Find the modulus and argument of the complex number (1 + 2i) / (1 - 3i)

Solution:

Let us convert the given complex number to the form a + ib.

(1 + 2i) / (1 - 3i) = (1 + 2i) / (1 - 3i) · (1 + 3i) / (1 + 3i)

= (1 + 2i + 3i + 6i²) /  (1 - 9i²)

= (1 + 5i - 6) / (1 + 9) (because i² = -1)

= (-5 + 5i) / 10

= -1/2 + i/2

Let us assume that -1/2 + i/2= = r (cosθ + i sinθ) (Polar form) 

Let r cosθ = - 1/2 and r sinθ = 1/2

On squaring and adding, we obtain

r² cos² θ + r² sin² θ = (- 1/2)² + (1/2)²

⇒ r² (cos² θ + sin² θ) = 1/4 + 1/4 = 1/2

⇒ r = 1/√2 [∵ Conventionally, r > 0]

Therefore,

(1/√2) cosθ = - 1/2 and (1/√2) sinθ = 1/2

⇒ cosθ = - 1/√2 and sinθ = 1/√2

Since, θ lies in the quadrant II, θ = π - π/4 = 3π/4

Hence, the modulus and argument of the given complex number are 1/√2 and 3π/4, respectively

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NCERT Solutions Class 11 Maths Chapter 5 Exercise ME Question 13

Find the modulus and argument of the complex number (1 + 2i) / (1 - 3i)

Summary:

The modulus and argument of the given complex number are 1/√2 and 3π/4, respectively