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

Question

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?

Answer 1

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.