2.10

2.10 Consider the function \(f:\mathbb{C} \backslash \{ 0 \} \rightarrow \mathbb{C}\) given by \(f(z)=\frac{1}{z}\). Apply the definition of the derivative to give a direct proof that \(f'(z)=-\frac{1}{z^2}\). ———————————————————————————————–
Proof:
Let the function \(f:\mathbb{C} \backslash \{ 0 \} \rightarrow \mathbb{C}\) be given by \(f(z)=\frac{1}{z}\) and let \(z \in \mathbb{C} \backslash \lbrace 0 \rbrace\). To find the derivative, we apply the difference quotient limit: \[\begin{aligned} f'(z)&=\lim\limits_{z \rightarrow z_0} \frac{f(z+h)-f(z)}{h}\\ &=\lim\limits_{z \to x_0} \frac{\frac{1}{(z+h)}-\frac{1}{z}}{h}\\ &=\lim\limits_{z \to z_0} \frac{\frac{z}{z(z+h)}-\frac{z+h}{z(z+h)}}{h}\\ &=\lim\limits_{z \to z_0} \frac{\frac{z-z-h}{z^2+hz}}{h}\\ &=\lim\limits_{z \to z_0} \frac{\frac{-h}{z^2+hz}}{h}\\ &=\lim\limits_{z \to z_0} \frac{-h}{h(z^2+hz)}\\ &=\lim\limits_{z \to z_0} \frac{-1}{(z^2+hz)}\\ &=\frac{-1}{z^2+0z} &\text{apply the limit}\\ &=\frac{-1}{z^2}\end{aligned}\] Thus the derivative of \(f:\mathbb{C} \backslash \{ 0 \} \rightarrow \mathbb{C}\) given by \(f(z)=\frac{1}{z}\) is \(-\frac{1}{z^2}\).