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Math Help - [SOLVED] quotient rule on holomorphic functions

  1. #1
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    [SOLVED] quotient rule on holomorphic functions



    Clearly, this is not fun.

    If we decompose f(z) and g(z) into real valued functions u + iv and s + it, it seems like a reasonable first step to multiply the numerator and denominator by the complex conjugate of g (s - it), which yields something that looks very similar to the desired result. However, when we take the derivative of these real valued functions, the denominator is no longer g^2 and the numerator looks more like f' multiplied by the complex conjugate of g, minus g'f.

    Any hints on better steps to take?

    Thank you!
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    I suggest you first solve the chain rule problem. Then you can show that \frac{d}{dz}\frac{1}{g(z)} = \frac{-g'(z)}{g(z)^2} and then you can apply the product rule to \frac{f(z)}{g(z)} = f(z) \times \frac{1}{g(z)}.
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  3. #3
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    Quote Originally Posted by Bruno J. View Post
    I suggest you first solve the chain rule problem. Then you can show that \frac{d}{dz}\frac{1}{g(z)} = \frac{-g'(z)}{g(z)^2} and then you can apply the product rule to \frac{f(z)}{g(z)} = f(z) \times \frac{1}{g(z)}.
    Thank you very much.

    This question was actually ordinally before the chain rule question. Is there any other way to approach it? Thanks for your help!
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    MHF Contributor Bruno J.'s Avatar
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    Once again you are very welcome!

    There might be a way to prove it without the chain rule, but it would just be equivalent to proving the chain rule for the specific case F \circ g where F(z)=z^{-1}. Therefore I think it really makes more sense to do the other problem first in all its generality.
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