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.
Best Answer
Sorry if I was not too clear: the existence of partial derivatives is a necessary (tho not sufficient) condition for f(x,y) to be real-differentiable. Given that you showed that the partials don't exist, f cannot be real-differentiable at (0,0), let alone be analytic there.