number of zeros of a complex polynomial

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.

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Alpha's solution – Bill Cook Nov 12 '11 at 1:29

1 Answer

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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And to add one thing: to apply Rouché's theorem, you want to write your function as the sum of two functions, one of which is always bigger than the other. For your polynomial, it makes sense to first look at $z^4$ or $5z$ or $1$ as one of those two functions. Which one is biggest when $|z|=1$? when $|z|=2$? – Greg Martin Nov 12 '11 at 2:01
@Potato: I have added to my answer. could you please check if it's right? Thanks. – Jon Nov 12 '11 at 2:16
It looks fine to me. – Potato Nov 12 '11 at 2:18