# Derivatives and the Cauchy-Riemann equations

• Nov 1st 2007, 08:04 PM
Liav Teichner
Derivatives and the Cauchy-Riemann equations
I need to decide whether the following statement is true or not:
let f be a complex function, f(z)=f(x+iy)=u(x,y)+iv(x,y).if u and v satisfy the Cauchy-Riemann equations in a neighborhood of z_0 then f is differentiable at z_0.
• Nov 1st 2007, 08:10 PM
ThePerfectHacker
False.

This is Mine 75:):)th Post!!!
• Nov 1st 2007, 08:17 PM
Jhevon
Quote:

Originally Posted by Liav Teichner
I need to decide whether the following statement is true or not:
let f be a complex function, f(z)=f(x+iy)=u(x,y)+iv(x,y).if u and v satisfy the Cauchy-Riemann equations in a neighborhood of z_0 then f is differentiable at z_0.

no. consider $f(z) = u(x,y) + iv(x,y)$, where $u(x,y) = \sqrt{|xy|}$, $v(x,y) = 0$ and consider the derivative at z_0 = 0

i got this example from a text book.

can you come up with a nicer example, TPH?
• Nov 1st 2007, 08:25 PM
Liav Teichner
I don't think the example is correct:
"if u and v satisfy the Cauchy-Riemann equations in a neighborhood of z_0" - It means there's an open circle (disk) around z0 where the Cauchy-Rieamann equations are satisfied. In your example there is no such circle (neighborhood). Every circle around (0,0) in your example contain positive real numbers where it is easy to show the Cauchy-Riemann equations aren't satisfied.