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I was thinking about the following problem:

Let $f$ be analytic on $D=\{z \in \mathbb C: |z|<1\}$ and $f(0)=0.$ Define $g(z)=f(z)/z;z \neq 0$ and $g(z)=f'(0);z=0.$ Then which of the following options are true:

(a)$ g$ is discontinuous at $z=0$
(b)$g$ is continuous but not analytic at $z=0$ for all $f$
(c)$g$ is analytic at $z=0$ for all $f$
(d)$g$ is analytic at $z=0$ only if $f'(0)=0$.

My attempts: Clearly $g$ is continuous at $z=0$, since we see that $\lim_{z\to\ 0}g(z) =\lim_{z\to\ 0} f(z)/z=\lim_{z\to\ 0}f'(z)$[by L'hospital's rule as $0/0$ form]=$f'(0)=g(0).$ But here i know that i have to use Schwarz lemma but i could not use it effectively enough to make other conclusion about the other options. Also i see that
$f$ is analytic in some neighbourhood of $0$, so i can write it as a power series in $z.$ that is $f(z)=a_0+a_1(z)+a_2(z^2)+..... $and so $g(z)=f(z)/z=a_0/z+a_1+a_2(z)+...$but now i lost the track, as see if i take $f'(0)=g(0)=0$ ,then $g'(0)={g(z)-g(0)}/(z-0)=a_0/z^2+a_1/z+a_2+...$ which clearly does not exist at $z=0$ even if we take $f'(0)=0.$ Am i going in the right direction? I need help about options $(b),(c),(d).$Thanks in advance for your time.

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Just adding to Nameless' answer, I would like to point out that Riemann's removable singularity theorem helps you right away (the answer being (c)). But I would also like to point out that this function $f(z)/z$ is what one considers in trying to prove Schwarz's Lemma. So you should not try to use that lemma for this question. Here is a proof for Schwarz's Lemma – Deon Dec 18 '12 at 20:14

1 Answer 1

I suggest using Riemann's Removable Singularity Theorem. This will make the problem rather simple

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