Is $f(x+iy)=x^2-y^2 + i\sqrt{|xy|}$ complex differentiable? I was solving a problem yesterday and it has bugged me for the whole night, im not sure whether if I got it correct or not.
First I was asked if $f(x+iy)=x^2-y^2 + i\sqrt{|xy|}$ satisfies the C-R equations at $0$. So I found $\frac{\partial u}{\partial x}$ $=$ $\frac{\partial v}{\partial y}$ & $-\frac{\partial u}{\partial y}$ $=$ $\frac{\partial v}{\partial x}$
$u=x^2-y^2$ 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 $\frac{\partial v}{\partial y}=\frac{1}{2}\frac{\sqrt{x}}{\sqrt{y}}$.
So obviously it does not satisfy the C-R equations.


*

*I was wondering if I have to do anything else because it asks if it satisfies the C-R equations at $0$?

*The next part of the question asks me if $f$ is differentiable at $0$. And it hints that I should consider $\mathrm{lim}_{r\rightarrow 0} \frac{f(re^{i\theta})}{re^{i\theta}}$. I assume they are meaning $f$ is complex differentiable? (not real differentiable?). How would I determine if f is differentiable at $0$? 
Because it says on wikipedia that the sole existence of partial derivatives satisfying the Cauchy-Riemann equations is not enough to ensure complex differentiability at that point. It is necessary to make sure that u and v are real differentiable, which is a stronger condition than the existence of the partial derivatives but it is not necessary to require continuity of these partial derivatives. Then f = u + iv is complex-differentiable at that point if and only if the partial derivatives of u and v satisfy the Cauchy–Riemann equations at that point. 


*

*The question I have is, since $f$ does not satisfy the C-R equations (see my calculations above), is there any need to do anything further (in question 2)? Can I just say its not differentiable at $0$? Why did they ask me to consider that limit?


Thanks alot, Im really stuck and slightly confused with all this..
 A: $$
f\;'(0) = \lim_{z\to0}\frac{f(z)-f(0)}{z-0} = \lim_{z\to0}\frac{f(z)}{z}.\tag{1}
$$
In order that this limit exist, it is necessary that it be equal to the limit as $z$ approaches $0$ along any path that passes through $0$.  Thus it must be the same as $$
\lim_{r\to0} \frac{f(re^{i\theta})}{re^{i\theta}},
$$
which is the limit in (1) as $z$ approaches $0$ along a certain line.  The angle between that line and the $x$-axis is $\theta$.  Since the function is expressed in terms of real and imaginary parts, write $x+iy=re^{i\theta}=r(\cos\theta+i\sin\theta)$.  Then $$f(re^{i\theta})=f(x+iy)= x^2-y^2+i\sqrt{|xy|}=r^2(\cos^2\theta-\sin^2\theta) + i|r|\sqrt{|\cos\theta\sin\theta|}.$$
So
$$
\frac{f(re^{i\theta})}{re^{i\theta}} = \frac{r(\cos^2\theta-\sin^2\theta) + i\operatorname{sign}(r)\sqrt{|\cos\theta\sin\theta|}}{e^{i\theta}}
$$
where $\operatorname{sign}(r)= \pm1$ according as $r$ is positive or negative.  So try taking the limit of that as $r\to0$.
A: Since $x^2−y^2$ is the real part of $z^2$, which is entire, the function cannot be analytic at (0,0) unless it agrees (in a 'hood of (0,0)) with $i2xy$, the imaginary part of $z^2$, which it does not, e.g., check along $y=x$ 
