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I would like to find how many zeros $z^4-5z+1$ have in the annulus $\{z | 1\lt |z| \lt 2\}$.

I think I have to apply Rouche's theorem, but I don't know how. I would like some help.

Edit:

First, consider the circle $|z|=2.$ Let $f(z)=z^4$ and $g(z)=-5z+1$. On the curve $|z|=2$, $|g(z)|=|-5z+1|\leq |-5z|+|3|\leq 13$, and $|f(z)|=2^4=16.$ Thus, the hypothesis of Rouche's Theorem are satisfied. Now, since $f(z)=z^4$ has four zeros inside $|z|=2$, by Rouche's Theorem, $f(z)+g(z)= z^4 -5z +1$ also has four zeros inside $|z|=2.$

Now, consider the circle $|z|=1$. Let $f(z)=-5z+1~,g(z)=z^4$. Then on $|z|=1$, $|g(z)|=|z^4|=1.$ But $|f(z)|=|-5z+1|\lt |-5z| +|1|=4.$ So again, the hypothesis is satisfied. But $f(z)$ has only one zero inside $|z|=1$, so, $f(z)+g(z)=z^4-5z+1$ also has only one zero in $|z|=1$.. Hence $z^4-5z+1$ has $(4-1)=3$ zeros in the the annulus $\{z | 1\lt |z| \lt 2\}$.

Please, is the above right?

thanks.

  • 0
    [Alpha's solution](http://www.wolframalpha.com/input/?i=z^4%E2%88%925z%2B1%3D0)2011-11-12

1 Answers 1

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We can break this into simpler problems. Look at the number of zeros with $|z|<1$. Then find the number of zeros with $|z|<2$. Subtract the first from the second, and you're done. Both sub-problems should be simple applications of Rouche.

Depending on how you argue, you may have to check that there are no zeros with |$z|=1$ or $|z|=2$. Make sure to do so if necessary.

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    It looks fine to me.2011-11-12