Theorem attached and proof.
I am stuck on
1) Where we get $|g(z)|\geq |a_m|/2 $ comes from so $a_{m}$ is the first non-zero fourier coeffient. So I think this term is $< |a_m|r^{m}$, from $r$ the radius of the open set, but I don't know how to take care of the rest of the higher terms through $a_{m}$ , is this some theorem or?
2) The conclusion thus $f(z)$ has only one zero at $z=z_0$ I think i'm being stupid but what is this being made from? We know $g(z_0) = a_{m} \neq 0 $ and $a_{0}=0$, but I dont understand.
Thanks
1) follows just from the definition of continuity. You don't need to worry about the other Taylor series coefficients. Note that $g(z_0) = a_m$, so by continuity $\lim_{z\to z_0}g(z) = a_m$. By definition, for every $\epsilon > 0$ there exists an $r > 0$ such that $\vert z - z_0 \vert < r$ implies $$\vert g(z) - a_m \vert < \epsilon,$$ i.e. $$\vert a_m \vert - \epsilon < \vert g(z) \vert < \vert a_m \vert + \epsilon.$$ Setting $\epsilon = \tfrac{\vert a_m \vert}{2},$ we conclude that $$\vert g(z) \vert > \frac{\vert a_m\vert}{2}$$ whenever $\vert z - z_0 \vert < r$.
2) Using 1), we conclude that throughout the disk $\vert z - z_0 \vert < r$, we have $$\vert f(z) \vert = \vert (z - z_0)^m g(z)\vert \geq \vert z - z_0\vert^m \frac{\vert a_m \vert}{2}.$$ The expression on the right clearly has only one zero in this disk (namely $z = z_0$), so the same is true of $f$.