NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2

Go back to  'Complex Numbers and Quadratic Equations'

Chapter 5 Ex.5.2 Question 1

Find the modulus and argument of the complex number \(z = - 1 - i\sqrt 3 \)

Solution

\(z = - 1 - i\sqrt 3 \)

Let \(r\cos \theta = - 1\) and \(r\sin \theta = - \sqrt 3 \)

On squaring and adding, we obtain

\[\begin{align} {{\left( r\cos \theta \right)}^{2}}+{{\left( r\sin \theta \right)}^{2}}&={{\left( -1 \right)}^{2}}+{{\left( -\sqrt{3} \right)}^{2}} \\ \Rightarrow {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right) &=1+3\ \ \ \ \ \ \ \ \ \left[ \because {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \right] \\  \Rightarrow {{r}^{2}}&=4 \\ \Rightarrow r=\sqrt{4} &=2\ \ \ \ \ \ \ \ \ \ \ \left[ \because \text{Conventionally},\ r>0 \right] \\\end{align}\]

Therefore, Modulus\( = 2\)

Hence, \(2\cos \theta = - 1\) and \(2\sin \theta = - \sqrt 3 \)

\( \Rightarrow \cos \theta = - \frac{1}{2}\) and \(\sin \theta = - \frac{{\sqrt 3 }}{2}\)

Since both the values of \(\sin \theta \) and \(\cos \theta \) are negative in III quadrant,

Argument\( = - \left( {\pi - \frac{\pi }{3}} \right) = \frac{{ - 2\pi }}{3}\)

Thus, the modulus and argument of the complex number \( - 1 - i\sqrt 3 \) are 2 and \(\frac{{ - 2\pi }}{3}\)respectively.

Chapter 5 Ex.5.2 Question 2

Find the modulus and argument of the complex number \(z = - \sqrt 3 + i\)

Solution

\(z = - \sqrt 3 + i\)

Let \(r\cos \theta = - \sqrt 3 \) and \(r\sin \theta = 1\)

On squaring and adding, we obtain

\({r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta = {\left( { - \sqrt 3 } \right)^2} + {1^2}\)

\[\begin{align} \Rightarrow {{r}^{2}}&=3+1=4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \right] \\  \Rightarrow r&=\sqrt{4}=2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \text{Conventionally},\ r>0 \right] \end{align}\]

Therefore, Modulus\( = 2\)

Hence, \(2\cos \theta = - \sqrt 3 \) and \(2\sin \theta = 1\)

\( \Rightarrow \cos \theta = - \frac{{\sqrt 3 }}{2}\) and \(\sin \theta = \frac{1}{2}\)

Since, \(\theta \) lies in the quadrant II, \(\theta = \pi - \frac{\pi }{6} = \frac{{5\pi }}{6}\)

Thus, the modulus and argument of the complex number \( - \sqrt 3  + i\) are \(2\)  and \(\frac{5\pi }{6}\)respectively.

Chapter 5 Ex.5.2 Question 3

Convert the given complex number in polar form: \(1 - i\)

Solution

\(z = 1 - i\)

Let \(r\cos \theta = 1\) and \(r\sin \theta = - 1\)

On squaring and adding, we obtain

\[\begin{align}  {{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta&={{1}^{2}}+{{\left( -1 \right)}^{2}} \\  \Rightarrow {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)&=1+1 \\  \Rightarrow {{r}^{2}}&=2 \\  \Rightarrow r&=\sqrt{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \text{Conventionally},\ r>0 \right] \\\end{align}\]

Therefore,

\(\sqrt 2 \cos \theta = 1\) and \(\sqrt 2 \sin \theta = - 1\)

\( \Rightarrow \cos \theta = \frac{1}{{\sqrt 2 }}\) and \(\sin \theta = - \frac{1}{{\sqrt 2 }}\)

Since, \(\theta \) lies in the quadrant IV, \(\theta = - \frac{\pi }{4}\)

Hence,

\[\begin{align}1 - i&= r\cos \theta + ir\sin \theta \\&= \sqrt 2 \cos \left( { - \frac{\pi }{4}} \right) + i\sqrt 2 \sin \left( { - \frac{\pi }{4}} \right)\\&= \sqrt 2 \left[ {\cos \left( { - \frac{\pi }{4}} \right) + i\sin \left( { - \frac{\pi }{4}} \right)} \right]\end{align}\]

Thus, this is the required polar form.

Chapter 5 Ex.5.2 Question 4

Convert the given complex number in polar form: \( - 1 + i\)

Solution

\(z = - 1 + i\)

Let \(r\cos \theta = - 1\) and \(r\sin \theta = 1\)

On squaring and adding, we obtain

\[\begin{align}{{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta&={{\left( -1 \right)}^{2}}+{{1}^{2}} \\  \Rightarrow {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)&=1+1 \\  \Rightarrow {{r}^{2}}&=2 \\  \Rightarrow r&=\sqrt{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \text{Conventionally},\ r>0 \right] \\\end{align}\]

Therefore,

\(\sqrt 2 \cos \theta = - 1\) and \(\sqrt 2 \sin \theta = 1\)

\( \Rightarrow \cos \theta = - \frac{1}{{\sqrt 2 }}\) and \(\sin \theta = \frac{1}{{\sqrt 2 }}\)

Since, \(\theta \) lies in the quadrant II, \(\theta = \pi - \frac{\pi }{4} = \frac{{3\pi }}{4}\)

Hence,

\[\begin{align} - 1 + i &= r\cos \theta + ir\sin \theta \\&= \sqrt 2 \cos \frac{{3\pi }}{4} + i\sqrt 2 \sin \frac{{3\pi }}{4}\\&= \sqrt 2 \left( {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right)\end{align}\]

Thus, this is the required polar form.

Chapter 5 Ex.5.2 Question 5

Convert the given complex number in polar form: \( - 1 - i\)

Solution

\(z = - 1 - i\)

Let \(r\cos \theta = - 1\) and \(r\sin \theta = - 1\)

On squaring and adding, we obtain

\[\begin{align}{{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta&={{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}} \\  \Rightarrow {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)&=1+1 \\ \Rightarrow {{r}^{2}}& =2 \\  \Rightarrow r& =\sqrt{2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \text{Conventionally},\ r>0 \right] \\\end{align}\]

Therefore,

\(\sqrt 2 \cos \theta = - 1\) and \(\sqrt 2 \sin \theta = - 1\)

\( \Rightarrow \cos \theta = - \frac{1}{{\sqrt 2 }}\) and \(\sin \theta = - \frac{1}{{\sqrt 2 }}\)

Since, \(\theta \) lies in the quadrant III, \(\theta = - \left( {\pi - \frac{\pi }{4}} \right) = - \frac{{3\pi }}{4}\)

Hence,

\[\begin{align} - 1 - i&= r\cos \theta + ir\sin \theta \\&= \sqrt 2 \cos \frac{{ - 3\pi }}{4} + i\sqrt 2 \sin \frac{{ - 3\pi }}{4}\\&= \sqrt 2 \left( {\cos \frac{{ - 3\pi }}{4} + i\sin \frac{{ - 3\pi }}{4}} \right)\end{align}\]

Thus, this is the required polar form.

Chapter 5 Ex.5.2 Question 6

Convert the given complex number in polar form: \( - 3\)

Solution

\(z = - 3\)

Let \(r\cos \theta = - 3\) and \(r\sin \theta = 0\)

On squaring and adding, we obtain

\[\begin{align}& \ \ \ \ {{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta ={{\left( -3 \right)}^{2}}+{{\left( 0 \right)}^{2}} \\ & \Rightarrow {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)=9 \\ & \Rightarrow {{r}^{2}}=9 \\ & \Rightarrow r=3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \text{Conventionally},\ r>0 \right] \\\end{align}\]

Therefore,

\(3\cos \theta = - 3\) and \(3\sin \theta = 0\)

\( \Rightarrow \cos \theta = - 1\) and \(\sin \theta = 0\)

Since the \(\theta \) lies in the quadrant II, \(\theta = \pi \)

Hence,

\[\begin{align} - 3&= r\cos \theta + ir\sin \theta \\&= 3\cos \pi + i3\sin \pi \\&= 3\left( {\cos \pi + i\sin \pi } \right)\end{align}\]

Thus, this is the required polar form.

Chapter 5 Ex.5.2 Question 7

Convert the given complex number in polar form: \(\sqrt 3 + i\)

Solution

\(z = \sqrt 3 + i\)

Let \(r\cos \theta = \sqrt 3 \) and \(r\sin \theta = 1\)

On squaring and adding, we obtain

\[\begin{align}{r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta&= {\left( {\sqrt 3 } \right)^2} + {1^2}\\ \Rightarrow {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) &= 3 + 1\\ \Rightarrow {r^2}&= 4\\ \Rightarrow r&= \sqrt 4 = 2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{Conventionally}},\;r > 0} \right]\end{align}\]

Therefore,

\(2\cos \theta = \sqrt 3 \)and \(2\sin \theta = 1\)

\( \Rightarrow \cos \theta = \frac{{\sqrt 3 }}{2}\)and \(\sin \theta = \frac{1}{2}\)

Since, \(\theta \) lies in quadrant I, \(\theta = \frac{\pi }{6}\)

Hence,

\[\begin{align}\sqrt 3 + i &= r\cos \theta + ir\sin \theta \\&= 2\cos \frac{\pi }{6} + i2\sin \frac{\pi }{6}\\&= 2\left( {\cos \frac{\pi }{6} + i\sin \frac{\pi }{6}} \right)\end{align}\]

Thus, this is the required polar form.

Chapter 5 Ex.5.2 Question 8

Convert the given complex number in polar form: \(i\)

Solution

\(z = i\)

Let \(r\cos \theta = 0\) and \(r\sin \theta = 1\)

On squaring and adding, we obtain

\[\begin{align}{r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta&= {0^2} + {1^2}\\ \Rightarrow {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)&= 1\\ \Rightarrow {r^2} &= 1\\ \Rightarrow r &= \sqrt 1 = 1\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{Conventionally}},\;r > 0} \right]\end{align}\]

Therefore,

\(\cos \theta = 0\) and \(\sin \theta = 1\)

Since, \(\theta \) lies in quadrant I, \(\theta = \frac{\pi }{2}\)

Hence,

\[\begin{align}i&= r\cos \theta + ir\sin \theta \\&= \cos \frac{\pi }{2} + i\sin \frac{\pi }{2}\end{align}\]

Thus, this is the required polar form.

  
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
0