showing removable singularity at origin. Let $f$ be holomorphic in the punctured disk $\{z:0< \vert z \vert<2\}$ such that
$$\vert f(z) \vert \leq \bigg(\log \frac{1}{\vert z \vert}\bigg)^{100}, \space \text{in $\vert z \vert \leq \frac{1}{2}$}$$
and
$$\vert f(z) \vert = 1, \space \text{on $\vert z \vert = 1$}$$
I need to show $f$ has a removable singularity at the origin. Does this mean I need to find an analytic function $g$ defined on an $\epsilon$ ball about the origin agrees with $f$ for $0< \vert z \vert < \epsilon$? So is this by construction? Am I contracting such an analytic $g$?

Note that$$\lim_{z\to0}\bigl|zf(z)\bigr|=\lim_{z\to0}|z|\left(\log|z|\right)^{100}=0.$$Therefore, if $g(z)=zf(z)$, then $g$ has a removable singularity at $0$ (by Riemann's theorem) and, if you extend it to $0$, defining $g(0)=0$, $g$ is analytic (by the same theorem). Since $g(0)=0$, you can write $g(z)$ near $0$ as $a_1z+a_2z^2+\cdots$, and therefore near $0$ you have $f(z)=a_1+a_2z+\cdots$.
