Derivatives and the Cauchy-Riemann equations

Liav Teichner · #1 $Old$ November 1st, 2007, 09:04 PM

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.

ThePerfectHacker · #2 $Old$ November 1st, 2007, 09:10 PM

False.

This is Mine 75

th Post!!!

Jhevon · #3 $Old$ November 1st, 2007, 09:17 PM

Quote:

Originally Posted by Liav Teichner $View Post$

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?

Liav Teichner · #4 $Old$ November 1st, 2007, 09:25 PM

"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.