Proving Analyticity: Cauchy Riemann Equations & Mixed Partial Derivatives

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Homework Statement


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.


Homework Equations


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


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
 
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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.
 
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.
 
Use the chain rule and remember that when you are taking the partial with respect to one variable, the other variable is just constant.
 
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?
 
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.
 
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?
 
obviously I will have to re-label some stuff
 
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.
 
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