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.
Solution :-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)(i2) = (-5/8) x (-1) = 5/8 (ii) (-i)(2i)(-i/8)3 = (-2)(i2)(-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.
Solution :-By using Identity: (z₁ - z₂)3 = z₁3 - z₂3 - 3 z₁2.z₂ + 3 z₁.z₂2 We have, (5 – 3i)3 = 53 – 3 × 52 × (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.
Solution :-We have, (-√3 + √-2)(2√3 - i) = (-√3 + √2i)(2√3 - i) = -√3 x 2√3 + √3i + √2i x 2√3 - √2i2 = -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.
Solution :-Let z = 2 – 3i
Conjugate z = 2 + 3i |z|2 = 22 + (-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)/[12 - (√2i)2] = (5 + 5√2i + √2i + 2i2)/(1 - 2i2) = (5 + 6√2i - 2 )/(1 + 2) = (3 + 6√2i)/3 = 3(1 + 2√2i)/3 = 1 + 2√2i
(ii) We have, i-35 = 1/i35 = 1/(i2)17.i = 1/(-1)17.i = -1/i = -1/i x i/i = -i/i2 = -i/(-1) = i
Example-7 :- Represent the complex number z = 1 + i√3 in the polar form.
Solution :-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 = 12 + (√3)2 r2 (cos2 θ + sin2 θ) = 1 + 3 r2 (cos2 θ + sin2 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.
Solution :-We have, z = -16/(1 + i√3) z = -16/(1 + i√3) x (1 - i√3)/(1 - i√3) z = -16(1 - i√3)/[12 - (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 r2 cos2 θ + r2 sin2 θ = (-4)2 + (4√3)2 r2 (cos2 θ + sin2 θ) = 16 + 48 r2 (cos2 θ + sin2 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 x2 + 2 = 0.
Solution :-We have, x2 + 2 = 0 x2 = -2 x = ±√-2 = ±√2i
Example-10 :- Solve x2 + x + 1= 0.
Solution :-We have, x2 + x + 1= 0 Now, a = 1; b = 1; c = 1 b2 - 4ac = 12 - 4 x 1 x 1 = 1 - 4 = -3 Therefore, the solutions are given by x = (-1 ± √-3)/2 = (-1 ± √3i)/2
Example-11 :- Solve √5x2 + x + √5 = 0.
Solution :-We have, √5x2 + x + √5 = 0 Now, a = √5; b = 1; c = √5 b2 - 4ac = 12 - 4 x √5 x √5 = 1 - 20 = -19 Therefore, the solutions are given by x = (-1 ± √-19)/2√5 = (-1 ± √19i)/2√5