# Let $f$ be analytic on $D = \{z \in \Bbb{C} : |z| < 1\}$ and $f(0) = 0$.

I was stuck on the following problem:

Let $f$ be analytic on $D = \{z \in \Bbb{C} : |z| < 1\}$ and $f(0) = 0$.

Define $$g(z) = \begin{cases} \displaystyle \frac{f(z)}{z} & z \neq 0, \\ f'(0) & z = 0. \end{cases}$$

Then which of the following option(s) is/are correct?

1. $g$ is discontinuous at $z = 0$ for all $f$.

2. $g$ is continuous, but not analytic, at $z = 0$ for all $f$.

3. $g$ is analytic at $z = 0$ for all $f$.

4. $g$ is analytic at $z = 0$ only if $f'(0) = 0$.

Can someone point me in the right direction with some explanation? Thanks in advance for your time.

EDIT: I have posted an answer .Feel free to comment if I missed anything in my answer.

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If $f(z) = z^n$ and $n \geq 1$ then $g(z) = z^{n-1}$ on $D$. – Antonio Vargas Nov 19 '12 at 18:08
@AntonioVargas Then g should be analytic at z=0 only if f'(0)=0.Is not it? – learner Nov 19 '12 at 18:11
$f$ is analytic in some neighbourhood of $0$, so you can write it as a power series in $z$. This should make it pretty obvious which of those statements are true, and which not. – Rhys Nov 19 '12 at 18:12
@Rhys thank you – learner Nov 19 '12 at 18:14

We have, $$g(z) = \begin{cases} \displaystyle \frac{f(z)}{z} & z \neq 0, \\ f'(0) & z = 0. \end{cases}$$.
Since $f$ is analytic in some neighbourhood of $0$, so I can write it as a power series in $z$. So,$$f(z)=a_0+a_1z+ \cdots \cdots +a_nz^n$$ for $|z|<1$. Now,$f(0)=0$ gives $a_0=0.$ Hence \begin{align}f(z) &=a_1z+ \cdots \cdots +a_nz^n \\ \implies {f(z) \over z} &=g(z) =a_1+a_2z+ \cdots +a_nz^{n-1} \\ \lim_{z \to 0} g(z) &= a_1=f'(0)=g(0)\\ \text{Since},f'(z) &=a_1+ \cdots +na_nz^{n-1} \implies f'(0)=a_1\end {align} Hence $g$ is continuous at $z=0.$ This eliminates option 1.
Also,we notice $$\lim_{z \to 0}{g(z)-g(0) \over z-0}=a_2$$, which exists for all $f$. This eliminates options 2 and 4.