Why is it clear from this formulation that f is continuous wherever it is holomorphic? Hi I am new on here so not sure if this is right place to post but quick and presumably easy question:
So holomorphic at a point $z_0 \in \Omega$ is defined as the limit as $h\rightarrow 0$ of 
$\frac{f(z_0+h)-f(z_0)}{h}$. My textbook says that we can rewrite this as $f(z_0+h)-f(z_0)-ah=h\psi(h)$, where $\psi$ is a function defined for all small $h$ and the limit as $h\rightarrow 0$ of $\psi(h)=0$ and $a=f'(z_0)$. Why does this reformulation that $f$ is continuous wherever it is holomorphic?
Thanks
JG
 A: A function which is differentiable at a point in any usual sense of the word (including holomorphic, which is, after all, another name for complex differentiability) will be continuous at that point.
This is because the reformulation $f(z_0+h)-f(z_0)-ah=h\psi(h)$ is equivalent to $f(z_0+h) = f(z_0) + ah + h\psi(h)$. 
Letting $h \rightarrow 0$ we have that the limit on the right hand side is $f(z_0)$, and so $\displaystyle\lim_{h \rightarrow 0}f(z_o+h) = f(z_0)$, which is the definition of continuity. 
A: This is the same "differentiability implies continuity" argument as for real variables.
$$\lim_{h\to 0} f(z_0+h)=\lim_{h\to 0} f(z_0)+ah-h\psi(h)=f(z_0)$$
A: Continuity means (by definition) that $f(z_0+h)-f(z_0)=o_{h \rightarrow 0}(1)$.
It should then be clear from the definition that holomorphic implies continuous.
In fact, the definition of holomorphic is equivalent to asking real differentiability and a certain form of the differential (namely, it must be a similitude). Since differentiable functions are continuous, this is another way to look at it. What it really remarkable is that this definition actually implies much more regularity : holomorphic functions are analytic, and in particular must be $C^\infty$.
