If Cauchy-Riemann hold and the first order partials are continuous does that imply it is differentiable? Suppose we have a function $f(z)$ where $f(z)$ satisfies Cauchy-Riemann only one point say $z=0$ then if the  first order partials are continuous does that imply it is differentiable at $z=0$ or do we need to use the definition of differentiability at a point? I believe it has something to do with the set we are talking about (here it is just the point $z=0$) and whether the set is open or not?
So basically in summary I am asking which is true:
Let $A \subset \Bbb{C}$ where C-R hold and partials are continuous $\implies$ $f$ is differentiable on $A$.
or
Let $A \subset \Bbb{C}$ be an open set where C-R hold and partials are continuous $\implies$ $f$ is differentiable on $A$.
I ask because sometimes when it on a point (or line) my professor uses the definition of differentiability whereas if the C-R equations hold on $\Bbb{C}$ he just says the partials are continuous and then concludes it must be differentiable?
 A: A function $f(z)$ is differentiable at a point $z$, if we have the following conditions:


*

*$f_x$ and $f_y$ exist in a neighbourhood-an open set- of $z$.

*$f_x$ and $f_y$ are continuous in $z$ and $f_y=if_x$ in $z$ (Cauchy-Riemann conditions hold).

A: Yes, it is enough that the partial derivatives are continuous at $0$ (which of course requires that they exist in some neighborhood of $0$).
Without loss of generality we can assume that $f(0)=0$ and that the partial derivatives at $z=0$ are all $0$ (otherwise subtract the appropriate first-degree complex polynomial from $f$; thanks to the C-R equation holding, this can kill all four partial derivatives at once). To show that $f$ is complex differentiable we then need to establish that
$$ \frac{f(z)}{z} \to 0 \quad\text{when }z\to 0 $$
Let $\varepsilon>0$ be given and choose $\delta$ small enough that all of the partial derivatives are absolutely less than $\varepsilon/4$ on $B_\delta(0)$.
Then given $z=x+yi$ we have $|x|\le |z|$ and $|y|\le |z|$, and if $|z|<\delta$ we have, say, for the real part of $f(z)$:
$$ u(x,y) = (u(x,y) - u(x,0)) + (u(x,0) - u(0,0)) $$
and each of the two terms on the right of this is absolutely less than $|z|\varepsilon/4$ thanks to the Mean Value Theorem. Similarly for the imaginary part.
So both the real and the imaginary part of $f(z)$ are absolutely less than $|z|\varepsilon/2$, which means that $|f(z)|<|z|\varepsilon$ and therefore $ \left|\frac{f(z)}{z}\right| = \frac{|f(z)|}{|z|} < \varepsilon$ as required.
