# Cauchy Riemann conditions

I am a bit confused about differentiability in complex analysis. We showed in class that if a function $f$ is differentiable at $z_0$, then the Cauchy-Riemann equations hold at $z_0$. We also showed that if a function has continuous first partial derivatives at $z_0$, and the Cauchy-Riemann equations hold at $z_0$, then $f$ is differentiable at $z_0$. So apparently just satisfying Cauchy-Riemann at a point is not sufficient to determine differentiability.

So my question is: if one is given some function, what is a good way to see where this function is differentiable without going to the definition of the derivative?

-
You could always compute the first partials and write down the CR equations. If either one of those fails at some point, then the function is not differentiable there. –  John Martin Dec 8 '12 at 5:30
A good way is what you described. Show that $f$ has continuous partial derivative at $z_0$ and that Cauchy-Riemann equations are satisfied at $z_0$. –  P.. Dec 8 '12 at 5:44
But there may be points where $f$ does not have continuous partials, in which case we have to go back to the definition of the derivative, no? –  nigelvr Dec 8 '12 at 5:56

Well, if a function $f$ is holomorphic (i.e. $\mathbb{C}$-differentiable), it is real analytic, then it has continuous partial derivatives. Therefore, checking the continuity of partial derivatives and then the CR-equations is, in some sense, a perfectly general method in order to determine whether a function is holomorphic or not.
By the way, it is absolutely true that satisfying the CR-equations is not enough: set $f(z)=e^{-z^{-4}}$ for $z\neq0$ and $f(0)=0$. Then $f$ has partial derivatives everywhere, also in the origin, and they satisfy the CR-equations everywhere, but $f$ is not $\mathbb{C}-$differentiable in $0$. Indeed, the partial derivatives are not continuous in $0$.
I an wondering whether $z^2$ or $z^3$ also works instead of $z^4$ in the exponentiation. –  Groups Dec 31 '14 at 9:46