Exercise 4.1 (Revised) - Chapter 5 - Complex Numbers & Quadratic Equations - Ncert Solutions class 11 - Maths

You can Download the Exercise 4.1 (Revised) - Chapter 5 - Complex Numbers & Quadratic Equations - Ncert Solutions class 11 - Maths with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

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

Express each of the complex numbers given in the exercises 1 to 10 in the form a+ib:
Ex 4.1 Question 1.

$(5 i)\left(\frac{-3}{5} i\right)$

Answer.

Here $(5 i)\left(\frac{-3}{5} i\right)=-3 i^2$
$
\begin{aligned}
& =-3 x-1 \\
& =3
\end{aligned}
$
Ex 4.1 Question 2.

$i^9+i^{19}$

Answer.

Here $i^9+i^{19}$
$
=\left(i^2\right)^4 i+\left(i^2\right)^9 \cdot i=(-1)^4 i+(-1)^9 \cdot i=i-i=0
$

Ex 4.1 Question 3.

$i^{-39}$

Answer.

$i^{-39}=\frac{1}{i^{39}}=\frac{1}{\left(i^2\right)^{19} i}=\frac{1}{(-1)^{19} i}=\frac{1}{-i}$
$
=\frac{-1}{i} \times \frac{i}{i}=\frac{-i}{i^2}=\frac{-i}{-1}=i
$
Ex 4.1 Question 4.

$3(7+i 7)+i(7+i 7)$

Answer.

Here $3(7+i 7)+i(7+i 7)$

$
=21+21 i+7 i+7 i^2=21+28 i-7=14+28 i
$
Ex 4.1 Question 5.

$(1-i)-(-1+i 6)$

Answer.

Here $(1-i)-(-1+i 6)$
$
=1-i+1-i 6=2-7 i
$
Ex 4.1 Question 6.

$\left(\frac{1}{5}+\frac{2}{5} i\right)-\left(4+\frac{5}{2} i\right)$

Answer.

Here $\left(\frac{1}{5}+\frac{2}{5} i\right)-\left(4+\frac{5}{2} i\right)$
$
\begin{aligned}
& =\frac{1}{5}+\frac{2}{5} i-4-\frac{5}{2} i \\
& =\left(\frac{1}{5}-4\right)+\left(\frac{2}{5}-\frac{5}{2}\right) i=\frac{-19}{5}-\frac{21}{10} i
\end{aligned}
$
Ex 4.1 Question 7.

$\left[\left(\frac{1}{3}+\frac{7}{3} i\right)+\left(4+\frac{1}{3} i\right)\right]-\left[\frac{-4}{3}+i\right]$

Answer.

Here $\left[\left(\frac{1}{3}+\frac{7}{3} i\right)+\left(4+\frac{1}{3} i\right)\right]-\left[\frac{-4}{3}+i\right]$

$
\begin{aligned}
& =\left[\left(\frac{1}{3}+4\right)+\left(\frac{7}{3}+\frac{1}{3} i\right)\right]-\left[\frac{-4}{3}+i\right] \\
& =\frac{13}{3}+\frac{8}{3} i+\frac{4}{3}-i=\left(\frac{13}{3}+\frac{4}{3}\right)+\left(\frac{8}{3}-1\right) i=\frac{17}{3}+\frac{5}{3} i
\end{aligned}
$
Ex 4.1 Question 8.

$(1-i)^4$

Answer.

Here $(1-i)^4=\left[(1-i)^2\right]^2$

$
\begin{aligned}
& =\left[1+i^2-2 i\right]^2=(1-1-2 i)^2=(-2 i)^2 \\
& =4 i^2=-4
\end{aligned}
$
Ex 4.1 Question 9.

$\left(\frac{1}{3}+3 i\right)^3$

Answer.

Here $\left(\frac{1}{3}+3 i\right)^3$
$=\left(\frac{1}{3}\right)^3+(3 i)^3+3 \times\left(\frac{1}{3}\right)^2 \times 3 i+3 \times \frac{1}{3} \times(3 i)^2$
$=\frac{1}{27}+27 i^3+i+9 i^2$
$=\frac{1}{27}-27 i+i-9$
$=\frac{1}{27}-9-26 i$
$=\frac{-242}{27}-26 i$

Ex 4.1 Question 10.

$\left(-2-\frac{1}{3} i\right)^3$

Answer.

Here $\left(-2-\frac{1}{3} i\right)^3=-\left(2+\frac{1}{3} i\right)^3$
$
\begin{aligned}
& =-\left[(2)^3+\left(\frac{1}{3} i\right)^3+3 \times(2)^2 \times \frac{1}{3} i+3 \times 2 \times\left(\frac{1}{3} i\right)^2\right] \\
& =-\left[8+\frac{1}{27} i^3+4 i+\frac{2}{3} i^2\right]
\end{aligned}
$

$\begin{aligned}
& =-\left[8-\frac{1}{27} i+4 i-\frac{2}{3}\right] \\
& =-\left[8-\frac{2}{3}-\frac{1}{27} i+4 i\right]=\frac{-22}{3}-\frac{107}{27} i
\end{aligned}$

Find the multiplicative inverse of each of the complex numbers given in the exercises 11 to 13.
Ex 4.1 Question 11.

$4-3 i$

Answer.

Multiplicative Inverse of $4-3 i$
$
\begin{aligned}
& =\frac{1}{4-3 i} \times \frac{4+3 i}{4+3 i} \\
& =\frac{4+3 i}{(4)^2-(3 i)^2} \\
& =\frac{4+3 i}{16-9 i^2} \\
& =\frac{4+3 i}{16+9}=\frac{1}{25}(4+3 i) \\
& =\frac{4}{25}+i \frac{3}{25}
\end{aligned}
$

Ex 4.1 Question 12.

$\sqrt{5}+3 i$

Answer.

Multiplicative Inverse of $\sqrt{5}+3 i$
$
\begin{aligned}
& =\frac{1}{\sqrt{5}+3 i} \times \frac{\sqrt{5}-3 i}{\sqrt{5}-3 i} \\
& =\frac{\sqrt{5}-3 i}{(\sqrt{5})^2-(3 i)^2}
\end{aligned}
$

$
\begin{aligned}
& =\frac{\sqrt{5}-3 i}{5-9 i^2} \\
& =\frac{\sqrt{5}-3 i}{5+9}=\frac{1}{14}(\sqrt{5}-3 i) \\
& =\frac{\sqrt{5}}{14}-i \frac{3}{14}
\end{aligned}
$
Ex 4.1 Question 13.

$-i$

Answer.

Multiplicative Inverse of $-i$
$
\begin{aligned}
& =\frac{1}{-i} \times \frac{i}{i} \\
& =\frac{i}{-i^2} \\
& =\frac{i}{-(-1)}=i
\end{aligned}
$

Ex 4.1 Question 14.

Express the following expression in the form of $a+i b: \frac{(3+\sqrt{5} i)(3-\sqrt{5} i)}{(3+\sqrt{2} i)-(3-\sqrt{2} i)}$

Answer.

Here $\frac{(3+\sqrt{5} i)(3-\sqrt{5} i)}{(3+\sqrt{2} i)-(3-\sqrt{2} i)}$
$
\begin{aligned}
& =\frac{(3)^2-(\sqrt{5 i})^2}{\sqrt{3}+\sqrt{2} i-\sqrt{3}+\sqrt{2} i} \\
& =\frac{9-5 i^2}{2 \sqrt{2 i}} \\
& =\frac{9+5}{2 \sqrt{2} i}=\frac{14}{2 \sqrt{2} i} \\
& =\frac{7}{\sqrt{2} i}=\frac{7}{\sqrt{2} i} \times \frac{i}{i}=\frac{7 i}{\sqrt{2} i^2}=\frac{-7 i}{\sqrt{2}}=\frac{-7 \sqrt{2} i}{2}
\end{aligned}
$