Complex problem f(z) = sqrt(|xy|) in x + iy form?

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The discussion centers on the function f(z) = sqrt(|xy|) and its compliance with the Cauchy-Riemann equations at the point z=0. Participants conclude that while the partial derivatives of u and v may equal zero at this point, the function is not differentiable there. The necessity of separating the function into its real and imaginary components is debated, with the consensus that Cauchy-Riemann conditions do not hold due to the nature of the function at z=0.

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Homework Statement



In the title: f(z) = sqrt(|xy|)...show that this satisfies the Cauchy-Riemann equations at z=0, but is not differentiable there.

Homework Equations



Cauchy-Riemann just states that partial u partial x = partial v partial y and partial u partial y = - partial v partial x.

The Attempt at a Solution



I think all the partials du/dx, du/dy, dv/dx, and dv/dy are 0. Because f(0) is 0 in this case, right? Maybe that's wrong. Maybe I'm assumign too much by saying that if z=0, x and y also are 0.

Anyway, then I just need to show the limit as h->0 for differentiation doesn't exist, right? If what I said above is right, than it would just be plugging in some x_0 and y_0 (if h = x_0 + iy_0) for x and y and than finding the limit for when x_0 = 0 and when y_0 = 0, and as long as those aren't equal then it's not differentiable.

Does this sound right, or totally off?
 
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saraaaahhhhhh said:
I think all the partials du/dx, du/dy, dv/dx, and dv/dy are 0. Because f(0) is 0 in this case, right? Maybe that's wrong.

You're supposed to take the partial derivatives first and then evaluate them at z=0; not the other way around!:smile:
 
Wouldn't that mean the Cauchy-Riemann equations don't hold? I'm a little unsure on what u would be in this case.

Do I need to separate sqrt(|xy|) into the real and imaginary parts? Or can I just assume all is real and then take the partial derivates of u, and the partials of v would just be 0, since there's no complex part?

This is why I think Cauchy-Riemann wouldn't hold, but I'm not sure.

Thanks so much for the response! I do appreciate it.
 
Just to go ahead and try this: would partial u partial x be y(xy)^(-1/2)?
And then partial u partial y be x(xy)^(-1/2)?
And both partials of v be 0? This is assuming that sqrt(xy) is just the 'real' part...if f(z) takes the form u + iv.

I have a feeling this is wrong, since Cauchy-Riemann is supposed to hold and doesn't hold using this method. But if it's wrong, how am I supposed to figure out what u and v are in cases like this?
 
saraaaahhhhhh said:
Do I need to separate sqrt(|xy|) into the real and imaginary parts? Or can I just assume all is real and then take the partial derivates of u, and the partials of v would just be 0, since there's no complex part?

Yes, x and y are real numbers, so |xy| is a positive, real number and so sqrt(|xy|) is real.

This is why I think Cauchy-Riemann wouldn't hold, but I'm not sure.

Careful, what do you get for \frac{\partial u}{\partial x} and \frac{\partial u}{\partial y}...where do they equal zero?
 
I'm not sure what you're saying here. I think you get that they don't exist or are undefined, because there would be 0 on the bottom of a fraction? But I'm not sure where you're going with this.
 

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