# Analytic Functions

1. Apr 28, 2010

### g1990

1. The problem statement, all variables and given/known data
The problem is that I need to prove that if f(z) is an analytic function, then g(z)=f*(z*) is also an analytic function. The fact that the mixed partial derivatives are continuous is given.

2. Relevant equations
Cauchy Riemann: if f(z)=u(x,y)+iv(x,y) is analytic, then du/dx=dv/dy and du/dy=-dv/dx

3. The attempt at a solution
I know that I should prove that the Cauchy Riemann equations hold for the new function as well. I have g(z)=u(z*)-iv(z*) ( where the stars are conjugates) but I don't know where to go from there

2. Apr 28, 2010

### hgfalling

Before, you wrote your equation as:

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

Now you have:

f*(z*) = u(z*) - iv(z*)

What is z*?

Try writing your equation in the same form as before, and then evaluate the partial derivatives for the new function.

3. Apr 28, 2010

### g1990

I know that z* is x-iy if z=x+iy. Thus, g(z)=f*(z*)=u(x,-y)-iv(x,-y). I am having trouble taking the partial derivatives of that.

4. Apr 28, 2010

### hgfalling

Use the chain rule and remember that when you are taking the partial with respect to one variable, the other variable is just constant.

5. Apr 28, 2010

### g1990

I am unsure how to use the chain rule. I could say that du/dx is just that, and du/dy=du/d(-y)*d(-y)/dy=-du/d(-y) but then I don't know what du/d(-y) is. Can you help me use the chain rule?

6. Apr 28, 2010

### hgfalling

Sure, let w = (-y). Now $$\frac{\partial u}{\partial y} = \left(\frac{\partial u}{\partial w} \right) \left(\frac{\partial w}{\partial y} \right)$$, right? And $$\frac{\partial u}{\partial w}$$ is the same as $$\frac{\partial u}{\partial y}$$ from the original (non-conjugated) version.

7. Apr 28, 2010

### g1990

why is du/dw the same as du/dy in the non conjugated version?
So I would get du/dy=-du/dy, du/dx=du/dx, dv/dx=-dv/dx, and dv/dy=-(-dv/dy)=dv/dy?

8. Apr 28, 2010

### g1990

obviously I will have to re-label some stuff

9. Apr 28, 2010

### hgfalling

Yes, you seem to have it:

If $$\frac{\partial u}{\partial y} u(x,y) = w$$, then $$\frac{\partial u}{\partial y} u(x,-y) = -w$$.

Now just verify that the Cauchy-Riemann equations hold for f(x,-y).

EDIT: By the way that w has nothing to do with the w i suggested before.