Exercise 2.2-Additional Problems - Chapter 2 - Complex Numbers - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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AdditionalProblems
Question 1.
Find the real values of $x$ and $y$, if
(i) $(3 x-7)+2 i y=-5 y+(5+x) i$
(ii) $(1-\mathrm{i}) \mathrm{x}+(1+\mathrm{i}) \mathrm{y}=1-3 \mathrm{i}$
(iii) $(\mathrm{x}+\mathrm{iy})(2-3 \mathrm{i})=4+\mathrm{i}$
(iv) $\frac{x-1}{3+i}+\frac{y-1}{3-i}=i$
Solution:
$
\begin{aligned}
& \text { (i) We have }(3 x-7)+2 \text { iy }=5 y+(5+x) i \\
& \Rightarrow 3 x-7=5 y \text { and } 2 y=5+x \\
& \Rightarrow 3 x+5 y=7 \text { and } x-2 y=-5 \\
& \Rightarrow x=-1 y=2
\end{aligned}
$
(ii) We have, $(1-i) x+(1+i) y=1-3 i$
$
\begin{aligned}
& \Rightarrow(\mathrm{x}+\mathrm{y})+\mathrm{i}(-\mathrm{x}+\mathrm{y})=1-3 \mathrm{i} \\
& \Rightarrow \mathrm{x}+\mathrm{y}=1
\end{aligned}
$
and $-x+y=3$ [On equating real and imaginary parts] $\Rightarrow \mathrm{x}=2$ and $\mathrm{y}=-1$

(iii) We have $(x+i y)(2-3 i)=4+i$
$
\begin{array}{lrl}
\Rightarrow & (2 x+3 y)+i(-3 x+2 y)=4+i \\
\Rightarrow & 2 x+3 y=4 \text { and }-3 x+2 y=1 \\
\Rightarrow & x=\frac{5}{13}, y=\frac{14}{13}
\end{array}
$
[On equating real and imaginary parts]
(iv) We have, $\frac{x-1}{3+i}+\frac{y-1}{3-i}=i$
$
\begin{aligned}
& \Rightarrow \quad \frac{(x-1)(3-i)+(y-1)(3+i)}{(3+i)(3-i)}=i \\
& \Rightarrow \frac{(3 x+3 y-6)+i(y-x)}{9-i^2}=i \\
& \Rightarrow\left(\frac{3 x+3 y-6}{10}\right)+i\left(\frac{y-x}{10}\right)=0+i \\
& \Rightarrow \quad \frac{3 x+3 y-6}{10}=0 \text { and } \frac{y-x}{10}=1 \\
& \Rightarrow \mathrm{x}+\mathrm{y}-2=0 \text { and } \mathrm{y}-\mathrm{x}=10 \\
& \Rightarrow \mathrm{x}=-4, \mathrm{y}=6 .
\end{aligned}
$
[On equating real and imaginary parts]Question 2.
Find the real values of $x$ and $y$ for which the complex numbers $-3+i x^2 y$ and $x^2+y+4 i$ are conjugate of each other.
Solution:
Since $-3+i^2 y$ and $x^2+y+4 i$ are complex conjugates.
$\therefore-3+\mathrm{ix}^2 y=\mathrm{x}^2+\mathrm{y}+4 \mathrm{i}$
... (0 and, $x^2 y=-4$.... (ii)
$\Rightarrow-3=x^2-\frac{4}{x^2}$ $\Rightarrow x^4+3 x^2-4=0$ $\Rightarrow\left(x^2+4\right)\left(x^2-1\right)=0$ $\Rightarrow x^2-1=0$ $\Rightarrow x=\pm 1$ From $($ ii $), y=-4$, when $x=\pm 1$.
$\left[\right.$ Putting $y=-4 / x^2$ from $(i i)$ in $\left.(i)\right]$
Hence $x=1, y=-4$ or, $x=-1, y=-4$.
Question 3.
Given $x=2-3 i$ and $y=4+i$
Find (i) $x y$
(ii) $x+y$
(iii) $x-y$
(iv) $\frac{x}{y}$
(v) $\frac{\bar{x}}{y}$
$(v i)(x-y)^2$
Solution:
Do it yourself