three. What am I doing wrong?

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Take $g(z) = z^4-4 z^3$. Clearly if $|z|=3$ we have $g(z) \neq 0$.

If $|z| = 3$, then $|f(z)-g(z)| = |8 z-2| \le 26 < 27 = |z|^3 (4-3) \le |z|^3|z-4| = |g(z)|$.

Hence $f,g$ have the same zeros in $|z|<3$.

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  • $\begingroup$ so what was my mistake? $\endgroup$ – user65985 Oct 30 '14 at 2:56
  • $\begingroup$ Why did you pick your particular $g$? Actually, I'm not exactly sure what you are doing in the question. For example, I would say that $z^4$ is the dominant factor. $\endgroup$ – copper.hat Oct 30 '14 at 2:57
  • $\begingroup$ because if $|z|<1$ then I switched $z^4$ in $z$ and removed the minuses. According to MATLAB, it has one root in $|z|<1$ so I think the mistake is with $h(z)$ but alas I can't find it. $\endgroup$ – user65985 Oct 30 '14 at 3:00
  • $\begingroup$ Why are you dealing with $|z|<1$??? The question asks about $|z|<3$. $\endgroup$ – copper.hat Oct 30 '14 at 3:00
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    $\begingroup$ Look at the 'symmetric version' in en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem. Generally you are looking for some $g$ such that $|f(z)-g(z)| < |f(z)|+|g(z)| $ for $z$ on the boundary. $\endgroup$ – copper.hat Oct 30 '14 at 3:08

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