NCERT Class 11 Maths Complex Numbers and Quadratic Equations

NCERT Class 11 Maths 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

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

Complex Numbers and Quadratic Equations | NCERT Solutions
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