Proving quotient rule in the complex plane

Question

Using complex limits, prove that if $f(z)$ is differentiable at $z = a$ and $f(a) \neq 0$, then the reciprocal function $1/f(z)$ is differentiable at $z = a$ and the derivative at that point is $\frac{−f'(a)}{(f(a))2}$.

I'm not quite sure how to approach this proof. I assume we substitute the limit definition of a derivative in the numerator, but that's as far as I understand how to take this.

Answer 1

we know that $$\frac{d}{dz}(g(z))^{-1}=-\frac{g'(z)}{(g(z))^2}$$ Substituting this gives $$\frac{d}{dz}(h(z))=\frac{f'(z)}{g(z)}-\frac{g'(z)\cdot f(z)}{(g(z))^2}.$$ To find a common denominator one can multiply the first fraction by $\frac{g(z)}{g(z)}$ to get $$\frac{d}{dz}(h(z))=\frac{f'(z)\cdot g(z)}{(g(z))^2}-\frac{g'(z)\cdot f(z)}{(g(z))^2}.$$ Subtracting these fractions and rearranging the products in the numerator via the commutative property of multiplication gives

$$\frac{d}{dz}(h(z))=\frac{g(z)\cdot f'(z)-f(z)\cdot g'(z)}{(g(z))^2}.$$ Finally, substituting $h(z)=\frac{f(z)}{g(z)}$ gives the product rule $$\frac{d}{dz}(\frac{f(z)}{g(z)})=\frac{g(z)\cdot f'(z)-f(z)\cdot g'(z)}{(g(z))^2}.$$

Answer 2

You simply have to prove that if $\iota(z)=\frac1z$, then $\iota'(z)=-\frac1{z^2}$. After that, use the chain rule and the fact that $\frac1f=\iota\circ f$.