# A quick question about complex function differentiation

• Apr 28th 2011, 02:20 AM
Glitch
A quick question about complex function differentiation
Say I have a complex function,

f(z) = u(x,y) + iv(x,y)

If I were to find the derivative, is the answer merely finding the partial derivatives du/dx, du/dy, dv/dx, dv/dy and adding the results? This site seems to suggest that this is the case:
Derivative Of Complex Function | Tutorvista.com

Am I correct? Thanks.
• Apr 28th 2011, 11:53 PM
Opalg
Quote:

Originally Posted by Glitch
Say I have a complex function,

f(z) = u(x,y) + iv(x,y)

If I were to find the derivative, is the answer merely finding the partial derivatives du/dx, du/dy, dv/dx, dv/dy and adding the results? This site seems to suggest that this is the case:
Derivative Of Complex Function | Tutorvista.com

Am I correct? Thanks.

Warning: That Tutorvista page is complete gibberish from start to finish. Do not be misled by it.

If you have a differentiable function of a complex variable, in the form f(z) = u(x,y) + iv(x,y) (where z = x+iy, of course), then its derivative is given by the formula $\displaystyle \boxed{f'(z) = \tfrac\partial{\partial x}u(x,y) + i\tfrac\partial{\partial x}v(x,y)}.$

That formula may look odd, because why should you differentiate both functions u(x,y) and v(x,y) partially with respect to x, and not with respect to y? The answer is that you could equally well differentiate both functions with respect to iy, and you would get the same answer for f'(z) as in the above formula, because of the Cauchy–Riemann equations.
• Apr 29th 2011, 03:00 AM
Prove It
Actually f'(z) equals whatever the derivative of the function with respect to z is...

But the Cauchy-Riemann equations need to be satisfied in order for the function to be differentiable.
• Apr 29th 2011, 05:45 AM
Glitch
Thanks!