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I have a question. I am asked to find the number of zeros of a complex polynomial $P(z) = z^6 + (1 + i)z + 1$ in the annulus $\frac{1}{2} < |z| < \frac{5}{4}$ by using Rouche's Theorem.

I got 6 zeros in the annulus. Since I argued by Rouche's THM that $P(z)$ has the same number of zeros as $z^6$ in the closed disk $|z|\leq \frac{5}{4}$ and same number of zeros as $g(z) = 1$ (i.e. no zero) in the closed disk $|z| \leq \frac{2}{3} > \frac{1}{2}$, so $P(z)$ has 6 zeros in the annulus.

Am I doing this right? If not, can someone show me some details of how to do it?

Thank you

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1 Answer 1

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On $|z|= \frac{1}{2}$, $$| z^6 + (1+i)z| \leq |z^6| + |(1+i)z| = |z|^6 + |1+i| |z| = \frac{1}{2^6} + \frac{\sqrt{2}}{2} < 1$$

so $f(z)=1$ and $P(z)$ have the same number zeros inside $|z| < \frac{1}{2}$.

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