Examples (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

If $4 x+i(3 x-y)=3+i(-6)$, where $x$ and $y$ are real numbers, then find the values of $x$ and $y$.

Solution

We have
$
4 x+i(3 x-y)=3+i(-6)
$

Equating the real and the imaginary parts of (1), we get
$
4 x=3,3 x-y=-6 \text {, }
$
which, on solving simultaneously, give $x=\frac{3}{4}$ and $y=\frac{33}{4}$.

Example 2

Express the following in the form of $a+b i$ :
(i) $(-5 i)\left(\frac{1}{8} i\right)$
(ii) $(-i)(2 i)\left(-\frac{1}{8} i\right)^3$

Solution
(i) $(-5 i)\left(\frac{1}{8} i\right)=\frac{-5}{8} i^2=\frac{-5}{8}(-1)=\frac{5}{8}=\frac{5}{8}+i 0$
(ii) $(-i)(2 i)\left(-\frac{1}{8} i\right)^3=2 \times \frac{1}{8 \times 8 \times 8} \times i^5=\frac{1}{256}\left(i^2\right)^2 i=\frac{1}{256} i$.

Example 3

Express $(5-3 i)^3$ in the form $a+i b$.
Solution

We have, $(5-3 i)^3=5^3-3 \times 5^2 \times(3 i)+3 \times 5(3 i)^2-(3 i)^3$
$
=125-225 i-135+27 i=-10-198 i \text {. }
$

Example 4

Express $(-\sqrt{3}+\sqrt{-2})(2 \sqrt{3}-i)$ in the form of $a+i b$

Solution

We have,
$
\begin{aligned}
& (-\sqrt{3}+\sqrt{-2})(2 \sqrt{3}-i)=(-\sqrt{3}+\sqrt{2} i)(2 \sqrt{3}-i) \\
& =-6+\sqrt{3} i+2 \sqrt{6} i-\sqrt{2} i^2=(-6+\sqrt{2})+\sqrt{3}(1+2 \sqrt{2}) i
\end{aligned}
$

Example 5

Find the multiplicative inverse of $2-3 i$.
Solution

Let $z=2-3 i$
Then $\quad \bar{z}=2+3 i$ and $\quad|z|^2=2^2+(-3)^2=13$
Therefore, the multiplicative inverse of $2-3 i$ is given by
$
z^{-1}=\frac{\bar{z}}{|z|^2}=\frac{2+3 i}{13}=\frac{2}{13}+\frac{3}{13} i
$

The above working can be reproduced in the following manner also,
$
\begin{aligned}
z^{-1} & =\frac{1}{2-3 i}=\frac{2+3 i}{(2-3 i)(2+3 i)} \\
& =\frac{2+3 i}{2^2-(3 i)^2}=\frac{2+3 i}{13}=\frac{2}{13}+\frac{3}{13} i
\end{aligned}
$

Example 6

Express the following in the form $a+i b$
(i) $\frac{5+\sqrt{2} i}{1-\sqrt{2} i}$
(ii) $i^{-35}$

Solution
(i) We have,
$
\begin{aligned}
& \frac{5+\sqrt{2} i}{1-\sqrt{2} i}=\frac{5+\sqrt{2} i}{1-\sqrt{2} i} \times \frac{1+\sqrt{2} i}{1+\sqrt{2} i}=\frac{5+5 \sqrt{2} i+\sqrt{2} i-2}{1-(\sqrt{2} i)^2} \\
& =\frac{3+6 \sqrt{2} i}{1+2}=\frac{3(1+2 \sqrt{2} i)}{3}=1+2 \sqrt{2} i
\end{aligned}
$

$\text { (ii) } i^{-35}=\frac{1}{i^{35}}=\frac{1}{\left(i^2\right)^{17} i}=\frac{1}{-i} \times \frac{i}{i}=\frac{i}{-i^2}=i$