Let $p(z)=2z^4-2z^3+2z^2+7$. I'm trying to determine the number of roots of $p(z)$ that lie inside $\{z\in\mathbb{C}:|z|<2\}$
I used Rouche's theorem and I'm trying to find find $f(z)$ such that $|f(z)-g(z)|<|f(z)|$.
Let $f(z)=z^4$, $g(z)=2z^4-2z^3+2z^2+7$. Let $|z|=2$.
I get
$|f(z)-g(z)|\\=|-z^4+2z^3-2z^2-7|\\ \le|z|^4+2|z|^3+2|z|^2+|7|=47\not\le z^4=16$
But if I let $f(z)=2z^4$, $g(z)=2z^4-2z^3+2z^2+7$ then I get $|f(z)-g(z)|\\=|2z^3-2z^2-7|\\ \le2|z|^3+2|z|^2+|7|=31\le 2z^4=32.$
So is the number of roots of $p(z)$ that lie inside $\{z\in\mathbb{C}:|z|<2\}$, $4$? If so, whats the difference between $f(z)=2z^4$ and $f(z)=z^4$?
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$\begingroup$As in your other question, make sure you are doing proofs in the correct direction.
As to the difference between $2z^4$ and $z^4$: In general, Rouché's theorem is not bidirectional, by which I mean that even if $|f(z)-g(z)|\nless|f(z)|$ for appropriate $z$ it might still be the case that $f$ and $g$ have the same number of zeros inside the region.
$\endgroup$ 2 $\begingroup$Take $f(z)=z^4, g(z)=z^4-2z^3+2z^2+7$, Then on $|z|=2, |f(z)|>|g(z)|$, so inside $|z|=2$, $f$ and $f+g=p(z)$ has same number of roots.
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