Problem
Show that $(\alpha + \beta) + \lambda = \alpha + (\beta + \lambda)$ for all $\alpha, \beta, \lambda \in \mathbb{C}$
Solution
Let $\alpha = a + bi$, $\beta = c + di$ and $\lambda = e + fi$. Then by the associativity of real numbers: $$(\alpha + \beta) + \lambda = \left((a + bi) + (c + di)\right) + (e + fi)$$ $$\stackrel{D1.1}{=} ((a + c) + (b + d)i) + (e + fi) \stackrel{D1.1}{=} ((a + c) + e) + ((b + d) + f)i$$ $$= (a + (c + e)) + (b + (d + f))i \stackrel{D1.1}{=} (a + bi) + ((c + e) + (d + f)i)$$ $$= \alpha + (\beta + \lambda)$$ Proving the associativity complex numbers under addition.