Recall that \sum_{n=0}^\infty z^n = 1/(1-z) if |z|<1.
Question 7, 2019: How many zeros of the function h(z) = 1 + z^m + z^{2m} lie inside the annulus \{z : 1/2 < z < 2\}?
We should be thinking Rouche’s theorem. It states that if f(z) and g(z) are both analytic in and on C, and if |g(z)|<|f(z)| (i.e. g is dominated by f) on C, then f and f+g have the same number of zeros inside C.
On the outside circle or radius 2, z^{2m} dominates and has 2m zeros by the fundamental theorem of algebra. Then, f(z) = z^{2m} and g(z) = 1 + z^m, nothing that both are entire (analytic everywhere). On C_1, |z|=2 and we note that |g(z)| = |1 + z^m| \le 1 + 2^m < 2^{m+1} < 2^{2m} = |f(z)|. Therefore, f+g has 2m zeros inside C_1, viz a 2m-fold zero at 0.
Applying Rouche’s theorem again with |z^{m} + z^{2m}|\le 2^{-m} + 2^{-2m} \le 1/2 + 1/4 < |1| on the circle of radius 1/2, we see there are no zeros within this circle.
Finally, h has 2m zeros inside the given annulus.
For functions differentiable only in some subset, use C/R.