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Exercise - 5.2

Complex Numbers And Quadratic Equations

**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 r^{2}cos^{2}θ + r^{2}sin^{2}θ = (-1)^{2}+ (-√3)^{2}r^{2}(cos^{2}θ + sin^{2}θ) = 1 + 3 r^{2}(cos^{2}θ + sin^{2}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 r^{2}cos^{2}θ + r^{2}sin^{2}θ = (-√3)^{2}+ 1^{2}r^{2}(cos^{2}θ + sin^{2}θ) = 3 + 1 r^{2}(cos^{2}θ + sin^{2}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 r^{2}cos^{2}θ + r^{2}sin^{2}θ = 1^{2}+ (-1)^{2}r^{2}(cos^{2}θ + sin^{2}θ) = 1 + 1 r^{2}(cos^{2}θ + sin^{2}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 r^{2}cos^{2}θ + r^{2}sin^{2}θ = (-1)^{2}+ 1^{2}r^{2}(cos^{2}θ + sin^{2}θ) = 1 + 1 r^{2}(cos^{2}θ + sin^{2}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 r^{2}cos^{2}θ + r^{2}sin^{2}θ = (-1)^{2}+ (-1)^{2}r^{2}(cos^{2}θ + sin^{2}θ) = 1 + 1 r^{2}(cos^{2}θ + sin^{2}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 r^{2}cos^{2}θ + r^{2}sin^{2}θ = (-3)^{2}+ 0 r^{2}(cos^{2}θ + sin^{2}θ) = 9 + 0 r^{2}(cos^{2}θ + sin^{2}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 r^{2}cos^{2}θ + r^{2}sin^{2}θ = (√3)^{2}+ 1^{2}r^{2}(cos^{2}θ + sin^{2}θ) = 3 + 1 r^{2}(cos^{2}θ + sin^{2}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 r^{2}cos^{2}θ + r^{2}sin^{2}θ = 0 + 1^{2}r^{2}(cos^{2}θ + sin^{2}θ) = 0 + 1 r^{2}(cos^{2}θ + sin^{2}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)]

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