# Converse of the Cauchy-Riemann condition

The Cauchy-Riemann condition states that an analytic function satisfies: \begin{split} \frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y}; \frac{\partial u}{\partial y} &= -\frac{\partial v}{\partial x} \end{split} The converse of the statement requires additional conditions: the first partial derivatives of $u, v$ exist and are continuous.

My question is: When a function satisfies the Cauchy-Riemann condition, shouldn't it have already satisfied the above additional conditions? Otherwise, it wouldn't satisfy the Cauchy-Riemann condition if its partial derivatives do not exist. Am I missing something fundamental here?

• the partials may exist but not be continuous. – Angina Seng Mar 9 '18 at 18:23
• Doesn’t it mean we only need the second additional condition? But my textbook says we need both conditions. – A Slow Learner Mar 9 '18 at 18:25
• It assumes that the function is continuous on a neighborhood of $z_0$, not only at the point. – Mathmath Mar 9 '18 at 18:27

Let $f(x+iy)=\sqrt {|xy|}$. You can verify that the partial derivatives all exist at $0$ and satisfy the C-R equations but f is not differentiable at 0.
The converse of the Cauchy-Riemman conditions require that $u,v$ be differentiable. This implies in particular that the partial derivatives exist, but does not require its continuity.
• I believe the converse of the C-R statement requires $u,v$ to be continuous at the point under consideration. – A Slow Learner Mar 10 '18 at 20:32
• I am sorry. In the last comment, I meant the converse of C-R requires the partial derivative of $u, v$ to be continuous at the point under consideration (according to my textbook). – A Slow Learner Mar 10 '18 at 23:06
• If the partial derivatives are continuous, then $u$ and $v$ are differentiable. But to carry on the proof of the converse of the Cauchy-Riemann conditions, differentiability is all that is needed. – Julián Aguirre Mar 11 '18 at 18:36