Question-1 :- Find the modulus and the arguments of the complex numbers :
z = -1 - i√3
We have, z = -1 - i√3 Now, -1 = r cos θ, -√3 = r sin θ By squaring and adding, we get r2 cos2 θ + r2 sin2 θ = (-1)2 + (-√3)2 r2 (cos2 θ + sin2 θ) = 1 + 3 r2 (cos2 θ + sin2 θ) = 4 r x 1 = √4 r = 2 Modulus = 2 Therefore, -1 = r cos θ and -√3 = r sin θ cos θ = -1/2 and sin θ = -√3/2, Since both the values of sin θ and cos θ are negative and sinθ and cosθ are negative in III quadrant, So, which gives θ = -(π - π/3) = -2π/3 Arguments = -2π/3
Question-2 :- Find the modulus and the arguments of the complex numbers :
z = -√3 + i
We have, z = -√3 + i Now, -√3 = r cos θ, 1 = r sin θ By squaring and adding, we get r2 cos2 θ + r2 sin2 θ = (-√3)2 + 12 r2 (cos2 θ + sin2 θ) = 3 + 1 r2 (cos2 θ + sin2 θ) = 4 r x 1 = √4 r = 2 Modulus = 2 Therefore, -√3 = r cos θ and 1 = r sin θ cos θ = -√3/2 and sin θ = 1/2, Since cos θ are negative and cosθ are negative in II quadrant, So, which gives θ = (π - π/6) = 5π/6 Arguments = 5π/6
Question-3 :- Convert the complex numbers in the polar form:
1 - i
We have, z = 1 - i Now, 1 = r cos θ, -1 = r sin θ By squaring and adding, we get r2 cos2 θ + r2 sin2 θ = 12 + (-1)2 r2 (cos2 θ + sin2 θ) = 1 + 1 r2 (cos2 θ + sin2 θ) = 2 r x 1 = √2 r = √2 Therefore, 1 = r cos θ and -1 = r sin θ cos θ = 1/√2 and sin θ = -1/√2, which gives θ = -π/4 [lies on IV Quadrant] Therefore, required polar form is z = r[cosθ + i sinθ] = √2[cos(-π/4) + i sin(-π/4)]
Question-4 :- Convert the complex numbers in the polar form:
-1 + i
We have, z = -1 + i Now, -1 = r cos θ, 1 = r sin θ By squaring and adding, we get r2 cos2 θ + r2 sin2 θ = (-1)2 + 12 r2 (cos2 θ + sin2 θ) = 1 + 1 r2 (cos2 θ + sin2 θ) = 2 r x 1 = √2 r = √2 Therefore, -1 = r cos θ and 1 = r sin θ cos θ = -1/√2 and sin θ = 1/√2, which gives θ = π - π/4 = 3π/4 [lies on II Quadrant] Therefore, required polar form is z = r[cosθ + i sinθ] = √2[cos(3π/4) + i sin(3π/4)]
Question-5 :- Convert the complex numbers in the polar form:
-1 - i
We have, z = -1 - i Now, -1 = r cos θ, -1 = r sin θ By squaring and adding, we get r2 cos2 θ + r2 sin2 θ = (-1)2 + (-1)2 r2 (cos2 θ + sin2 θ) = 1 + 1 r2 (cos2 θ + sin2 θ) = 2 r x 1 = √2 r = √2 Therefore, -1 = r cos θ and -1 = r sin θ cos θ = -1/√2 and sin θ = -1/√2, which gives θ = -(π - π/4) = -3π/4 [lies on III Quadrant] Therefore, required polar form is z = r[cosθ + i sinθ] = √2[cos(-3π/4) + i sin(-3π/4)]
Question-6 :- Convert the complex numbers in the polar form:
-3
We have, z = -3 + 0.i Now, -3 = r cos θ, 0 = r sin θ By squaring and adding, we get r2 cos2 θ + r2 sin2 θ = (-3)2 + 0 r2 (cos2 θ + sin2 θ) = 9 + 0 r2 (cos2 θ + sin2 θ) = 9 r x 1 = 9 r = 3 Therefore, -3 = r cos θ and 0 = r sin θ cos θ = -3/3 = -1 and sin θ = 0, which gives θ = π Therefore, required polar form is z = r[cosθ + i sinθ] = 3[cos(π) + i sin(π)]
Question-7 :- Convert the complex numbers in the polar form:
√3 + i
We have, z = √3 + i√3 Now, √3 = r cos θ, 1 = r sin θ By squaring and adding, we get r2 cos2 θ + r2 sin2 θ = (√3)2 + 12 r2 (cos2 θ + sin2 θ) = 3 + 1 r2 (cos2 θ + sin2 θ) = 4 r x 1 = √4 r = 2 Therefore, √3 = r cos θ and 1 = r sin θ cos θ = √3/2 and sin θ = 1/2, which gives θ = π/6 [lies on I quadrant] Therefore, required polar form is z = r[cosθ + i sinθ] = 2[cos(π/6) + i sin(π/6)]
Question-8 :- Convert the complex numbers in the polar form:
i
We have, z = 0 + 1.i Now, 0 = r cos θ, 1 = r sin θ By squaring and adding, we get r2 cos2 θ + r2 sin2 θ = 0 + 12 r2 (cos2 θ + sin2 θ) = 0 + 1 r2 (cos2 θ + sin2 θ) = 1 r x 1 = √1 r = 1 Therefore, 0 = r cos θ and 1 = r sin θ cos θ = 0 and sin θ = 1, which gives θ = π/2 Therefore, required polar form is z = r[cosθ + i sinθ] = [cos(π/2) + i sin(π/3)]