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Complex conjugate of a complicated function

  1. Feb 25, 2014 #1

    I know if we have a complex number z written as z = x +iy , with a and real, the complex conjugate is z* = x - iy. Also if we write a complex function f(z) = u(x,y) + iv(x,y), with u and v real valued, then similarly the complex conjugate of this function is f(z)* = u(x,y) - iv(x,y).

    And if we have some complex numbers around a circle like z = i + exp(iΘ) then the conjugate of these numbers is z* = exp(-iΘ) - i.

    It always seems like the way to conjugate something complex is to literally just 'put a minus in front of the i's ' ...but is this always the case?

    Surely if we had some complicated function like g(x + iy) = [sin(x+iy) + iexp(cos(x-y))]^(-1) , we couldn't just replace the i's with (-i)'s? Seems like we'd have to convert the above to the form a + ib first.

    If not - if it really is that elegant - is there a proof or something that it works in every case?


  2. jcsd
  3. Feb 25, 2014 #2


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    2017 Award

    Staff: Mentor


    Not always.

    As an example, (1/z)* = 1/(z*), you don't have to calculate real and imaginary parts of 1/z here.
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