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Let m and n be positive integers, and let $p(z)=1+z+\frac{z^2}{2}+...+\frac{z^m}{2^{m-1}}+3z^n$. How many zeroes does $p$ have counting multiplicities in the unit disk.

How I solved it: On the unit disk $|1+z+\frac{z^2}{2}+...+\frac{z^m}{2^{m-1}}|<|3z^n|$ by employing triangle inequality and then showing the left side is a series that converges to 3 from below. Thus by Rouche's theorem we know $p$ has n zeroes counting muliplicities.

IS this correct?

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Your idea works but you did not prove rigorously that the inequality holds. Just note that $|1+...+\frac {z^{m}} {2^{m-1}} | \leq 1+\frac 1 2+...+\frac 1{2^{m-1}} <3=|3z^{n}|$ to complete the argument.

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    $\begingroup$ You missed a $+1$ in the beginning. Thank you for the response! $\endgroup$
    – 2132123
    Jul 3, 2020 at 0:10
  • $\begingroup$ @2132123 You are right. $\endgroup$ Jul 3, 2020 at 0:12

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