Question: Show that (πΌπ½)π = πΌ(π½π) for all πΌ, π½, π β π.
Answer: Let $\alpha = a + ib$, $\beta = c + id$ and $\lambda = e + if$ be given.
Then, $$ (\alpha + \beta)\lambda = ((a + ib)(c + id))(e + if) = (ac - bd) + i(ad + bc))(e + if) = (ace - bde - bdf + bcf) + i(ade + bce + bcf + bde) = (a + ib)((ce - df) + i(de + cf)) = \alpha (\beta \lambda) $$
Where the middle equalites come from the properties of $\mathbb{R}$.