Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Complex Analysis

  1. Jan 21, 2012 #1
    I am trying to decipher what this means:

    [tex]F(z) = \overline{f(\bar{z})}[/tex]

    Thanks for the help.
  2. jcsd
  3. Jan 21, 2012 #2

    I like Serena

    User Avatar
    Homework Helper

    Welcome to PF, fauboca! :smile:

    The overbar notation indicates the complex conjugate.

    Apparently the function F(z) is defined by taking the complex conjugate of z, applying the function f, followed by taking the conjugate of the result.

    Note that for regular functions F(z) is equal to f(z).
  4. Jan 23, 2012 #3
    I was hoping with this information I would have been able to solve the problem, but I am still unsure on how to work it.

    The question is:

    If [itex]f:\mathbb{C}\to\mathbb{C}[/itex] is entire, i.e. holomorphic at every point, then [itex]F(z) = \overline{f(\bar{z})}[/itex].

    I am not sure how to do this problem. So we know f is differentiable everywhere. How does this help?
  5. Jan 23, 2012 #4
    You'll need to prove that:

    [tex] f(z) = \overline{f({\overline{z}}) }, [/tex]

    if you write out f in a real and imaginary part and apply the CR conditions you'll find this quick and easy.
  6. Jan 23, 2012 #5
    I understand what you are saying but I still don't quite see it.

    So if we let [itex]u(x,y)[/itex] be the real and [itex]v(x,y)[/itex] be the imaginary parts of F. How do I define them for [itex]\overline{f({\overline{z}})}[/itex]
  7. Jan 23, 2012 #6
    [tex] f(z) = f(x,y) = u(x,y) + iv(x,y) \rightarrow f( \overline{z}) =u(x,y) - iv(x,y)[/tex]
  8. Jan 23, 2012 #7
    I suppose the question is actually to show that F(z)=f(z) ?

    Anyway suppose you write z = x+ iy then the conjugate is x - iy

    so take u(x,y) to u(x,-y) and v(x,y) to v(x,-y). Then (and of course taking the conjuagte of the function also means another minus sign. The the Cauchy Riemann relation will provide the answer.
  9. Jan 23, 2012 #8
    This is exactly what needs proving. Where you use the fact that the function is entire.
  10. Jan 23, 2012 #9
    Then [itex]\overline{f(\bar{z})} = u(x,-y)-iv(x,-y)[/itex]??
  11. Jan 23, 2012 #10

    I like Serena

    User Avatar
    Homework Helper

  12. Jan 23, 2012 #11
    Since f is holomorphic, we know:

    [tex]u_x = v_y \ \text{and} \ u_y = -v_x[/tex]

    Now, how do I use [itex]F(z)=\overline{f(\bar{z})}=u(x,-y)-iv(x,-y)[/itex].

    I read that I need to use C.R. equations. I need to show:

    [tex]u_x(x,-y) = v_y(x,-y) \ \text{and} \ -u_y(x,-y) = v_x(x,-y)[/tex]

    Now, what should I do or how should I approach it?
  13. Jan 23, 2012 #12

    I like Serena

    User Avatar
    Homework Helper

    With just f(z) being holomorphic, that is not enough to show that F(z)=f(z).

    So I assume you're supposed to show that F(z) is holomorphic?
    If so, according to C.R. you should check that ##\Re(F_x)=\Im(F_y)## and that ##\Re(F_y)=-\Im(F_x)##, starting from the fact that f(z) is holomorphic...
  14. Jan 23, 2012 #13
    How do I check that?

    I wrote that in terms of u and v in the post just above yours.
  15. Jan 23, 2012 #14

    I like Serena

    User Avatar
    Homework Helper

    Well, if ##u_x(x,y)=v_y(x,y)## for all x and y, don't you also have ##u_x(x,-y)=v_y(x,-y)##?
  16. Jan 23, 2012 #15
    I need to prove the second part. So can I say [itex]y=-y[/itex]
  17. Jan 23, 2012 #16

    I like Serena

    User Avatar
    Homework Helper

    Hmm, no you can't say y=-y.

    But since ##u_x(x,y)=v_y(x,y)## is true for all x and y, you could also say ##u_x(x,w)=v_y(x,w)## is true for all x and w.
    Now define y=-w...
  18. Jan 23, 2012 #17
    I am not sure what you are attempting to allude to from the ellipses.
  19. Jan 23, 2012 #18

    I like Serena

    User Avatar
    Homework Helper

    Since f(z) is holomorphic, ##u_x(x,w)=v_y(x,w)## is true for all x and w.
    In particular, if we define y=-w, that means that ##u_x(x,-y)=v_y(x,-y)## is true for all x and y.

    Put otherwise, the relation holds true for any point in the complex plane.
    It does not matter if we put a minus sign in front of the imaginary part of a point, as long as we do so consistently.
    The resulting points (x,-y) are all in the same complex plane for which the relation holds true.
  20. Jan 23, 2012 #19
    Thanks. Sorry for not being able to decipher that myself.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook