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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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Be careful to include your $|\cdot|$ signs, you should be saying $47\not\leq|z|^4$ and $31\leq 2|z|^4$. The field $\mathbb{C}$ is not ordered, so it rarely makes sense to say some number is less than $z\in\mathbb{C}$. – Adelaide Dokras May 25 '12 at 20:55
Thanks @Adeal , quick question -- so is what I've done above correct? As in when I used $f(z)=2z^4$, I have showed that $p(z)$ has 4 roots inside $|z|<2$? – Derrick May 25 '12 at 21:13
Yes, that is a correct application of Rouche's theorem, and $4$ roots would be the correct answer. – Adelaide Dokras May 25 '12 at 21:16
Okay, thanks for your help again! :) – Derrick May 25 '12 at 21:21
up vote 2 down vote accepted

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.

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Thanks @EricStucky , so is what was done in both cases above (with $2z^4$ and $z^4$) correct? As in, when I used $f(z)=2z^4$ in my proof above, can I then conclude that the no. of roots is 4? – Derrick May 25 '12 at 21:11
Yup! And thanks for the choice answers; I can make comments now :) – Eric Stucky May 26 '12 at 1:25

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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