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Consider a polynomial $$p(z) = z^6 + 9z^4 + z^3 + 2z + 4 $$

I need to find which quadrant of the complex plane contains how many zeros that lie in unit circle. Also, I need to find which quadrant contains the two zeros of $p(z)$ that lie outside the unit circle.

By using Rouche's theorem, I have found that there are four zeros of $p(z)$ that lie inside the unit circle. But I am not sure how to assign them quadrants. Likewise for the zeros that are outside the unit circle.

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  • $\begingroup$ There are real roots, count them out first. The rest are conjugates and should be equally split between upper and lower halves of real axis. What remains is to find how many roots are in any one quadrant and you have everything. For the last, check math.stackexchange.com/questions/174832 $\endgroup$ – Macavity Apr 29 '16 at 18:03

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