Exercise 2.3-Additional Problems - Chapter 2 - Complex Numbers - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Problems
Question 1.
If $z_1=4-7 i, z_2=2+3 i$ and $z_3=1+i$ show that.
(i) $z_1+\left(z_2+z_3\right)=\left(Z_1+z_2\right)+Z_3$
(ii) $\left(Z_1 z_2\right) z_3=Z_1\left(z_2 z_3\right)$
Solution:
$
\begin{aligned}
& \text { (i) } Z_1+\left(z_2+z_{13}\right)=4-7 i+(2+3 i+1+i)=4-7 i+(3+4 i)=7-3 i \\
& \left(z_1+z_2\right)+z_3=(4-7 i+2+3 i)+(1+i)=(6-4 i)+(1+i) \\
& =7-3 i \ldots \text { (2) }
\end{aligned}
$
From (1) and (2) we get, $z_1+\left(z_2+z_3\right)=\left(z_1+z_2\right)+z_3$
(ii) $\left(z_1 z_2\right) z_3=(4-7 i)+(2+3 i)(1+i)=(8+12 i-14 i+21)(1+i)$
$=(29-2 \mathrm{i})(1+\mathrm{i})=29+29 \mathrm{i}-2 \mathrm{i}+2$
$=31+27 i \ldots(1)$
$\mathrm{Z}_1\left(\mathrm{z}_2 \mathrm{z}_3\right)=(4-7 \mathrm{i})[(2+3 \mathrm{i})(1+\mathrm{i})]=4-7 \mathrm{i}[2+2 \mathrm{i}+3 \mathrm{i}-3]$
$=4-7 i[5 \mathrm{i}-1]=20 \mathrm{i}-4+35+7 \mathrm{i}$
$=31+27 \mathrm{i} \ldots(2)$
From (1) and (2) we get, $\left(Z_1 Z_2\right) z_3=z_1\left(z_2 z_3\right)$
Question 2.
Given $z_1=1+i, z_2=4-3 i$ and $z_3=2+5 i$ verify that.
$
\mathrm{Z}_1\left(\mathrm{Z}_2 \mathrm{Z}_3\right)=\mathrm{Z}_1 \mathrm{Z}_2-\mathrm{z}_1 \mathrm{Z}_3
$
Solution:
$
\begin{aligned}
& Z_1=1+i \\
& z_2=4-3 i \\
& z_3=2+5 i \\
& Z_1\left(z_2-z_3\right)=1+i[(4-3 i)-(2+5 i)]=1+i[2-8 i]=2-8 i+2 i+8 \\
& =10-6 i \ldots(1) \\
& Z_1 z_2=(1+i)(4-3 i)=4-3 i+4 i+3=7+i \\
& Z_1 Z_3=(1+i)(2+5 i)=2+5 i+2 i-5=7 i-3 \\
& Z_1 Z_2-Z_1 z_3=7+i-(7 i-3)=7+i-7 i+3 \\
& =10-6 i \ldots \text { (2) }
\end{aligned}
$
From (1) and (2) we get, $Z_1\left(z_2-z_3\right)=z_1 z_2-z_1 z_3$
Question 3.
Given $z_1=4-7 i$ and $z_2=5+6 i$ find the additive and multiplicative inverse of $z_1+z_2$ and $Z_1-Z_2$.
Solution:

$
\begin{aligned}
& Z_1=4-7 i, z_2=5+6 i \\
& Z_1+z_2=4-7 i+5+6 i=9+i
\end{aligned}
$
Additive inverse of $z_1+z_2=-\left(z_1+z_2\right)=-(9-i)=i-9$
Multiplicative inverse of $z_1+z_2=\frac{1}{z_1+z_2}=\frac{1}{9-i}$
$
z_1-z_2=4-7 i-(5+6 i)=4-7 i-5-6 i=-13 i-1=-(1+13 i)
$
Additive inverse of $z_1-z_2=-\left(z_1-z_2\right)=-[-(1+13 i)]=1+13 i$
Multiplicative inverse of $z_1-z_2=\frac{1}{z_1-z_2}=\frac{1}{-(1+13 i)}=-\frac{1}{1+13 i}$