Proving Analyticity: Cauchy Riemann Equations & Mixed Partial Derivatives

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In summary, we need to prove that if f(z) is an analytic function, then g(z)=f*(z*) is also an analytic function. The Cauchy-Riemann equations must hold for the new function, and we can use the chain rule to evaluate the partial derivatives. By re-labeling variables, we can verify that the Cauchy-Riemann equations hold for f(x,-y).
  • #1
g1990
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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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  • #2
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
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
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
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
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.
 
  • #7
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
obviously I will have to re-label some stuff
 
  • #9
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.
 

What is analyticity and why is it important in mathematics?

Analyticity is a property of a mathematical function or equation that describes how it is differentiable at every point within its domain. It is important because it allows us to use techniques and tools from calculus to analyze and solve problems involving these functions. It also has many applications in fields such as physics, engineering, and economics.

What are the Cauchy-Riemann equations and how do they relate to analyticity?

The Cauchy-Riemann equations are a set of conditions that must be satisfied for a complex-valued function to be analytic. They relate the partial derivatives of a complex-valued function to each other, and if these conditions are met, then the function is said to be analytic.

What is the significance of mixed partial derivatives in proving analyticity?

Mixed partial derivatives are important in proving analyticity because they allow us to check if the Cauchy-Riemann equations are satisfied. If the mixed partial derivatives of a complex-valued function are equal, then the function is analytic. This is because equal mixed partial derivatives imply that the function's first-order partial derivatives are continuous, which is a necessary condition for analyticity.

How can the Cauchy-Riemann equations be used to simplify complex functions?

The Cauchy-Riemann equations provide a useful tool for simplifying complex functions. By using these equations, we can transform a complex function into two real-valued functions, which are often easier to work with. This simplification allows us to apply common techniques from calculus, such as integration and differentiation, to solve problems involving complex functions.

Can the Cauchy-Riemann equations be used to prove analyticity for all complex-valued functions?

No, the Cauchy-Riemann equations are only applicable to functions that are differentiable at every point within their domain. There are some complex-valued functions that are not differentiable at certain points, such as the absolute value function, and therefore cannot be proven to be analytic using the Cauchy-Riemann equations.

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