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Let $\Omega$ be an open subset of $\mathbb{C}$ and $z_0\in\Omega$. Show that a function $f:\Omega\rightarrow \mathbb{C}$ is analytic at $z_0$ iff

$$g(z) = \begin{cases} \dfrac{f(z)-f(z_0)}{z-z_0}-f'(z_0) & \text{if $z\neq z_0$}\\[6px] 0 & \text{if $z=z_0$} \end{cases} $$ is continuous at $z_0$.

I think the forward direction is clear enough, but I am at a loss for a strategy for the reverse direction. I am new to the subject and would like to approach the problem from the ``ground level" if possible.

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    $\begingroup$ $g$ is continuous at $z_0$ if and only if $f$ is complex differentiable at $z_0$, it need not be analytic at $z_0$. ($f$ is analytic at $z_0$ if it is representable by a power series on some neighbourhood of $z_0$.) $\endgroup$ – Daniel Fischer Feb 18 '16 at 17:15
  • $\begingroup$ What do you mean by "analytic at $z_0\>$"? Maybe you have an $f$ in mind which is analytic in $\Omega\setminus\{z_0\}$ to start with. $\endgroup$ – Christian Blatter Feb 18 '16 at 17:16
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    $\begingroup$ Isn't $|z|^2$ a counter-example? It is complex differentiable at $0$, thus $g$ is continuous at 0, but it is not analytic. $\endgroup$ – johnnycrab Feb 18 '16 at 17:23
  • $\begingroup$ Christian Blatter- Yes, I think "analytic at $z_0$" means as you say. $\endgroup$ – Earl Feb 18 '16 at 18:22
  • $\begingroup$ In my course we have not yet been introduced to the power series representation perspective yet. It may be that we are abusing the the definition of analytic in this exercise. $\endgroup$ – Earl Feb 18 '16 at 19:04
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As shown above, $f$ need not be analytic at $z_0$, but the statement is true assuming $f$ is complex differenciable.

Assume first that $f$ is analytic at $z_0$. Then

$$\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}$$ exists, and the limit equals to $f'(z_0)$, so

$$\left( \lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}\right)-f'(z_0)=\lim_{z\to z_0}\left(\frac{f(z)-f(z_0}{z-z_0}-f'(z_0)\right)=0$$

Hence, $g$ is continous at $z_0$.

The other implication follows from the fact that since $g$ is continous at $z_0$, then

$$\lim_{z\to z_0}\left(\frac{f(z)-f(z_0)}{z-z_0}-f'(z_0)\right)=0$$ So $$\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}=f'(z_0)$$

Therefore, $f$ is complex differentiable at $z_0.$

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