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