ROUGH DRAFT authorea.com/102742
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  • Problem 2.10

    [2.10] Consider the function \( f:\mathbb{C}/\{{0}\}\to\mathbb{C}\) given by \(f(x)=\frac{1}{x}.\) Apply the definition of the derivative to give a direct proof that \(f'(z)=-\frac{1}{z^2}.\)

    Let \(f:\mathbb{C}/\{{0}\}\to\mathbb{C}\) be given by \(f(x)=\frac{1}{x}.\) Then by the definition of a derivative we have the following:

    \[\begin{aligned} f'(x) &=\lim_{h\to 0} \frac{f(z_0+h)-f(z_0)}{h}\\ &=\lim_{h\to0}\frac{\frac{1}{(z_0+h)}-\frac{1}{z_0}}{h}\\ &=\lim_{h\to 0}\frac{\frac{z_0}{z_0(z_0+h)}-\frac{(z_0+h)}{z_0(z_0+h)}}{h}\\ &=\lim_{h\to 0}\frac{\frac{z_0-z_0-h}{z_0(z_0+h)}}{h}\\ &=\lim_{h\to 0}\frac{\frac{-h}{z_0(z_0+h)}}{h}\\ &= \lim_{h\to 0}\frac{-1}{z_0(z_0+h)}\\ &= \lim_{h\to 0}\frac{-1}{(z_0)^2+z_0h}\\ &= \frac{-1}{(z_0)^2+z_0(0)}\\\end{aligned}\]

    Therefore, \(f'(z) = \frac{-1}{(z_0)^2}\), such that \(z_0 \neq 0.\)

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