Class 11 NCERT Solutions Chapter 5 Complex Numbers And Quadratic Equations Exercise 5.2 (original) (raw)

Last Updated : 25 Jan, 2021

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

r2 (cos 2θ + sin 2θ) = (-1)2 + (-√3)2

r2 = 1 + 3

r = √4

Since r has to positive, Therefore r = 2

Putting 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 2 and - 2π / 3 respectively.

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

r2 (cos 2θ + sin 2θ) = (-√3)2 + (1)2

r2 = 3 + 1 (Since, cos 2θ + sin 2θ = 1)

r2 = 3 + 1

r = √4

Since r has to positive, Therefore r = 2

Putting 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 2 and 5π / 6 respectively.

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

r2 ( cos 2θ + sin 2θ ) = (1)2 + (-1)2

r2 = 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

r2 (cos 2θ + sin 2θ) = (-1)2 + (1)2

r2 = 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

r2 (cos 2θ + sin 2θ) = (-1)2 + (-1)2

r2 = 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

r2 (cos 2θ + sin 2θ) = (-3)2 + (0)2

r2 = 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 = π

Hence, 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

r2 (cos 2θ + sin 2θ) = (√3)2 + (1)2

r2 = 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

r2 (cos 2θ + sin 2θ) = (0)2 + (1)2

r2 = 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 / 2

Hence, z in polar form: z = r cosθ + i r sinθ = cos (π / 2) + i sin (π / 2)