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Use Rouche's theorem to find the number of roots of the polynomial $z^5+3z^2+1$ in the anulus $1<|z|<2$.

I am looking for a solution to this problem.


My thoughts:

This is a topic that was explained rather quickly at the end of class last week, and I'm struggling on how to answer it. I know that Rouche's Theorem says: If $f$ and $h$ are each functions that are analytic inside and on a simple closed contour C and if the strict inequality $|h(z)|<|f(z)|$ holds at each point on C, then $f$ and $f+h$ must have the same total number of zeros inside C.

My confusion with this problem begins with the fact that the theorem is describing two functions and I've only been given one. I believe that I can split up the function I've been given into two analytic functions and approach it with the theorem that way; but I'm not sure how to decide what to split the function into.

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1 Answer 1

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I don't think that there is a fixed method how you can always proceed. Generally, one tries to find a "dominant" term. In this case, the trick is to consider your polynomial in the circles $|z| < 1 $ and $|z| < 2 $ separately.

For $|z| = 1 $, $$ |z^5 + 1| \le 1 + 1 = 2 < 3 = |3z^2| \, , $$ so you can choose $h(z) = z^5+1$ and $f(z) = 3z^2$ and conclude that $z^5+3z^2+1$ has 2 zeros in $|z| < 1 $.

For $|z| = 2 $, $$ |3z^2 + 1| \le 3 \cdot 4 + 1 = 13 < 32 = |z^5| \, , $$ so you can choose $h(z) = 3z^2 + 1$ and $f(z) = z^5$ and conclude that $z^5+3z^2+1$ has 5 zeros in $|z| < 2 $.

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    $\begingroup$ You seem to have a $z^2$ becoming $z^3$ in your first part. Otherwise I agree entirely $\endgroup$
    – Sten
    May 7, 2015 at 21:17
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    $\begingroup$ @Sten: Oops – yes – fixed – thanks! $\endgroup$
    – Martin R
    May 7, 2015 at 21:19
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    $\begingroup$ @MartinR -- that looks great thanks! Given that the polynomial has 2 zeros in one circle and 5 in the other, would it be okay to assume that the polynomial has a total of 7 zeros in the annuls? $\endgroup$
    – Kristin
    May 7, 2015 at 21:56
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    $\begingroup$ @Kristin: No. The annulus is the difference between the disk $|z| < 2$ and the disk $|z| < 1$ ... $\endgroup$
    – Martin R
    May 7, 2015 at 22:06
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    $\begingroup$ @ab123: Yes, with multiplicity. $\endgroup$
    – Martin R
    Nov 13, 2017 at 18:46

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