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Complex valued functions

  1. Nov 29, 2008 #1
    Let f(z) be some complex valued function of complex variable z=x+iy. Since f(z) is (in general) complex, we can write it as f(z) = u(z)+iv(z), where u and v are real. But how does one prove that we can also write it as f(z) = u(x,y)+iv(x,y), i.e. shouldn't x and y always appear in the form "x+iy"?

    Thanks
     
  2. jcsd
  3. Nov 29, 2008 #2

    nicksauce

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    Well u and v are just real valued functions of two real variables. x and y uniquely specify z, therefore we can write u(z) = u(x,y).
     
  4. Nov 29, 2008 #3
    ok, but if I have any u(x,y) and v(x,y), can I always find some f(z) so that f(z) = u(x,y) + iv(x,y), where z=x+iy? For example, if I have u(x,y) = x sin(y) and v(x,y) = cos(x), what would f(z) be?
     
  5. Nov 29, 2008 #4

    Office_Shredder

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    That would be f(z) = Re(z)*sin(Im(z)) + icos(Re(z))

    Where if x+iy=z, Re(z):= x, Im(z):= y
     
  6. Nov 29, 2008 #5

    lurflurf

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    Trivially yes
    x=Re(z)
    y=Im(z)
    so
    u(x,y)+iv(x,y)=u(Re(z),Im(z))+iv(Re(z),Im(z))=f(z)
    That said
    Re(z) and Im(z) are not considered proper functions of z as they break z apart and do not treat it as whole variable
    So we are led to the concept of an analystic function that maps z as a whole
    The Cauchy-Riemann Equations allow us to check if a function is analystic
    for a function in
    u+iv form the condition is written
    Dx(u)=Dy(v)
    Dy(u)=-Dx(v)
    where Dx and Dy are partial derivatives with respect to x and y
    so for your example
    Dx(x sin(y))=Dy(cos(x))
    sin(y)=0
    Dy(x sin(y))=-Dx(cos(x))
    x cos(y)=sin(x)

    so we se your function is not analytic and cannot be written as f(z) in a proper way
    we could write it as f(z,z*) (where z* is the conjugate of z and z*z=|z|^2)
    when written in this form the condition is
    Dz*(f)=0 where Dz* is the partial derivative with respect to z*
     
  7. Nov 29, 2008 #6
    Yes, that's it! Thank you!
     
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