# 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.

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