NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2

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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.