Miscellaneous Example (Revised) - Chapter 5 - Complex Numbers & Quadratic Equations - Ncert Solutions class 11 - Maths

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Chapter 4 - Complex Numbers & Quadratic Equations | NCERT Solutions Class 11 Maths

Example 1 

Find the conjugate of $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$.
Solution

We have, $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$
$
\begin{aligned}
& =\frac{6+9 i-4 i+6}{2-i+4 i+2}=\frac{12+5 i}{4+3 i} \times \frac{4-3 i}{4-3 i} \\
& =\frac{48-36 i+20 i+15}{16+9}=\frac{63-16 i}{25}=\frac{63}{25}-\frac{16}{25} i
\end{aligned}
$

Therefore, conjugate of $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$ is $\frac{63}{25}+\frac{16}{25} i$.
Example 2

If $x+i y=\frac{a+i b}{a-i b}$, prove that $x^2+y^2=1$.
Solution

We have,
$
x+i y=\frac{(a+i b)(a+i b)}{(a-i b)(a+i b)}=\frac{a^2-b^2+2 a b i}{a^2+b^2}=\frac{a^2-b^2}{a^2+b^2}+\frac{2 a b}{a^2+b^2} i
$

So that, $x-i y=\frac{a^2-b^2}{a^2+b^2}-\frac{2 a b}{a^2+b^2} i$
Therefore,
$
x^2+y^2=(x+i y)(x-i y)=\frac{\left(a^2-b^2\right)^2}{\left(a^2+b^2\right)^2}+\frac{4 a^2 b^2}{\left(a^2+b^2\right)^2}=\frac{\left(a^2+b^2\right)^2}{\left(a^2+b^2\right)^2}=1
$