Question 2, Exercise 1.2

Solutions of Question 2 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Use the algebraic properties of complex numbers to prove that $$\left(z_{1} z_{2}\right)\left(z_{3} z_{4}\right)=\left(z_{1} z_{3}\right)\left(z_{2} z_{4}\right)=z_{3}\left(z_{1} z_{2}\right) z_{4}$$

Solution.

$$\begin{align} &(z_1 z_2)(z_3 z_4) \\ =&(z_1 z_2)z_5 \quad \text {Let }z_5=z_3 z_4 \\ =&z_1 (z_2 z_5) \quad \text{Multiplicative assocative law}\\ =&z_1\left(z_2 (z_3 z_4) \right) \quad \because\,\, z_5=z_3 z_4 \\ =&z_1 \left((z_2 z_3) z_4 \right) \quad \text{Multiplicative assocative law}\\ =&z_1 \left((z_3 z_2) z_4 \right) \quad \text{Multiplicative comutative law}\\ =&z_1 \left(z_3 (z_2 z_4) \right) \quad \text{Multiplicative assocative law}\\ =&(z_1 z_3) (z_2 z_4) \quad \text{Multiplicative assocative law} \end{align}$$ That is, we have proved $$(z_1 z_2)(z_3 z_4)=(z_1 z_3) (z_2 z_4) ... (i)$$ Now $$\begin{align} &(z_1 z_3) (z_2 z_4) \\ =&(z_3 z_1) (z_2 z_4)\quad \text{Multiplicative commutative law} \\ =&z_3 \left(z_1 (z_2 z_4)\right)\quad \text{Multiplicative associative law} \\ &z_3 \left((z_1 (z_2) z_4\right)\quad \text{Multiplicative associative law} \\ &z_3 (z_1 z_2) z_4 \quad \text{Multiplicative associative law} \end{align}$$ That is, we have proved $$(z_1 z_3) (z_2 z_4)=z_3 (z_1 z_2) z_4 ... (ii)$$ From (i) and (ii), we have the required result.

Remark: For any three complex numbers $z_1$, $z_2$ and $z_3$, we have $$z_1 (z_2 z_3) = (z_1 z_2)z_3 = z_1 z_2 z_3.$$ Logically, z_1 z_2 z_3 has no meaning as three number cannot be multiplies simultanously, but associate law tells us that the order in which we multiply three complex numbers doesn't matter; we will always end up with the same product. This property ensures consistency and helps simplify calculations involving complex numbers.

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