Problem 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 $$z_0 \neq 0.$$