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Math Help - [SOLVED] Proof of the chain rule on holomorphic function

  1. #1
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    [SOLVED] Proof of the chain rule on holomorphic function



    Okay. A) This just doesn't work. That's a pretty huge problem.

    B (probably the answer to the problem))

    We know any function on a complex variable can be decomposed:

    f(z) = Re(f(z)) + i[Im(f(z))] = u + iv

    Where u(x,y) and v(x,y) are real-valued functions.

    Now, let g(z) = s + it similarly.

    Thus, g(f(z)) = s(u(x,y) + i[v(x,y)]) + i[t(u(x,y) + i[v(x,y)])].

    so (g(f(z))' = (s(u(x,y) + i[v(x,y)]))' + (i[t(u(x,y) + i[v(x,y)])])'

    = s'(u(x,y))u'(x,y) + s'(i[v(x,y)])iv'(x,y) + i[t'(u(x,y))u'(x,y) + t'(i[v(x,y)])iv'(x,y)

    Right? But then what happens to the i's inside the real-valued functions?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Not right! Clearly there should be no imaginary numbers inside your real-valued functions. If f(z) = f(x,y)=u(x,y)+iv(x,y),\, g(z)= g(x,y)=s(x,y)+it(x,y), then f(g(z)) = f(s,t)=u(s,t)+iv(s,t)=u(s(x,y),t(x,y))+iv(s(x,y),t  (x,y)). See if that helps!
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    This is what I was thinking, but I wasn't sure if I could do that. Thank you!
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by davismj View Post
    This is what I was thinking, but I wasn't sure if I could do that. Thank you!
    You're welcome! But make sure you understand why what I wrote makes sense. It's just basic function composition! Think about it and report back if you're still not sure.
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