Exercise 2.3 - Chapter 2 Complex Numbers 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $2.3$
Question 1.
If $z_1=1-3 i, z_2=-4 i$, and $z_3=5$, show that
(i) $\left(z_1+z_2\right)+z_3=z_1+\left(z_2+z_3\right)$
(ii) $\left(z_1 z_2\right) z_3=z_1\left(z_2 z_3\right)$
Solution:
$
\begin{aligned}
& \text { (i) } z_1=1-3 i, z_2=-4 i, z_3=5 \\
& \left(z_1+z_2\right)+z_3 \\
& =(1-3 i-4 i)+5 \\
& =(1-7 i)+5 \\
& =6-7 i \ldots \ldots(1) \\
& z_1+\left(z_2+z_3\right) \\
& =(1-3 i)+(-4 i+5) \\
& =6-7 i \ldots \ldots \text { (2) }
\end{aligned}
$
from (1) \& (2) we get
$
\therefore\left(\mathrm{z}_1+\mathrm{z}_2\right)+\mathrm{z}_3=\mathrm{z}_1+\left(\mathrm{z}_2+\mathrm{z}_3\right)
$
(ii) $\left(z_1 z_2\right) z_3=[(1-3 i)(-4 i] 5$
$
\begin{aligned}
& =\left[-4 i+12 i^2\right] 5 \\
& =[-12-4 i] 5 \\
& =-60-20 i \ldots \ldots(1) \\
& z_1\left(z_2 z_3\right)=(1-3 i)[(-4 i)(5)] \\
& =\left(-20 i+60 i^2\right) \\
& =-20 i-60 \ldots \ldots(2)
\end{aligned}
$
from (1) \& (2) we get
$
\therefore\left(\mathrm{z}_1 \mathrm{z}_2\right)\left(\mathrm{z}_3\right)=\mathrm{z}_1\left(\mathrm{z}_2 \mathrm{z}_3\right)
$
Question 2.
If $z_1=3, z_2=-7 i$, and $z_3=5+4 i$, show that

(i) $z_1\left(z_2+z_3\right)=z_1 z_2+z_1 z_3$
(ii) $\left(z_1 z_2\right) z_3=z_1 z_3+z_2 z_3$
Solution:
$
\begin{aligned}
& \text { (i) } z_1=3, z_2=-7 i, z_3=5+4 i \\
& z_1\left(z_2+z_3\right)=3(-7 i+5+4 i) \\
& =3(5-3 i) \\
& =15-9 i \ldots \ldots(1) \\
& z_1 z_2+z_1 z_3=3(-7 i)+3(5+4 i) \\
& =-21 i+15+12 i \\
& =15-9 i \ldots \ldots \text { (2) }
\end{aligned}
$
from (1) \& (2), we get
$
\therefore \mathrm{z}_1\left(\mathrm{z}_2+\mathrm{z}_3\right)=\mathrm{z}_1 \mathrm{z}_2+\mathrm{z}_1 \mathrm{z}_3
$
(ii) $\left(z_1+z_2\right) z_3=(3-7 i)(5+4 i)$
$
=15+12 \mathrm{i}-35 \mathrm{i}-28 \mathrm{i}^2
$
$
\begin{aligned}
& =15-23 \mathrm{i}+28 \\
& =43-23 \mathrm{i} \ldots \ldots(1) \\
& \mathrm{z}_1 \mathrm{z}_3+\mathrm{z}_2 \mathrm{z}_3=3(5+4 \mathrm{i})-7 \mathrm{i}(5+4 \mathrm{i}) \\
& =15+12 \mathrm{i}-35 \mathrm{i}-28 \mathrm{i}^2 \\
& =15-23 \mathrm{i}+28 \\
& =43-23 \mathrm{i} \ldots .(2)
\end{aligned}
$
from (1) \& (2), we get
$
\therefore\left(z_1+z_2\right) z_3=z_1 z_3+z_2 z_3
$
Question 3.
If $z_1=2+5 i, z_2=-3-4 i$, and $z_3=1+i$, find the additive and multiplicative inverse of $z_1, z_2$, and $z_3$.
Solution:
$
\mathrm{z}_1=2+5 \mathrm{i}
$
(a) Additive inverse of $\mathrm{z}_1=-\mathrm{z}_1=-(2+5 \mathrm{i})=-2-5 \mathrm{i}$
(b) Multiplicative inverse of
$
\begin{aligned}
& z_1=\frac{1}{z_1}=\frac{1}{2+5 i} \times \frac{2-5 i}{2-5 i}=\frac{(2-5 i)}{4+25}=\frac{2-5 i}{29} \\
& \mathrm{z}_2=-3-4 \mathrm{i}
\end{aligned}
$
(a) Additive inverse of $z_2=-z_2=-(-3-4 i)=3+4 i$
(b) Multiplicative inverse of
$
\begin{aligned}
& z_2=\frac{1}{z_2} \frac{1}{-3-4 i} \times \frac{-3+4 i}{-3+4 i}=\frac{-3+4 i}{9+16}=\frac{-3+4 i}{25} \\
& \mathrm{z}_3=1+\mathrm{i}
\end{aligned}
$
(a) Additive inverse of $z_3=-z_3=-(1+i)=-1-i$
(b) Multiplicative inverse of
$
z_3=\frac{1}{z_3} \frac{1}{1+i} \times \frac{1-i}{1-i}=\frac{1-i}{2}
$