Proving Analyticity: Cauchy Riemann Equations & Mixed Partial Derivatives

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Homework Help Overview

The problem involves proving that if f(z) is an analytic function, then g(z)=f*(z*) is also analytic. The context includes the Cauchy-Riemann equations and the continuity of mixed partial derivatives.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the need to prove the Cauchy-Riemann equations for the new function g(z). There is exploration of the representation of z* and its implications for the function's derivatives.

Discussion Status

Participants are actively engaging with the problem, attempting to apply the chain rule and clarify the relationships between the derivatives of the original and conjugated functions. Some guidance has been offered regarding the application of the chain rule, but uncertainty remains about specific derivative evaluations.

Contextual Notes

There is a noted difficulty in taking partial derivatives and a need for clarity on the relationship between variables in the context of conjugation.

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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 [tex]\frac{\partial u}{\partial y} = \left(\frac{\partial u}{\partial w} \right) \left(\frac{\partial w}{\partial y} \right)[/tex], right? And [tex]\frac{\partial u}{\partial w}[/tex] is the same as [tex]\frac{\partial u}{\partial y}[/tex] 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 [tex]\frac{\partial u}{\partial y} u(x,y) = w[/tex], then [tex]\frac{\partial u}{\partial y} u(x,-y) = -w[/tex].

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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