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Theorem attached and proof.enter image description here

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 Answers 1

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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 $f$.

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    apologies actually, how do we know both f and g are continous? edit: oh holomorphic implies continous at every point where the function is defined? – 2017-01-24
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    correct, if $f$ is holomorphic at a point $z$, then $f$ is continuous at $z$. – 2017-01-27