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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?

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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 $z=-4$.

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    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?2012-12-20