If $f(z)=f(x+iy)=\sqrt {|xy|}$
a) Show that the Cauchy-Riemann equations hold for (0,0), namely that $f_x(0,0)=-if_y(0,0)$.
b)Show that the derivative of f at z=0 does not exist.
I'm trying to solve it by taking the limits for the partial derivatives of f at 0 while setting the other (constant) variable as 0.
(e.g. $f_x(0,0)=lim_{x\to 0} {{\sqrt {|xy_0|}}-\sqrt {|0y_0|}\over {x-x_0}}$ keeping y=y0 constant at 0 and x0=0)
I think this is the right approach but having trouble with the limits.
edit: Tried another approach without any success:
Writing $|xy|=\sqrt {(xy)^2}$ we get:
$f(x,y)=\sqrt {\sqrt {(xy)^2}}$ so taking the partial derivative of x:
$f_x(x_0,y_0)= $${1}\over {2 \sqrt {\sqrt {(xy)^2}}}$$ {1} \over {\sqrt {(xy)^2}}$$2xy^2$
simplifying: $f_x(x_0,y_0)=$$ xy^2 \over 2|xy|^{2\over 3}$ which is again undefined at (0,0).
Best Answer
Well, you have $f(0+yi)=f(x+0i)=0$ for every $x$ and $y$. Thus the partial derivatives are both zero at the origin, and C-R is trivially satisfied. On the other hand, $$df_{0}=\lim_{|z| \to 0}\frac{\sqrt{|xy|}}{\sqrt{x^{2}+y^{2}}}$$ does not exist, because along the line $x=y$, the directional derivative (the limit along that line) is $\frac{|x|}{\sqrt{2}|x|}=\frac{1}{\sqrt{2}} \ne 0$
Edit for clarification: The reason the partial derivatives are $0$ at the origin, is that in the definition: $$\frac{\partial f}{\partial x}(0,0)=\lim_{x \to 0} \frac{f(x,0)-f(0,0)}{x}$$ But the fraction is $0$ for every $x \ne 0$. Therefore, the limit is $0$.