1. Not finding help here? Sign up for a free 30min 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!

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:
    with x,y real, what do you get if you take the complex conjugate of the variable z?

    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:


    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.


    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


    User Avatar
    Science Advisor
    Homework Helper

    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,
  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,
    Last edited: Oct 20, 2004
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Complex conjugates
  1. Conjugate Hyperbola? (Replies: 3)

  2. Complex Conjugates (Replies: 1)