Prove that if $ f : D(0,1) \to D(0,1) $ is analytic with $ f(0) = 0 $, then $\frac{f(z)}{z} $ has a removable singularity at 0 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
 A: Yes, that's one way to see it.  In general, if $f$ has a zero of order $m$ at $a$, then $f(z)=(z-a)^mg(z)$ for some analytic function $g$ with $g(a)\neq 0$, and $g$ is the analytic extension of $\frac{f(z)}{(z-a)^m}$ whose domain includes $a$.  The statement that $f$ maps into the unit disk is irrelevant.
In this case, you could also use the definition of the derivative to observe that $\displaystyle{\lim_{z\to 0}g(z)=f'(0)}$.
You write, "it's already been removed." That is not quite accurate, but this is just a technicality due to the fact that $\frac{f(z)}{z}$ can't be evaluated directly at $0$; you would get $\frac{0}{0}$.  It is removable because there is a limit at $0$, and defining $g(0)$ to be that limit gives the unique analytic extension to all of $D(0,1)$.
A: Let $g(z)= \frac{f(z)}{z}$. By a known result we can say $0$ is a removable singularity for $g(z)$ if and only if $\lim_{z\rightarrow0} g(z) $ finite. Here we have $\lim_{z\rightarrow0} g(z)=f'(0)$ and is finite as f is analytic at zero, so $0$ is a removable singularity of $\,g$ .  
A: I think what you are missing is that, basically, $ f(z) =1$ and $g(z)=z/z$ are not the same functions ("though" they agree everywhere except at $0$). $g$ is not defined at $0$ because its value must be evaluated as $0/0$ at $0$ which is not defined.
However, clearly, $g$ has a removable singularity at $0$ and its analytic extension is $f$.
A: The function $f(z)/z$ is holomorphic on the punctured unit ball and so the singularity at $0$ is either removable, a pole of essential. It can't be essential by Big Picard, and it can't be a pole because the Schwarz lemma gives that $|f(z)| \leq |z|$ for all $z \in D$. Thus the singularity is removable.
