TOPICS

Unit-5(Examples)

Complex Numbers And Quadratic Equations

**Example-1 :-** If 4x + i(3x – y) = 3 + i (– 6), where x and y are real numbers, then find the values of x and y.

We have, 4x + i (3x – y) = 3 + i (–6) ... (1) Equating the real and the imaginary parts of (1), we get 4x = 3, 3x – y = – 6, Now, 4x = 3; x = 3/4 and 3 X 3/4 - y = -6; 9/4 - y = -6 ; y = 9/4 + 6; y = 33/4.

**Example-2 :-** Express the following in the form of a + bi:

(i) (-5i)(i/8) (ii) (-i)(2i)(-i/8)^{3}

(i) (-5i)(i/8) = (-5/8)(i^{2}) = (-5/8) x (-1) = 5/8 (ii) (-i)(2i)(-i/8)^{3}= (-2)(i^{2})(-i/8)^{3}= (-2)(-1)[(-i)^{3}/512] = 2 x i/512 = i/256.

**Example-3 :-** Express (5 – 3i)^{3} in the form a + ib.

By using Identity: (z₁ - z₂)^{3}= z₁^{3}- z₂^{3}- 3 z₁^{2}.z₂ + 3 z₁.z₂^{2}We have, (5 – 3i)^{3}= 5^{3}– 3 × 5^{2}× (3i) + 3 × 5 (3i)^{2}– (3i)^{3}= 125 – 225i – 135 + 27i = – 10 – 198i.

**Example-4 :-** Express (√-3 + √-2)(2√3 - i) in the form of a + ib.

We have, (-√3 + √-2)(2√3 - i) = (-√3 + √2i)(2√3 - i) = -√3 x 2√3 + √3i + √2i x 2√3 - √2i^{2}= -3 x 2 + √3i + 2√6i - √2 x (-1) = -6 + (√3 + 2√6)i + √2 = (-6 + √2) + √3(1 + 2√2)i

**Example-5 :-** Find the multiplicative inverse of 2 – 3i.

Let z = 2 – 3i

Conjugatez= 2 + 3i |z|^{2}= 2^{2}+ (-3)^{2}= 4 + 9 = 13 Therefore, the multiplicative inverse of 2 - 3i is given by z^{-1}=z/|z|^{2}z^{-1}= (2 + 3i)/13 = 2/13 + 3i/13

**Example-6 :-** Express the following in the form a + ib

(i) (5 + √2i)/(1 - √2i) (ii) i^{-35}

(i) We have, (5 + √2i)/(1 - √2i) = (5 + √2i)/(1 - √2i) x (1 + √2i)/(1 + √2i) = (5 + √2i) x (1 + √2i)/[1^{2}- (√2i)^{2}] = (5 + 5√2i + √2i + 2i^{2})/(1 - 2i^{2}) = (5 + 6√2i - 2 )/(1 + 2) = (3 + 6√2i)/3 = 3(1 + 2√2i)/3 = 1 + 2√2i

(ii) We have, i^{-35}= 1/i^{35}= 1/(i^{2})^{17}.i = 1/(-1)^{17}.i = -1/i = -1/i x i/i = -i/i^{2}= -i/(-1) = i

**Example-7 :-** Represent the complex number z = 1 + i√3 in the polar form.

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 Therefore, 1 = r cos θ and √3 = r sin θ cos θ = 1/2 and sin θ = √3/2, which gives θ = π/3 Therefore, required polar form is z = r[cosθ + i sinθ] = 2[cos(π/3) + i sin(π/3)]

**Example-8 :-** Convert the complex number -16/(1 + i√3) into polar form.

We have, z = -16/(1 + i√3) z = -16/(1 + i√3) x (1 - i√3)/(1 - i√3) z = -16(1 - i√3)/[1^{2}- (i√3)^{2}] z = (-16 + i 16√3)/(1 + 3) z = (-16 + i 16√3)/4 z = -4 + i 4√3 Now, -4 = r cos θ, 4√3 = r sin θ By squaring and adding, we get r^{2}cos^{2}θ + r^{2}sin^{2}θ = (-4)^{2}+ (4√3)^{2}r^{2}(cos^{2}θ + sin^{2}θ) = 16 + 48 r^{2}(cos^{2}θ + sin^{2}θ) = 64 r x 1 = √64 r = 8 Therefore, -4 = r cos θ and 4√3 = r sin θ cos θ = -4/8 = -1/2 and sin θ = 4√3/8 = √3/2, which gives θ = π - π/3 = 2π/3 Therefore, required polar form is z = r[cosθ + i sinθ] = 8[cos(2π/3) + i sin(2π/3)]

**Example-9 :-** Solve x^{2} + 2 = 0.

We have, x^{2}+ 2 = 0 x^{2}= -2 x = ±√-2 = ±√2i

**Example-10 :-** Solve x^{2} + x + 1= 0.

We have, x^{2}+ x + 1= 0 Now, a = 1; b = 1; c = 1 b^{2}- 4ac = 1^{2}- 4 x 1 x 1 = 1 - 4 = -3 Therefore, the solutions are given by x = (-1 ± √-3)/2 = (-1 ± √3i)/2

**Example-11 :-** Solve √5x^{2} + x + √5 = 0.

We have, √5x^{2}+ x + √5 = 0 Now, a = √5; b = 1; c = √5 b^{2}- 4ac = 1^{2}- 4 x √5 x √5 = 1 - 20 = -19 Therefore, the solutions are given by x = (-1 ± √-19)/2√5 = (-1 ± √19i)/2√5

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