Here is one of the comp questions I need to solve.

Find the smallest integer N such that the polynomial $p(z)=2z^5-9z+2012$ has a zero in the open disk of radius N centered at the origin. How many zeros does $P(z)$ have in this open disk?

So I started with the smallest integer 1 and ended up with the integer 4 where I chose $f(z)=2z^5-9z$ and $g(z)=2012$. Plugging the value N=4 gives me $|f(z)|>=2012$ and $|g(z)|=2012$ on $|z|=4$. From there I can not use Rouche's Theorem. My understanding is that to use Rouche's Theorem you have to have strict inequality hold. Since we are looking for the minimum integer, I do not think we can choose N=5 in this case.

Is there any other way than Rouche's Method? Or am I missing something?


1 Answer 1


Hint: Since $2*4^5=2048$, notice that $z=-4$ is a root of your polynomial. Now, the function $-9z+2012$ on the circle of radius $4$ is maximized when $z=-4$, so we see that we have a strict inequality everywhere except at this point. Making a slight pertubation in the contour allows us to conclude that $4$ roots lie within the disk of radius $4$, and one root lies on the disk at $z=-4$.

  • $\begingroup$ I think you mean $ 2.4^5 =2048 $. So you are still using the Rouche but in a slightly bigger circle, may be of radius 4+e, for e>0? $\endgroup$
    – Deepak
    Commented Dec 19, 2012 at 19:36
  • $\begingroup$ @Deepak: I was thinking of something slightly different. Imagine the circle of radius $4$ around the origin, and a circle of radius $\epsilon$ around the point $z=-4$. Take the outer figure created by joining these two. That is, it is a circle of radius $4$ with a very small bump around $z=-4$. You can use Rouche's theorem on this curve, and the $2z^5$ term will dominate everywhere. This proves what I claimed. $\endgroup$ Commented Dec 19, 2012 at 20:41
  • $\begingroup$ I think this make sense now. I see the point what you are trying to make here. Thanks a lot. By the way what goes wrong with my idea of having circle of slightly bigger than 4? I think that also works too because the dominating term still dominates in this bigger circle too. Am I correct? $\endgroup$
    – Deepak
    Commented Dec 20, 2012 at 6:09

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