Prove that if $ f : D(0,1) \to D(0,1) $ is analytic with $ f(0) = 0 $, then $ g(z) = \frac{f(z)}{z} $ has a removable singularity at 0.
My thoughts so far:
Is this even a question? If $f$ is analytic, then we can write $ f(z) = \sum_{n=0}^\infty a_n z^n $ valid for all $ z \in D(0,1) $. Then $ f(0) = 0 $ gives that $ a_0 = 0 $. Then we can write a Laurent series $ g(z) = \frac{a_0}{z} + \sum_{n = 0}^\infty a_{n+1} z^n $, and as $ a_0 = 0 $, the point 0 is necessarily a removable singularity (and it's already been removed by setting $ f(0) = 0 $!) Am I missing something?
Thanks