Question 2, Exercise 1.2

Solutions of Question 2 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

$z_1=-1+i$, $z_2=3-2i$ and ${{z}_{3}}=2-2i$, then verify associative property w.r.t. addition and multiplication.

Given ${{z}_{1}}=-1+i$, ${{z}_{2}}=3-2i$ and ${{z}_{3}}=2-2i$. First, we prove associative property under addition, that is, $$(z_1+z_2)+z_3=z_1+(z_2+z_3).$$ Take \begin{align} {{z}_{1}}+{{z}_{2}}&=\left( -1+i \right)+\left( 3-2i \right)\\ &=2-i\end{align} So \begin{align} \left( {{z}_{1}}+{{z}_{2}} \right)+{{z}_{3}}&=\left( 2-i \right)+\left( 2-2i \right)\\ &=4-3i \ldots (1)\end{align} Now \begin{align} {{z}_{2}}+{{z}_{3}}&=\left( 3-2i \right)+\left( 2-2i \right)\\ &=5-4i\end{align} So \begin{align} {{z}_{1}}+\left({{z}_{2}}+{{z}_{3}} \right)&=\left(-1+i \right)+\left( 5-4i \right)\\ &=4-3i \ldots (2) \end{align} From (1) and (2), we have the required result.

Now, we prove associative property under multiplication, that is, $$z_1 (z_2 z_3)=(z_1 z_2) z_3.$$ Take \begin{align} z_2 z_3&=(3-2i)\cdot(2-2i)\\ &=(6-4)+(-4-6)i\\ &=2-10i \end{align} So \begin{align} z_1(z_2 z_3)&=(-1+i)\cdot (2-10i)\\ &=(-2+10)+(2+10)i\\ &=8+12i \ldots (3)\end{align} Now, we take \begin{align} z_1 z_2 &=(-1+i)\cdot (3-2i)\\ &=(-3+2)+(3+2)i\\ &=-1+5i\end{align} So \begin{align} (z_1 z_2) z_3&=(-1+5i)\cdot (2-2i)\\ &=(-2+10)+(10+2)i\\ &=8+12i \ldots (4)\end{align} From (3) and (4), we get the required result.

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