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Analytic Functions Cauchy & Riemann Equations

  1. Oct 23, 2011 #1
    Hi, this is fairly fundamental and basic, but I cannot seem to make sense of it

    I know z = x + iy

    and hence a function of this variable would be in the form h = f(z). BUT I do not understand why

    f(z) = u(x,y) + iv(x,y)

    why so? in z = x + iy, x is the real part and iy is the imaginary part, so why does y have influence in real part and x have influence in the imaginary part in f(z)?

    Thanks
    Thomas
     
  2. jcsd
  3. Oct 23, 2011 #2

    I like Serena

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    Homework Helper

    Hi thomas49th! :smile:

    Consider for instance f(z)=z2.
    What is (x+iy)2?
     
  4. Oct 23, 2011 #3

    Mark44

    Staff: Mentor

    Mod note: Moved from Precalculus section.
     
  5. Oct 23, 2011 #4
    I had a bit of trouble grasping this when first exposed to it, as well. What convinced me was to see f(z) as just another complex number, w. And w has its own real and imaginary part, u + iv.

    Now if w = f(z) then we can say that w is (the result of) a function of z, and likewise that u and v are functions of z. So w(z) = u(z) + iv(z). But since the value of z is dependent of the values of its real and imaginary parts, x and y, we can think of w(z) = w(z(x,y)) = u(z(x,y)) + iv(z(x,y)). But there is no need for such ugly notation, since the root idea is that w is a function of both x and y. So we simplify to f(z) = w(x,y) = u(x,y) + iv(x,y).
     
  6. Oct 24, 2011 #5
    @ I like... thought someone might say that :)

    @ kru_ that has satisfied me for the time being

    all is good for now

    Thanks :)
     
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