Exercise 2.2 - Chapter 2 Complex Numbers 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $2.2$
Question 1.
Evaluate the following if $z=5-2 \mathrm{i}$ and $w=-1+3 \mathrm{i}$
(i) $z+w$
(ii) $z-$ iw
(iii) $2 z+3 w$
(iv) $\mathrm{zW}$
(v) $z^2+2 z w+w^2$
(vi) $(z+w)^2$
Solution:
$
\begin{aligned}
& \text { (i) } z=5-2 i, w=-1+3 i \\
& z+w=(5-2 i)+(-1+3 i) \\
& =(5-1)+(-2 i+3 i) \\
& =4+i
\end{aligned}
$
(ii) $z-i w=(5-2 i)-i(-1+3 i)$
$=5-2 \mathrm{i}+\mathrm{i}+3$
$=(5+3)+(-2 \mathrm{i}+\mathrm{i})$
$=8-i$
(iii) $2 z+3 w=2(5-2 i)+3(-1+3 i)$
$=10-4 \mathrm{i}-3+9 \mathrm{i}$
$=7+5 i$
(iv) $\mathrm{zw}=(5-2 \mathrm{i})(-1+3 \mathrm{i})$
$=-5+15 \mathrm{i}+2 \mathrm{i}-6 \mathrm{i}^2$
$=-5+17 \mathrm{i}+6$
$=1+17 \mathrm{i}$
(v) $z^2+2 z w+w^2=(z+w)^2[$ from (i) $]$
$=(4+i)^2$
$=16-1+8 \mathrm{i}$
$=15+8 \mathrm{i}$
(vi) $(z+w)^2=15+8 z[$ from $(v)]$
Question 2.
Given the complex number $z=2+3 i$, represent the complex numbers in Argand diagram.
(i) $z$, iz, and $z+i z$

(ii) $z,-i z$, and $z-i z$
Solution:
$
\begin{aligned}
& \text { (i) } z=2+3 i \\
& \text { iz }=i(2+3 i) \\
& =(2 i-3) \\
& =-3+2 i \\
& z+i z=(2+3 i)+(-3+2 i) \\
& =-1+5 i
\end{aligned}
$

$
\begin{aligned}
& \text { (ii) } z=2+3 i \\
& -i z=-i(2+3 i) \\
& =-2 i-3 i^2 \\
& =(3-2 i) \\
& z-i z=(2+3 i)+(3-2 i) \\
& =5+i
\end{aligned}
$

Question 3.
Find the values of the real numbers $\mathrm{x}$ and $\mathrm{y}$, if the complex numbers. $(3-i) x-(2-i) y+2 i+5$ and $2 x+(-1+2 i) y+3+2 i$ are equal

Solution:
$
\begin{aligned}
& (3-i) x-(2-i) y+2 i+5=2 x+(-1+2 i) y+3+2 i \\
& \Rightarrow 3 x-i x-2 y+i y+2 i+5=2 x-y+2 y i+3+2 i \\
& \Rightarrow(3 x-2 y+5)+1(-x+y+2)=(2 x-y+3)+i(2 y+2)
\end{aligned}
$
Equate real parts on both sides
$
3 \mathrm{x}-2 \mathrm{y}+5=2 \mathrm{x}-\mathrm{y}+3
$
$
x-y=-2
$
Equate imaginary parts on both sides
$
-x+y+2=2 y+2
$

$
\begin{aligned}
& -\mathrm{x}-\mathrm{y}=0 \\
& \mathrm{x}+\mathrm{y}=0 \ldots \ldots(2)
\end{aligned}
$
(1) $+(2) \Rightarrow 2 \mathrm{x}=-2$
$\mathrm{x}=-1$
Substituting $x=-1$ in (2)
$
\begin{aligned}
& -1+\mathrm{y}=0 \\
& \Rightarrow \mathrm{y}=1 \\
& \therefore \mathrm{x}=-1, \mathrm{y}=1
\end{aligned}
$