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Homework Help: Analytic Functions

  1. Apr 28, 2010 #1
    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. jcsd
  3. Apr 28, 2010 #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.
  4. Apr 28, 2010 #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.
  5. Apr 28, 2010 #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.
  6. Apr 28, 2010 #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?
  7. Apr 28, 2010 #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.
  8. Apr 28, 2010 #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?
  9. Apr 28, 2010 #8
    obviously I will have to re-label some stuff
  10. Apr 28, 2010 #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.
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