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
    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. $\endgroup$ 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
    Dec 20, 2012 at 6:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.