## 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

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

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.

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.

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.

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

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

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