NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Exercise 5.2

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Exercise 5.2

Here, you will find NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Exercise 5.2 that will be useful in solving problems in which your have difficulty. These NCERT Solutions for Class 11 Maths will guide you to achieve better marks in the examination. The solutions are accurate and to the point which can save your precious time and also understanding the basic concepts.

1. Find the modulus and the argument of the complex number z = -1-i√3

Answer

Let z = -1 – i√3
Here x = -1, y = -√3
∴ α = tan^-1 |y/x| = tan^-1 |-3/-1|
= tan^-1 tan π/3 = π/3
sience x < 0, and y< 0
∴ arg |i| = -(π - α ) = -(π - π/3) = -2π/ 3
|z| = √(-1)² + (-3)² = √1+3= 2

2. Find the modulus and the argument of the complex number z = - √3 + i

Answer

We know that the polar form of
z = r(cosθ + isinθ)
∴ Let - √3 + i = r(cosθ + isinθ)
=> rcosθ = -√3 and r sinθ = 1
By squaring and adding, we get
r²(cos²θ + sin²θ) = ( √3)² + 1
r².1 = 4 ∴ r = 2
by dividing, rsinθ/rcosθ = -1/ √3
i.e. θ lies in second quadrant
=> tanθ = 1/ √3
=> θ = 180 – 30 = 150 => θ= 5π/6
∴ |z|= 2 and arg z = 5π/6

3. Convert (1 - i) in the polar form.

Answer

We have 1 – i = r(cosθ + i sinθ)
=> rcosθ = 1, rsinθ = -1
By squaring and adding, we get
r²(cos² θ + sin²θ) = 1² + (-1)²
=> r². 1 = 1 + 1 => r²= 2
∴ r = √2, by dividing r sin θ/r cos θ = -1/1 = -1
tan θ = -1 i.e., θ lies in fourth quadrant.
=> θ = -45 => θ = -(π/4)
∴ Polar form of 1 – i
= 

4. Convert (-1 + i) in the polar form.

Answer

We have -1 + i = r(cos θ + i sin θ)
=> r cos θ = -1 and r sin θ = 1
By squaring and adding, we get
r²(cos² θ + sin² θ) = (-1)² + 1² => r² . 1 = 1 + 1
∴ r² = 2 ∴ r = √2
by dividing, r sin θ/r cos θ = 1/-1 = -1 => tan θ = - 1
∴ θ lies in second quadrant;
θ = 180° – 45° = 135° i.e. θ = 3π/4
∴ Polar form of -1 + i
= √2(cos(3π/4) + i sin(3π/4))

5. Convert (-1 – i) in the polar form.

Answer

We have -1 – I = r(cos θ + i sin θ)
=> r cos θ = -1 and r sin θ = -1
By squaring and adding, we get
r²(cos² θ + sin² θ ) = (-1)² + (- 1)²
=> r² . 1 = 1 + 1 => r² = 2
∴ r = √2
by dividing r sin θ/r cos θ = -1/-1 = 1 => tan θ = 1
∴ θ lies in IIIrd quadrant.
θ = -180 + 45 = -135 or θ = - 3π/4
∴ Polar form of -1 – i
= √2(cos (- 3π/4) + i sin(- 3π/4))

6. Convert each of the complex numbers given in Exercise 6 to 8 in the polar form.
-3

Answer

z = - 3 = r(cos θ + i sin θ)
∴ r cos θ = - 3, r sin θ = 0
squaring and adding r² = (-3)² ∴ r = 3
tan θ = 0 ∴ θ = π ∵ cos π = 0
∴ -3 = 3 (cosπ + i sinπ)

7. √3 + i

Answer

r = √3 + I = r(cos θ + i sin θ)
∴ r cos θ = √3, r sin θ = 1
squaring and adding r² = 3 + 1 = 4, r = 2
also tan θ = 1/√3 , sin θ and cos θ both are positive
∴ θ lies in the Ist quadrant
∴ θ = 30° = π/6
∴ Polar form of z is 2(cos (π/6) + i sin(π/6))

8. i

Answer

z = i = r(cos θ + i sin θ)
∴ r cos θ = 0 , r sin θ = 1
squaring and adding r² = 1, ∴ r = 1
now sin θ = 1, cos θ = 0 at θ = π/2
∴ Polar form of z is cos(π/2) + i sin(π/2)