# Finding if function satisfies Cauchy-Riemann Equations

I am trying to find if $f(x+iy)=x^2-y^2 + i\sqrt{|xy|}$ satisfy the C-R equations at 0.

i.e. $\frac{\partial u}{\partial x}$ $=$ $\frac{\partial v}{\partial y}$ & $-\frac{\partial u}{\partial y}$ $=$ $\frac{\partial v}{\partial x}$

$u=x^2-y^2$ and $v=\sqrt{|xy|}$

So I found $\frac{\partial u}{\partial x} = 2x$, $-\frac{\partial u}{\partial y}=2y$, $\frac{\partial v}{\partial x}=\frac{1}{2}\frac{\sqrt{y}}{\sqrt{x}}$ and $\frac{\partial v}{\partial y}=\frac{1}{2}\frac{\sqrt{x}}{\sqrt{y}}$. Obviously you can see that it does not satisfy the C-R equations above, but I was wondering if I have to do anything else since it said at 0? What does it mean/imply?

Because, later on I have to determine if f is differentiable at 0. Are they somehow linked? Should I show that u and v are real differentiable at zero?

Thanks.

-
Part of the condition for Cauchy-Riemann to be satisfied is that the partials must exist:en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations –  gary Nov 10 '11 at 2:50
Thia may also work, but I am not 100%: $x^2-y^2$ is the real part of $z^2$, which is entire, so I would think that the only complex, and complex-valued functions with real part $x^2-y^2$ that satisfy C-R with respect to $x^2-y^2$ are those that are equal to the complex part of $z^2$ plus a constanti.e., $2xy+C$. –  gary Nov 10 '11 at 2:58
I'm pretty sure the partials must at least exist (tho do not need to be continuous) for the function to be real-differentiable; if a function f: $\mathbb R^2 \rightarrow \mathbb R^2$ is real-differentiable, its differential is given by $J(f)(dx,dy)$ , where J(f) is the Jacobian matrix, and the dot product with (dx,dy). –  gary Nov 10 '11 at 3:15
Basically, if the real derivative existed, it should be given by the formula: en.wikipedia.org/wiki/… , where the Jacobian is used. –  gary Nov 10 '11 at 3:25
Sorry, John, I hope I did not make things too confusing, or that I was confused myself. –  gary Nov 10 '11 at 3:40