# Class 11 NCERT Solutions- Chapter 5 Complex Numbers And Quadratic Equations – Exercise 5.2

**Find the modulus and the arguments of each of the complex numbers i. Exercises 1 to 2.**

**Question 1. z = – 1 – i √3**

**Solution:**

We have,

z = -1 – i√3

We know that, z = r (cosθ + i sinθ)

Therefore,

r cosθ = -1 —(1)

r sinθ = -√3 —-(2)

On Squaring and adding (1) and (2), we obtain

r

^{2}(cos^{2}θ + sin^{2}θ) = (-1)^{2}+ (-√3)^{2}r

^{2}= 1 + 3r = √4

Since r has to positive, Therefore

r = 2Putting r = 2 on (1) and (2), we get

cosθ = -1 / 2 and sinθ = -√3 / 2

Therefore,

θ = – 2π / 3(Since cosθ and sinθ both are negative, therefore θ lies in third quadrant)Hence, modulus and argument of z = -1 – i√3 are

2and– 2π / 3respectively.

**Question 2. z = -√3 + i**

**Solution:**

We have,

z = -√3 + i

We know that, z = r (cosθ + i sinθ)

Therefore,

r cosθ = -√3 —(1)

r sinθ = 1 —-(2)

On Squaring and adding (1) and (2), we obtain

r

^{2}(cos^{ 2}θ + sin^{2}θ) = (-√3)^{2}+ (1)^{2 }r

^{2 }= 3 + 1^{ }(Since, cos^{2}θ + sin^{2}θ = 1)r

^{2}= 3 + 1r = √4

Since r has to positive, Therefore

r = 2Putting r = 2 on (1) and (2), we get

cosθ = -√3 / 2 and sinθ = 1 / 2

Therefore,

θ = 5π / 6(Since cosθ negative and sinθ positive, therefore θ lies on second quadrant)Hence, modulus and argument of z = -√3 + i are

2and5π / 6respectively.

**Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:**

**Question 3. 1 – i**

**Solution:**

We have z = 1 – i,

Let r cosθ = 1 —(1) and,

r sinθ = -1 —(2)

On Squaring and adding (1) and (2) , we obtain

r

^{2 }( cos^{ 2}θ + sin^{2}θ ) = (1)^{2}+ (-1)^{2 }r

^{2}= 2

r = √2( Since r has to be positive )Putting r = √2 on (1) and (2) , we get

cosθ = 1 / √2 and sinθ = -1 / √2

Therefore,

θ = – π / 4( Since cosθ positive and sinθ negative, therefore θ is negative as it lies on fourth quadrant)Hence , z in polar form: z = r cosθ + i r sinθ =

√2 (cos (- π / 4) + i sin (- π / 4)).

**Question 4. -1 + i **

**Solution:**

We have z = -1 + i,

Let r cosθ = -1 —(1) and,

r sinθ = 1 —(2)

On Squaring and adding (1) and (2), we obtain

r

^{2}(cos^{2}θ + sin^{2}θ) = (-1)^{2 }+ (1)^{2}r

^{2}= 2

r = √2(Since r has to be positive)Putting r = √2 on (1) and (2), we get

cosθ = -1 / √2 and sinθ = 1 / √2

Therefore,

θ = 3π / 4(Since cosθ negative and sinθ positive, therefore θ is positive as it lies on second quadrant)Hence, z in polar form: z = r cosθ + i r sinθ =

√2 (cos (3π / 4) + i sin (3π / 4)).

**Question 5. -1 – i**

**Solution:**

We have z = -1 – i,

Let r cosθ = -1 —(1) and,

r sinθ = -1 —(2)

On Squaring and adding (1) and (2), we obtain

r

^{2}(cos^{2}θ + sin^{2}θ) = (-1)^{2}+ (-1)^{2}r

^{2}= 2

r = √2( Since r has to be positive )Putting r = √2 on (1) and (2) , we get

cosθ = -1 / √2 and sinθ = -1 / √2

Therefore

, θ = -3π / 4(Since cosθ negative and sinθ negative, therefore θ is negative as it lies on third quadrant)Hence, z in polar form: z = r cosθ + i r sinθ =

√2 (cos (-3π / 4) + i sin (-3π / 4)).

**Question 6. -3 **

**Solution:**

We have z = -3,

Let r cosθ = -3 —(1) and,

r sinθ = 0 —(2)

On Squaring and adding (1) and (2), we obtain

r

^{2}(cos^{ 2}θ + sin^{2}θ) = (-3)^{2}+ (0)^{2 }r

^{2}= 9

r = 3(Since r has to be positive)Putting r = 3 on (1) and (2), we get

cosθ = -3 / 3 and sinθ = 0 / 3

Therefore,

θ = πHence, z in polar form: z = r cosθ + i r sinθ =

3(cos (π) + i sin (π)).

**Question 7. √3 + i **

**Solution:**

We have z = √3 + i,

Let r cosθ = √3 —(1) and,

r sinθ = 1 —(2)

On Squaring and adding (1) and (2) , we obtain

r

^{2 }(cos^{2}θ + sin^{2}θ) = (√3)^{2}+ (1)^{2}r

^{2}= 4

r = 2(Since r has to be positive )Putting r = 2 on (1) and (2), we get

cosθ = √3 / 2 and sinθ = 1 / 2

Therefore,

θ = π / 6( Since cosθ positive and sinθ positive, therefore θ is positive as it lies on first quadrant)Hence, z in polar form: z = r cosθ + i r sinθ =

2 (cos (π / 6) + i sin (π / 6)).

**Question 8. i**

**Solution:**

We have z = i,

Let r cosθ = 0 —(1) and,

r sinθ = 1 —(2)

On Squaring and adding (1) and (2) , we obtain

r

^{2}(cos^{2}θ + sin^{2}θ) = (0)^{2}+ (1)^{2 }r

^{2}= 1

r = 1(Since r has to be positive)Putting r = 1 on (1) and (2), we get

cosθ = 0 / 1 and sinθ = 1 / 1

Therefore,

θ = π / 2Hence, z in polar form: z = r cosθ + i r sinθ =

cos (π / 2) + i sin (π / 2)