Ch. - 5 Complex Numbers and Quadratic Equations

Ch. - 5 Complex Numbers and Quadratic Equations

Chapter 5 Ex.5.1 Question 1

Express the given complex number in the form \(a + ib:\)\(\left( {5i} \right)\left( { - \frac{3}{5}i} \right)\)

Solution

\[\begin{align} \left( 5i \right)\left( -\frac{3}{5}i \right)&=-5i\times \frac{3}{5}\times i \\& =-3{{i}^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because {{i}^{2}}=-1 \right] \\& =-3\left( -1 \right) \\& =3 \\& =3+i0\end{align}\]

Chapter 5 Ex.5.1 Question 2

Express the given complex number in the form \(a + ib:\)\({i^9} + {i^{19}}\)

Solution

\[\begin{align}{{i}^{9}}+{{i}^{19}}& ={{i}^{4\times 2+1}}+{{i}^{4\times 4+3}} \\& ={{\left( {{i}^{4}} \right)}^{2}}\times i+{{\left( {{i}^{4}} \right)}^{4}}\times {{i}^{3}} \\& =1\times i+1\times \left( -i \right)\ \ \ \ \ \ \ \ \ \ \ \ \left[ \because {{i}^{4}}=1,\ {{i}^{3}}=-i \right] \\& =i+\left( -i \right) \\& =0 \\& =0+i0\end{align}\]

Chapter 5 Ex.5.1 Question 3

Express the given complex number in the form \(a + ib:\)\({i^{ - 39}}\)

Solution

\[\begin{align} {{i}^{-39}}&={{i}^{4\times \left( -9 \right)-3}} \\& ={{\left( {{i}^{4}} \right)}^{-9}}\times {{i}^{-3}} \\& ={{\left( 1 \right)}^{-9}}\times {{i}^{-3}}\ \ \ \ \ \ \ \left[ \because {{i}^{4}}=1 \right] \\& =\frac{1}{{{i}^{3}}} \\& =\frac{1}{-i}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because {{i}^{3}}=-i \right] \\& =-\frac{1}{i}\times \frac{i}{i} \\& =-\frac{i}{{{i}^{2}}} \\& =\frac{-i}{-1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because {{i}^{2}}=-1 \right] \\& =i \\& =0+i1\end{align}\]

Chapter 5 Ex.5.1 Question 4

Express the given complex number in the form \(a + ib:\)\(3\left( {7 + i7} \right) + i\left( {7 + i7} \right)\)

Solution

\[\begin{align} 3\left( 7+i7 \right)+i\left( 7+i7 \right)&=21+21i+7i+7{{i}^{2}} \\& =21+28i+7\times \left( -1 \right)\ \ \ \ \ \left[ \because {{i}^{2}}=-1 \right] \\& =14+i28\end{align}\]

Chapter 5 Ex.5.1 Question 5

Express the given complex number in the form \(a + ib:\)\(\left( {1 - i} \right) - \left( { - 1 + i6} \right)\)

Solution

\[\begin{align}\left( {1 - i} \right) - \left( { - 1 + i6} \right) &= 1 - i + 1 - 6i\\&= 2 - i7\end{align}\]

Chapter 5 Ex.5.1 Question 6

Express the given complex number in the form \(a + ib:\)\(\left( {\frac{1}{5} + i\frac{2}{5}} \right) - \left( {4 + i\frac{5}{2}} \right)\)

Solution

\[\begin{align}\left( {\frac{1}{5} + i\frac{2}{5}} \right) - \left( {4 + i\frac{5}{2}} \right) &= \frac{1}{5} + \frac{2}{5}i - 4 - \frac{5}{2}i\\&= \left( {\frac{1}{5} - 4} \right) + i\left( {\frac{2}{5} - \frac{5}{2}} \right)\\&= \left( { - \frac{{19}}{5}} \right) + i\left( { - \frac{{21}}{{10}}} \right)\\&= - \frac{{19}}{5} - i\frac{{21}}{{10}}\end{align}\]

Chapter 5 Ex.5.1 Question 7

Express the given complex number in the form \(a + ib:\)\(\left[ {\left( {\frac{1}{3} + i\frac{7}{3}} \right) + \left( {4 + i\frac{1}{3}} \right)} \right] - \left( { - \frac{4}{3} + i} \right)\)

Solution

\[\begin{align}\left[ {\left( {\frac{1}{3} + i\frac{7}{3}} \right) + \left( {4 + i\frac{1}{3}} \right)} \right] - \left( { - \frac{4}{3} + i} \right) &= \frac{1}{3} + \frac{7}{3}i + 4 + \frac{1}{3}i + \frac{4}{3} - i\\&= \left( {\frac{1}{3} + 4 + \frac{4}{3}} \right) + i\left( {\frac{7}{3} + \frac{1}{3} - 1} \right)\\&= \frac{{17}}{3} + i\frac{5}{3}\end{align}\]

Chapter 5 Ex.5.1 Question 8

Express the given complex number in the form \(a + ib:\)\({\left( {1 - i} \right)^4}\)

Solution

\[\begin{align} {{\left( 1-i \right)}^{4}}&={{\left[ {{\left( 1-i \right)}^{2}} \right]}^{2}} \\& ={{\left[ {{1}^{2}}+{{i}^{2}}-2i \right]}^{2}} \\& ={{\left[ 1-1-2i \right]}^{2}} \\& ={{\left[ 2i \right]}^{2}} \\& =4{{i}^{2}}\text{ }\quad\left( \because {{i}^{2}}=-1 \right) \\& =-4\end{align}\]

Chapter 5 Ex.5.1 Question 9

Express the given complex number in the form \(a + ib:\)\({\left( {\frac{1}{3} + 3i} \right)^3}\)

Solution

\[\begin{align} {{\left( \frac{1}{3}+3i \right)}^{3}}&={{\left( \frac{1}{3} \right)}^{3}}+{{\left( 3i \right)}^{3}}+3\left( \frac{1}{3} \right)\left( 3i \right)\left( \frac{1}{3}+3i \right) \\& =\frac{1}{27}+27{{i}^{3}}+3i\left( \frac{1}{3}+3i \right) \\& =\frac{1}{27}+27\left( -i \right)+i+9{{i}^{2}}\ \ \ \ \ \ \ \ \ \left( \because {{i}^{3}}=-i \right) \\& =\frac{1}{27}-27i+i-9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \because {{i}^{2}}=-1 \right) \\& =\left( \frac{1}{27}-9 \right)-26i \\& =-\frac{242}{27}-i26\end{align}\]

Chapter 5 Ex.5.1 Question 10

Express the given complex number in the form \(a + ib:\)\({\left( { - 2 - \frac{1}{3}i} \right)^3}\)

Solution

\[\begin{align} {{\left( -2-\frac{1}{3}i \right)}^{3}}&={{\left( -1 \right)}^{3}}{{\left( 2+\frac{1}{3}i \right)}^{3}} \\& =-\left[ {{2}^{3}}+{{\left( \frac{i}{3} \right)}^{3}}+3\left( 2 \right)\left( \frac{i}{3} \right)\left( 2+\frac{i}{3} \right) \right] \\& =-\left[ 8+\frac{{{i}^{3}}}{27}+2i\left( 2+\frac{i}{3} \right) \right] \\& =-\left[ 8-\frac{i}{27}+4i+\frac{2}{3}{{i}^{2}} \right]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because {{i}^{3}}=-i \right] \\& =-\left[ 8-\frac{i}{27}+4i-\frac{2}{3} \right]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because {{i}^{2}}=-1 \right] \\& =-\left[ \frac{22}{3}+\frac{107i}{27} \right] \\& =-\frac{22}{3}-i\frac{107}{27}\end{align}\]

Chapter 5 Ex.5.1 Question 11

Find the multiplicative inverse of the complex number \(4 - 3i\)

Solution

Let\(z = 4 - 3i\)

Then, \(\bar z = 4 + 3i\) and

\[\begin{align} {{\left| z \right|}^{2}}&={{4}^{2}}+{{\left( -3 \right)}^{2}} \\& =16+9 \\& =25\end{align}\]

Therefore, the multiplicative inverse of \(4 - 3i\) is given by

\[\begin{align}{z^{ - 1}}& = \frac{{\bar z}}{{{{\left| z \right|}^2}}}\\&= \frac{{4 + 3i}}{{25}}\\&= \frac{4}{{25}} + i\frac{3}{{25}}\end{align}\]

Chapter 5 Ex.5.1 Question 12

Find the multiplicative inverse of the complex number \(\sqrt 5 + 3i\)

Solution

Let\(z = \sqrt 5 + 3i\)

Then, \(\bar z = \sqrt 5 - 3i\) and

\[\begin{align}{\left| z \right|^2} &= {\left( {\sqrt 5 } \right)^2} + {3^2}\\&= 5 + 9\\&= 14\end{align}\]

Therefore, the multiplicative inverse of \(\sqrt 5 + 3i\) is given by

\[\begin{align}{z^{ - 1}} A&= \frac{{\bar z}}{{{{\left| z \right|}^2}}}\\&= \frac{{\sqrt 5 - 3i}}{{14}}\\&= \frac{{\sqrt 5 }}{{14}} - \frac{3}{{14}}i\end{align}\]

Chapter 5 Ex.5.1 Question 13

Find the multiplicative inverse of the complex number \( - i\)

Solution

Let \(z = - i\)

Then,\(\bar z = i\) and

\[\begin{align}{\left| z \right|^2} &= {1^2}\\&= 1\end{align}\]

Therefore, the multiplicative inverse of\( - i\)is given by

\[\begin{align}{z^{ - 1}} &= \frac{{\bar z}}{{{{\left| z \right|}^2}}}\\&= \frac{i}{1}\\&= i\end{align}\]

Chapter 5 Ex.5.1 Question 14

Express the following expression in the form \(a + ib:\)

\[\frac{{\left( {3 + i\sqrt 5 } \right)\left( {3 - i\sqrt 5 } \right)}}{{\left( {\sqrt 3 + \sqrt 2 i} \right) - \left( {\sqrt 3 - \sqrt 2 i} \right)}}\]

Solution

\[\begin{align}\frac{\left( 3+i\sqrt{5} \right)\left( 3-i\sqrt{5} \right)}{\left( \sqrt{3}+\sqrt{2}i \right)-\left( \sqrt{3}-\sqrt{2}i \right)}&=\frac{{{\left( 3 \right)}^{2}}-{{\left( i\sqrt{5} \right)}^{2}}}{\sqrt{3}+\sqrt{2}i-\sqrt{3}+\sqrt{2}i}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} \right] \\& =\frac{9-5{{i}^{2}}}{2\sqrt{2}i} \\& =\frac{9-5\left( -1 \right)}{2\sqrt{2}i} \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because {{i}^{2}}=-1 \right] \\& =\frac{9+5}{2\sqrt{2}i} \\& =\frac{14}{2\sqrt{2}i}\times \frac{i}{i} \\& =\frac{7i}{\sqrt{2}{{i}^{2}}} \\& =\frac{7i}{\sqrt{2}\left( -1 \right)}\ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because {{i}^{2}}=-1 \right] \\& =\frac{-7i}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} \\& =\frac{-7\sqrt{2}i}{2} \\& =0+i\frac{-7\sqrt{2}}{2}\end{align}\]

Download FREE PDF of Chapter-5 Complex Numbers and Quadratic Equations
Complex Numbers and Quadratic Equations | NCERT Solutions
Download Cuemath NCERT App

Learn from the best math teachers and top your exams

Learn from the best

math teachers and top

your exams


Personalized Curriculum
Instant Doubts clarification
Cover latest CBSE Syllabus
Unlimited Mock & Practice tests
Covers CBSE, ICSE, IB curriculum

Instant doubt clearing with Cuemath Advanced Math Program
  
Download Cuemath NCERT App
Related Sections
Related Sections

Learn from the best math teachers and top your exams

Learn from the best

math teachers and top

your exams


Personalized Curriculum
Instant Doubts clarification
Cover latest CBSE Syllabus
Unlimited Mock & Practice tests
Covers CBSE, ICSE, IB curriculum

Instant doubt clearing with Cuemath Advanced Math Program