# Show function satisties C-R relations, but is not analytic

• Feb 6th 2010, 04:50 PM
platinumpimp68plus1
Show function satisties C-R relations, but is not analytic
Regarding the complex function f(z)=sqrt(|xy|), where z=x+iy, and its behavior at the origin.

I'm really confused - how can this possibly satisfy the Cauchy-Riemann relations? There is no imaginary part, so u(x,y)=sqrt(|xy|) and v(x,y)=0... all of v's partials will be 0, but u's are not, so they can't be equal.

Does the question mean to find the partials of u(0,0) and v(0,0) and show they are equal? ... This seems silly.

I know it can't be analytic at the origin, because the derivative is undefined there, but I don't understand how it satisfies C-R! Please help shed some light. (Shake)
• Feb 6th 2010, 05:10 PM
dedust
Quote:

Originally Posted by platinumpimp68plus1
Regarding the complex function f(z)=sqrt(|xy|), where z=x+iy, and its behavior at the origin.

I'm really confused - how can this possibly satisfy the Cauchy-Riemann relations? There is no imaginary part, so u(x,y)=sqrt(|xy|) and v(x,y)=0... all of v's partials will be 0, but u's are not, so they can't be equal.

Does the question mean to find the partials of u(0,0) and v(0,0) and show they are equal? ... This seems silly.

I know it can't be analytic at the origin, because the derivative is undefined there, but I don't understand how it satisfies C-R! Please help shed some light. (Shake)

$\displaystyle u(x,y)=\sqrt{(|xy|)}$, then

$\displaystyle u_{x} |_{(0,0)}= \lim_{\triangle x \to 0} \frac{u(\triangle x + x,y) - u(x,y)}{\triangle x}$

$\displaystyle = \lim_{\triangle x \to 0} \frac{\sqrt{(|(x + \triangle x)y|)} - \sqrt{(|xy|)}}{\triangle x}$

$\displaystyle =\sqrt{|y|} \left(\lim_{\triangle x \to 0} \frac{\sqrt{(|(x + \triangle x)|)} - \sqrt{(|x|)}}{\triangle x}\right)$

substitute $\displaystyle y = 0$ and we have

$\displaystyle u_{x} |_{(0,0)}=0 \cdot \left(\lim_{\triangle x \to 0} \frac{\sqrt{(|(x + \triangle x)|)} - \sqrt{(|x|)}}{\triangle x}\right) = 0 = v_y$