This is a rather rudimentary question, but I cannot seem to rigorously justify it other than appealing to intuition.
Suppose I am doing arithmetic in $\mathbb{C}$, and I come to $$ \frac{a - bi}{a^2 + b^2}. $$ (For simplicity, assume that at least one of $a$ or $b$ is non-zero). This is certainly equal to $$\frac{a}{a^2 + b^2} + \left(\frac{-b}{a^2 + b^2}\right)i.$$ What is the justification for this step other than "definition of complex addition"? Addition in $\mathbb{C}$ states to add the real components and the complex components but it doesn't seem that we're doing that. Instead, we're "applying the inverse" of $a^2 + b^2$ and using the fact that multiplication in $\mathbb{C}$ is commutative. If I wanted to rigorously justify this step by step, I might write: \begin{align*} \frac{a-bi}{a^2 + b^2} & = (a - bi)(a^2 + b^2)^{-1} & & \text{def of inverse in $\mathbb{C}$; commut of mult} \\ & = a(a^2 + b^2)^{-1} - bi(a^2 + b^2)^{-1} & & \text{distributivity in $\mathbb{C}$} \\ & = \frac{a}{a^2 + b^2} + \frac{-bi}{a^2 + b^2} & & \text{rewrite; associativity of scalar mult in $\mathbb{R}$} \\ & = \frac{a}{a^2 + b^2} + \left(\frac{-b}{a^2 + b^2}\right) i & & \text{associativity of mult in $\mathbb{R}$} \end{align*} Are either of these lines of reasoning correct?
Forget about $\Bbb C$ for the moment. How do you prove $\frac{x+y}{z}=\frac{x}{z}+\frac{y}{z}$? Your$$\frac{x+y}{z}=(x+y)z^{-1}=xz^{-1}+yz^{-1}=\frac{x}{z}+\frac{y}{z}$$technique (or change the $+$s to $-$ if you like) is the right one; it explains why multiplication's distributivity over addition implies division's behaviour.