Question -
Answer -
We know that the polarform of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ)
Where,
|Z| = modulus ofcomplex number = √(x2 +y2)
θ = arg (z) = argumentof complex number = tan-1 (|y| / |x|)
(i) 1 + i
Given: Z = 1 + i
So now,
|Z| = √(x2 + y2)
= √(12 + 12)
= √(1 + 1)
= √2
θ = tan-1 (|y|/ |x|)
= tan-1 (1/ 1)
= tan-1 1
Since x > 0, y >0 complex number lies in 1st quadrant and the value of θ is 00≤θ≤900.
θ = π/4
Z = √2 (cos (π/4) + i sin (π/4))
∴ Polar form of (1 + i)is √2 (cos (π/4) + i sin(π/4))
(ii) √3 + i
Given: Z = √3 + i
So now,
|Z| = √(x2 + y2)
= √((√3)2 + 12)
= √(3 + 1)
= √4
= 2
θ = tan-1 (|y|/ |x|)
= tan-1 (1/ √3)
Since x > 0, y >0 complex number lies in 1st quadrant and the value of θ is 00≤θ≤900.
θ = π/6
Z = 2 (cos (π/6) + isin (π/6))
∴ Polar form of (√3 +i) is 2 (cos (π/6) + i sin (π/6))
(iii) 1 – i
Given: Z = 1 – i
So now,
|Z| = √(x2 + y2)
= √(12 + (-1)2)
= √(1 + 1)
= √2
θ = tan-1 (|y|/ |x|)
= tan-1 (1/ 1)
= tan-1 1
Since x > 0, y <0 complex number lies in 4th quadrant and the value of θ is -900≤θ≤00.
θ = -π/4
Z = √2 (cos (-π/4) + i sin (-π/4))
= √2 (cos (π/4) – i sin (π/4))
∴ Polar form of (1 – i)is √2 (cos (π/4) – i sin(π/4))
(iv) (1 – i) / (1 + i)
Given: Z = (1 – i) /(1 + i)
Let us multiply anddivide by (1 – i), we get
= 0 – i
So now,
|Z| = √(x2 + y2)
= √(02 + (-1)2)
= √(0 + 1)
= √1
θ = tan-1 (|y|/ |x|)
= tan-1 (1/ 0)
= tan-1 ∞
Since x ≥ 0, y < 0complex number lies in 4th quadrant and the value of θ is -900≤θ≤00.
θ = -π/2
Z = 1 (cos (-π/2) + isin (-π/2))
= 1 (cos (π/2) – i sin(π/2))
∴ Polar form of (1 – i)/ (1 + i) is 1 (cos (π/2) – i sin (π/2))
(v) 1/(1 + i)
Given: Z = 1 / (1 + i)
Let us multiply anddivide by (1 – i), we get
So now,
|Z| = √(x2 + y2)
= √((1/2)2 + (-1/2)2)
= √(1/4 + 1/4)
= √(2/4)
= 1/√2
θ = tan-1 (|y|/ |x|)
= tan-1 ((1/2)/ (1/2))
= tan-1 1
Since x > 0, y <0 complex number lies in 4th quadrant and the value of θ is -900≤θ≤00.
θ = -π/4
Z = 1/√2 (cos (-π/4) + i sin (-π/4))
= 1/√2 (cos (π/4) – i sin (π/4))
∴ Polar form of 1/(1 +i) is 1/√2 (cos (π/4) – i sin(π/4))
(vi) (1 + 2i) / (1 – 3i)
Given: Z = (1 + 2i) /(1 – 3i)
Let us multiply anddivide by (1 + 3i), we get
So now,
|Z| = √(x2 + y2)
= √((-1/2)2 + (1/2)2)
= √(1/4 + 1/4)
= √(2/4)
= 1/√2
θ = tan-1 (|y|/ |x|)
= tan-1 ((1/2)/ (1/2))
= tan-1 1
Since x < 0, y >0 complex number lies in 2nd quadrant and the value of θ is 900≤θ≤1800.
θ = 3π/4
Z = 1/√2 (cos (3π/4) + i sin (3π/4))
∴ Polar form of (1 +2i) / (1 – 3i) is 1/√2 (cos(3π/4) + i sin (3π/4))
(vii) sin 120o –i cos 120o
Given: Z = sin 120o –i cos 120o
= √3/2 – i (-1/2)
= √3/2 + i (1/2)
So now,
|Z| = √(x2 + y2)
= √((√3/2)2 + (1/2)2)
= √(3/4 + 1/4)
= √(4/4)
= √1
= 1
θ = tan-1 (|y|/ |x|)
= tan-1 ((1/2)/ (√3/2))
= tan-1 (1/√3)
Since x > 0, y >0 complex number lies in 1st quadrant and the value of θ is 00≤θ≤900.
θ = π/6
Z = 1 (cos (π/6) + isin (π/6))
∴ Polar form of √3/2 + i (1/2) is 1 (cos (π/6) + i sin(π/6))
(viii) -16 / (1 + i√3)
Given: Z = -16 / (1 +i√3)
Let us multiply anddivide by (1 – i√3), we get
So now,
|Z| = √(x2 + y2)
= √((-4)2 + (4√3)2)
= √(16 + 48)
= √(64)
= 8
θ = tan-1 (|y|/ |x|)
= tan-1 ((4√3) / 4)
= tan-1 (√3)
Since x < 0, y >0 complex number lies in 2nd quadrant and the value of θ is 900≤θ≤1800.
θ = 2π/3
Z = 8 (cos (2π/3) + isin (2π/3))
∴ Polar form of -16 /(1 + i√3) is 8 (cos (2π/3) + i sin (2π/3))