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

Do all complex functions have orthogonal real and imaginary parts?

  1. Mar 25, 2013 #1
    z = h(x) + ig(x)
    True or False: By the definition of the complex plane, h(x) and ig(x) will always be orthogonal.

    If this was true, wouldn't that mean that we can find a 'very general' Fourier series representation of any function f(x) as an infinite series of An*h(x) + infinite series of Bn*ig(x) ?. I am aware that finding a Fourier series representation of f(x) doesn't mean that it will converge, and if it does converge, it won't necessary converge to f(x).

    for example h(x) = x^2 , g(x) = ln(x)

    Sorry if this is a stupid question, I'm just trying to understand some stuff...

    Thanks. :smile:
     
  2. jcsd
  3. Mar 25, 2013 #2

    marcusl

    User Avatar
    Science Advisor
    Gold Member

    No, only analytic functions are orthogonal. Most functions are not analytic. Look at the real and imaginary parts of z'(z) = z^2, as one example.
     
  4. Mar 25, 2013 #3
    I see, thank you sir!
     
  5. Mar 25, 2013 #4

    Bacle2

    User Avatar
    Science Advisor

    Maybe I misunderstand what you meant, but z^2 is analytic.
     
  6. Mar 25, 2013 #5

    marcusl

    User Avatar
    Science Advisor
    Gold Member

    Oops, sure enough that's my mistake!:blushing:
     
  7. Apr 2, 2013 #6

    Bacle2

    User Avatar
    Science Advisor

    No problem, happens to all of us.
     
  8. Apr 4, 2013 #7
    [itex] z^{1/2} [/itex]
     
  9. Apr 4, 2013 #8

    Bacle2

    User Avatar
    Science Advisor

    But z1/2 has a region where it is analytic; it is not entire ( it is

    actually a multi-function) , but you can find a region where it is analytic.

    Use, e.g., the inverse function theorem to see that there are points for which

    a local inverse exists. This local inverse is analytic in a 'hood of the point.

    An example of a nowhere-analytic function is z^ , the conjugate function;

    with z^(x+iy):= x-iy . Then z^:=U+iV , with U(x,y)=x and V(x,y)=-y

    Let's use C-R:

    U_x =1 , V_y=-1 , so U_x=V_y never holds.
     
    Last edited: Apr 4, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Do all complex functions have orthogonal real and imaginary parts?
Loading...