1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook