Complex%20Numbers

Answer :

          z = x + iy                                                                       (1)

                   3x + (3x – y)i = 4 – 6i

                   i.e. 3x = 4 and 3x – y = – 6 ----- equating real and imaginary parts

                             ∴ x = (4 / 3) and y = 3x + 6

                             ∴ x = (4 / 3) and y = 10

                             ∴ from (2), z = (4 / 3) + 10 i

Answer :

we know i² = – 1

                             ∴ i¹⁹ = (i²)⁹ ∙ i

                                i¹⁹ = (– 1)⁹ i = – i

                             also (1 / i) = – i

                             and (1 / i)²⁵ = (– i)²⁵

                   Hence

                             [i¹⁹ + (1 / i)²⁵]² = [– i + (– i)²⁵]²

                             = i² + 2(– i) (– i)²⁵ + [(– i)²⁵]²

                             = i² + 2 i²⁶ + (– i)⁵⁰

                             = i² + 2(i²)¹³ + (i²)²⁵

                             = – 1 + 2(– 1)¹³ + (– 1)²⁵

                             = – 1 – 2 – 1

                             = – 4

Answer :

i¹ + i² + i³ + i⁴ + -------- i¹⁰⁰⁰

                   This is G. P with first term a = i

                   and common ratio = r = i

                   Sum of n terms in G. P is given by

                                   S_n  = [{a(1 – rⁿ)} / (1 – r)]

                                      = [{(i) (1 – i¹⁰⁰⁰)} / (1 – i)]

                                      = [{(i) (1 – (i²)⁵⁰⁰)} / (1 – i)]

                                      = [{(i) (1 – ( – 1))⁵⁰⁰} / (1 – i)]

                                      = [i / (1 – i)] [1 – 1] = 0

Answer :

z = [1 / (1 + cos θ – i sin θ)]

                      = [{(1 + cos θ) + i sin θ} / {(1 + cos θ – i sin θ) (1 + cos θ + i sin θ)]

                      = [{2 cos² (θ/2) + i 2 sin (θ/2) ∙ cos (θ/2)} / {(1 + cos θ)² + sin² θ}]

                      = [{2 cos² (θ/2) + i ∙ 2 ∙ sin (θ/2) ∙ cos (θ/2)} / {1 + cos² θ + 2 cos θ + sin² θ}]

                      = [{2 cos (θ/2) {cos (θ/2) + i sin (θ/2)}} / {2(1 + cos θ)}]

                      = [{cos (θ/2) {cos (θ/2) + i sin (θ/2)}} / {2 cos² (θ/2)}]

                      = (1/2) [{{cos (θ/2)} / {cos (θ/2)}} + i {{sin (θ/2)} / {cos (θ/2)}}]

                      = (1/2) + (i/2) tan (θ/2)

Answer :

[{(1 + i)x – 2i} / (3 + i)] + [{(2– 3i)y + i} / (3 – i)] = i

                   ∴ (3 – i) (1 + i)x – 2i(3 – i) + (3 + i) (2 – 3i)y + (3 + i) (i) – i(3 + i) (3 – i)

                   ∴ (3 + 3i – i – i²)x – (6i – 2i²) + (6 – 9i + 2i – 3i²)y + 3i + i² = i(9 – i²)

                   put i² = – 1 we get

                                (2i + 4)x – (6i + 2) + (9 – 7i)y + 3i – 1 = i(10)

                             ∴ (4 + 2i)x + (9 – 7i)y – 3i – 3 = 10i

                             ∴ (4x + 9y – 3) + i(2x – 7y – 3) = 10i

                   ∴ equating real and imaginary part,

                                      4x + 9y – 3 = 0                                                    (1)

                                      2x – 7y – 3 = 10

                                      2x – 7y – 13 = 0                                                   (2)

                   Solving (1) & (2) we get x = 3, y = – 1 hence (x, y) = (3, – 1)