Exercise 2.4-Additional Problems - Chapter 2 Complex Numbers 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

AdditionalProblems
Question 1.
Express the following in the standard form $\mathrm{a}+\mathrm{ib}$.
(i) $\frac{2(i-3)}{(1+i)^2}$
(ii) $\frac{(1+i)(1-2 i)}{(1+3 i)}$
(iii) $(-3+i)(4-2 i)$
(iv) $\frac{i^4+i^9+i^{16}}{3-2 i^8-i^{10}-i^{15}}$
Solution:
(i)
$
\begin{aligned}
\frac{2(i-3)}{(1+i)^2} & =\frac{2(-3+i)}{1+i^2+2 i}=\frac{2(-3+i)}{2 i} \\
& =\frac{-3+i}{i}=(-3+i)(-i) \\
& =+3 i-i^2=3 i+1=1+3 i
\end{aligned}
$
(ii)
$
\begin{aligned}
\frac{(1+i)(1-2 i)}{(1+3 i)} & =\frac{1-2 i+i-2 i^2}{1+3 i}=\frac{1-i+2}{1+3 i} \\
& =\frac{3-i}{1+3 i}=\frac{3-i}{1+3 i} \times \frac{1-3 i}{1-3 i}=\frac{3-9 i-i+3 i^2}{1^2-3^2}=\frac{3-10 i-3}{10} \\
& =\frac{-10 i}{10}=-i=0+(-1) i
\end{aligned}
$
(iii) $\quad(-3+i)(4-2 i)=-12+6 i+4 i-2 i^2=-12+10 i+2=-10+10 i=-10+(10) i$

(iv) $\frac{i^4+i^9+i^{16}}{3-2 i^8-i^{10}-i^{15}}$
$
\begin{array}{cll}
i^4=1 & i^9=i^1=i & i^{16}=1 \\
i^8=1 & i^{10}=i^2=-1 & i^5=i^3=-i \\
\text { So; } \frac{i^4+i^9+i^{16}}{3-2 i^8-i^{10}-i^{15}}=\frac{1+i+1}{3-2(1)-(-1)-(-i)} \\
& =\frac{2+i}{3-2+1+i}=\frac{2+i}{2+i}=1=1+i(0)
\end{array}
$
Question 2.
Find the least positive integer $\mathrm{n}$ such that $\left(\frac{1+i}{1-i}\right)^n=\mathbf{1}$.

Solution:
$
\begin{aligned}
& \frac{1+i}{1-i}=\frac{1+i}{1-i} \times \frac{1+i}{1+i}=\frac{1+i+i+i^2}{1+1}=\frac{2 i}{2}=i \\
\therefore \quad\left(\frac{1+i}{1-i}\right)^n & =i^n=1 \Rightarrow \text { least value of } n=4
\end{aligned}
$
$\left[\because i^4=1\right]$

Question 3.
Find the real values of $\mathrm{x}$ and $\mathrm{y}$ for which the following equations are satisfied.
(i) $(1-i) x+(1+i) y=1-3 i$
(ii) $\frac{(1+i) x-2 i}{3+i}+\frac{(2-3 i) y+i}{3-i}=i$
(iii) $\sqrt{x^2+3 x+8}+(x+4) i=y(2+i)$
Solution:
(i) $(1-i) x+(1+i) y=x-i x+y+i y$ $=(x+y)+i(y-x)=1-3 i$ (given)
So, equating their RP and IP we get,
$
\Rightarrow \begin{aligned}
x+y & =1 \\
y-x & =-3 \\
x-y & =3
\end{aligned}
$
Solving (1) and (2),
$
\begin{aligned}
x+y & =1 \\
x-y & =3 \\
\Rightarrow 2 x & =4 \quad \Rightarrow x=2
\end{aligned}
$
Substituting $x=2$ in (1) we get,
$
2+y=1 \quad \Rightarrow y=1-2=-1
$
So, $x=2, y=-1$

$
\begin{aligned}
& \text { (ii) } \frac{x+i x-2 i}{3+i}+\frac{2 y-3 i y+i}{3-i}=i \\
& \text { i.e., } \frac{x+i(x-2)}{3+i}+\frac{2 y+i(1-3 y)}{3-i}=i \\
& \text { i.e., } \frac{(3-i)[x+i(x-2)]+(3+i)[2 y+i(1-3 y)]}{(3+i)(3-i)}=i \\
& \Rightarrow \quad(3-i)[x+i(x-2)]+(3+i)[2 y+i(1-3 y)]=10 i \quad[\because(3+i)(3-i)=1] \\
& \text { i.e., } 3 x+3 i(x-2)-i x-i^2(x-2)+6 y+ \\
& 3 i(1-3 y)+2 i y+i^2(1-3 y)=10 i \\
& \Rightarrow \quad R P=0 \text { and } I P=10 \\
&
\end{aligned}
$
Take real part, we get
$
\begin{aligned}
& \text { i.e., } 3 x+(x-2)+6 y-(1-3 y)=0 \\
& \Rightarrow 3 x+x-2+6 y-1+3 y=0 \\
& 4 x+9 y=3
\end{aligned}
$
Take imaginary part, we get
$
3(x-2)-x+3(1-3 y)+2 y=10
$

$\begin{aligned}
&\Rightarrow 3 x-6-x+3-9 y+2 y=10\\
&2 x-7 y=13
\end{aligned}$

(iii) $\sqrt{x^2+3 x+8}=2 y$
Comparing RP and IP, $y=x+4$
Substituting (2) in (1), $\sqrt{x^2+3 x+8}=2(x+4)$
Squaring on both sides, $x^2+3 x+8=4(x+4)^2$
$
\begin{aligned}
& \text { i.e., } \mathrm{x}^2+3 \mathrm{x}+8=4\left(\mathrm{x}^2+8 \mathrm{x}+16\right) \Rightarrow 4 \mathrm{x}^2+32 \mathrm{x}+64-\mathrm{x}^2-3 \mathrm{x}-8=0 \\
& 3 \mathrm{x}^2+29 \mathrm{x}+56=0 \\
& 3 \mathrm{x}^2+21 \mathrm{x}+8 \mathrm{x}+56=0 \\
& (\mathrm{x}+7)(3 \mathrm{x}+8)=0 \\
& \quad x=-7 \text { or } \frac{-8}{3}
\end{aligned}
$
When $x=-7, y=-3$
When $x=\frac{-8}{3}, y=\frac{4}{3}$