# problem applying Rouche's theorem for counting zeros

• Apr 17th 2011, 08:02 AM
hatsoff
problem applying Rouche's theorem for counting zeros
Hey guys. I have the following theorem:

Quote:

Originally Posted by Rouche
Let $\displaystyle f$ and $\displaystyle g$ be holomorphic inside and on a contour $\displaystyle \gamma$ and suppose that $\displaystyle |f(z)|>|g(z)|$ on the image $\displaystyle \gamma^*$ of $\displaystyle \gamma$. Then $\displaystyle f$ and $\displaystyle f+g$ have the same number of zeros inside $\displaystyle \gamma$.

I'm supposed to count the zeros of, for example, z^5+15z+1 on D(0;2) (the open disk about 0 of radius 2). But this gives me contradictory results.

Notice that |z^5+1|\geq ||z^5|-1|=31>30=|15z| on the contour $\displaystyle |z|=2$. It follows by Rouche that $\displaystyle z^5+1$ and z^5+15z+1 have the same number of zeros on $\displaystyle D(0;1)$.

But now observe that |z^5|=32>31=15|z|+1\geq|15z+1| on the same contour $\displaystyle |z|=2$ So again by Rouche, $\displaystyle z^5$ and z^5+15z+1 have the same number of zeros on $\displaystyle D(0;1)$.

Putting these together, we see $\displaystyle z^5+1$ and $\displaystyle z^5$ have the same number of zeros on D(0;2). But we can see that $\displaystyle z^5+1$ has five zeros on D(0;2), while $\displaystyle z^5$ only has one zero.

Does anyone have a couple minutes to show me where I've gone wrong?

Thanks!

NOTE: The forum's latex compiler is apparently acting up, so forgive the code.
• Apr 17th 2011, 08:19 AM
Opalg
$\displaystyle z^5$ has a zero of order 5 at the origin, and that counts as five zeros for the purposes of Rouché's theorem. You must always count zeros according to their multiplicity.
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