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Math Help - Derivative of complex function using definition

  1. #1
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    Derivative of complex function using definition

    The question

    Show that the definition of \frac{d}{dz}(\frac{1}{1 - z}) = \frac{1}{(1 - z)^2}

    My attempt
    I tried using:

    \lim_{\delta z \to 0} \frac{f(z + \delta z) - f(z)}{\delta z}

    But having fractions in the numerator in this case appears difficult to simplify. I tried substituting z = x + iy, but I ended up with a massive mess which still results in a division by 0. Is there a good method of solving this problem?

    Thanks.
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  2. #2
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    Quote Originally Posted by Glitch View Post
    Show that the definition of \frac{d}{dz}(\frac{1}{1 - z}) = \frac{1}{(1 - z)^2}
    \lim_{\delta z \to 0} \frac{f(z + \delta z) - f(z)}{\delta z}
    I ended up with a massive mess
    WHY?

    \dfrac{\frac{1}{1-z-dz}-\frac{1}{1-z}}{dz}=\dfrac{(1-z)-(1-z-dz)}{dz[(1-z)(1-z-dz)}
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Glitch View Post
    The question

    Show that the definition of \frac{d}{dz}(\frac{1}{1 - z}) = \frac{1}{(1 - z)^2}

    My attempt
    I tried using:

    \lim_{\delta z \to 0} \frac{f(z + \delta z) - f(z)}{\delta z}

    But having fractions in the numerator in this case appears difficult to simplify. I tried substituting z = x + iy, but I ended up with a massive mess which still results in a division by 0. Is there a good method of solving this problem?

    Thanks.
    With simple algebraic steps You obtain...

    \displaystyle \frac{f(z + \delta)- f(z)}{\delta} = \frac{1}{\delta} \ \{\frac{1}{1-z-\delta} - \frac{1}{1-z}\} =

    \displaystyle = \frac{1}{(1-z-\delta)\ (1-z)} (1)

    Now what is the \lim \delta \rightarrow 0 of the expression (1)?...

    Kind regards

    \chi \sigma
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  4. #4
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    Quote Originally Posted by Plato View Post
    WHY?

    \dfrac{\frac{1}{1-z-dz}-\frac{1}{1-z}}{dz}=\dfrac{(1-z)-(1-z-dz)}{dz[(1-z)(1-z-dz)}
    I'm such an idiot sometimes. Thanks.
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