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Homework Help: Complex conjugates

  1. Oct 19, 2004 #1
    Complex conjugates (new question in 1st post)

    I have a couple questions on complex variables:

    1.)If you have a complex function defined as follows:
    [tex]f(z)=u(x,y)+iv(x,y)[/tex]
    with x,y real, what do you get if you take the complex conjugate of the variable z?
    [tex]f(z^*)=?[/tex]

    I was thinking that it wouldn't change since the complex variable has been replaced with two real variables, but that doesn't seem right.

    If I take the conjugate of the entire function, is this what I should get:

    [tex]f^*(z)=u(x,y)-iv(x,y)[/tex]


    2.) If you are trying to prove that a function is not analytic at a specific point, is it sufficient to show that the Cauchy-Riemann conditions do not hold? I'm trying to show that the derivative of a function at zero is dependant on the direction that you approach zero. I've shown that the C-R conditions are not met, but I'm not sure how to show explicitly for that point that they are not met.


    EDIT: NEW QUESTION

    I figured I would just edit this topic instead of starting a new one. I'm trying to prove that if a function [tex]f(z)[/tex] is analytic then the function [tex]f^{*}(z^{*})[/tex] is also analytic.

    I'm not sure how to get this started. It makes sense that taking the conjugate wouldn't affect the differentiability, but I don't know how to prove that. Any hints on how to get this problem started?
     
    Last edited: Oct 20, 2004
  2. jcsd
  3. Oct 19, 2004 #2

    Tide

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    To get the complex conjugate of f(z) replace y with -y.
     
  4. Oct 19, 2004 #3
    Is that just a general rule, or is that something I should be trying to prove?
     
  5. Oct 19, 2004 #4
    [tex]z[/tex] is defined by [tex]x+iy[/tex]. Therefore, the function [tex]f(z)[/tex] is equivalent to a function [tex]f(x,y)[/tex] with two arguments. Since [tex]z^\ast=x-iy[/tex], [tex]f(z^\ast)[/tex] is similarly equivalent to a function [tex]f(x,-y)[/tex].

    Hope that this is the proof you are looking for.

    Best regards,
    Kenneth
     
  6. Oct 19, 2004 #5
    Thanks, that makes a lot of sense.
     
  7. Oct 20, 2004 #6
    I added a question to the first post. I'd appreciate any help you can offer.
     
  8. Oct 20, 2004 #7
    Given [tex]f(z)=u(x,y)+iv(x,y)[/tex] is analytic, we have

    [tex]\frac{\partial u(x,y)}{\partial x}=\frac{\partial v(x,y)}{\partial y}[/tex] and

    [tex]\frac{\partial u(x,y)}{\partial y}=-\frac{\partial v(x,y)}{\partial x}[/tex].

    Next, we have [tex]f^\ast(z^\ast)=u(x,-y)-iv(x,-y)[/tex]. Rewrite it as [tex]f^\ast(z^\ast)=u'(x,y')+iv'(x,y')[/tex] where [tex]u'(x,y')=u(x,-y)[/tex] and [tex]v'(x,y')=-v(x,-y)[/tex]. Now, we have the following:

    [tex]\frac{\partial u'(x,y')}{\partial x}=\frac{\partial u(x,y)}{\partial x}\,,[/tex]

    [tex]\frac{\partial u'(x,y')}{\partial y'}=-\frac{\partial u(x,y)}{\partial y}=\frac{\partial v(x,y)}{\partial x}\,,[/tex]

    [tex]\frac{\partial v'(x,y')}{\partial x}=-\frac{\partial v(x,y)}{\partial x}\,,[/tex]

    [tex]\frac{\partial v'(x,y')}{\partial y'}=\frac{\partial v(x,y)}{\partial y}=\frac{\partial u(x,y)}{\partial x}\,.[/tex]

    Isn't it the Cauchy-Riemann condition still be hold for [tex]f^\ast(z^\ast)[/tex]?

    Best regards,
    Kenneth
     
    Last edited: Oct 20, 2004
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