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Math Help - CRE form

  1. #1
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    CRE form

    Hi I got a question, which I think may be a trick question.

    "State Cauchy Riemann equations for real valued functions u(x,y), v(x,y).

    we know a function is generally of the form
    f(z) = u(x,y) + iv(x,y).

    but U is the real part and V is the imaginary. so why does the question say "state the CRE for REAL VALUED FUNCTIONS u(x,y), v(x,y)."
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  2. #2
    Super Member wingless's Avatar
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    When we separate f(z) as u(x,y) + i v(x,y), u and v are real valued functions. Notice that v is real valued and iv is imaginary.

    So if the book gives f(z)=u(x,y)+iv(x,y), then there's no trick.
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  3. #3
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    nice. i always was taught that V is the imaginary.

    but ill take it as you say since its easier hehe.
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  4. #4
    Super Member wingless's Avatar
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    Well, I should've been more precise.

    v(x,y) is a real function. It's also the imaginary part of f(z).

    For example, let z = 3 + 5i. The imaginary part of z is 5, which is a real number.

    For example, let f(z) = z^2. We can write f(z)=(x+yi)^2 = x^2 - y^2 + i(2xy). So here u(x,y) = x^2-y^2 and v(x,y) = 2xy. Now you say, is 2xy a real or complex function? (x and y are real values).
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  5. #5
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    ah, now that coincides with what i was taught aswell.. great job. thanks

    and yes, 2xy is a REAL function, and also the imaginary part of u(x,y)

    works well. thanks bro.
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