Complex Numbers

Answer :

      Z = [(4 + 3i) / (5 – 3i)]

                                                = [{(4 + 3i) (5 + 3i)} / {(5 – 3i) (5 + 3i)}]

                                                = [(20 + 12i + 15i + 9i²) / (5² – 9i²)]

                                           ∴ Z        = [(11 + 27i) / (25 + 9)] = (11 / 34) + (27 / 34)i

                                        ∴ Z^–1         = (1/Z) = [1 / {(11 / 34) + (27 / 34)i}]

                                                = [{(11 / 34) – (27 / 34)i} / {(11 / 34)² – (27 / 34)²i²}]

                                                = [{(11 – 27i) / (34)} / {(11 / 34)² + (27 / 34)²}]

                                                = [(11 – 27i) / (34)] × [(34)² / (850)]

                                                = [(11 – 27i) / (25)]

                                                = (11 / 25) – (27 / 25)i

Answer :

          z = x + iy

                   |3z| = |z – 4| hence |3x + 3iy| = |x + iy – 4|

                             ∴ |3x + 3iy| = |x – 4 + iy|

                             ∴ 3√(x² + y²) = √[(x – 4)² + y²]

                             ∴ 9x² + 9y² = x² – 8x + 16 + y²

                             ∴ 8x² + 8y² + 8x = 16

                                x² + y² + x = 2

Answer :

          |z – 2| = |z + 3 – 5|

                   we know |a + b| ≤ |a| + |b| hence

                             |z + 3 – 5| ≤ |z + 3| + |– 5|

                                            ≤ 8 + 5

                                   |z – 2| ≤ 13                                                               (1)

                   now

                             |z – 2| = |5 – (5 + 3)|

                             |5 – (5 – 3)| ≥ |5| – |z + 3|

                   ∴       |z – 2| ≥ |5| – 8 ------ |z + 3| ≤ 8

                   ∴       |z – 2| ≥ 3

                   but    |z + 3| ≤ 8

                   ∴ – 11 ≤ z ≤ 5 and z = 2 ∴ |z – 2| = 0

                   least value of |z – 2| is 0                                                          (2)

                   ∴ 0 ≤ |z – 2| ≤ 13 from (1) & (2) hence [0, 13]

Answer :

Let     z = 3√1

                         ∴          z³ = 1

                         ∴ z³ – 1 = 0

                         ∴ (z – 1) (z² + z + 1) = 0

                         ∴ z = 1, z = [(– 1 ± √3i) / 2]

                   1, w, w² are given roots hence

                             w = [(– 1 + √3i) / 2] and w² = [(– 1 – √3i) / 2]

                   ∴       1 + w + w² = 1 + [(– 1 + √3i) / 2] + [(– 1 – √3i) / 2]

                                             = 1 + (– 1)

                                             = 0

Answer :

          sin x + i cos 2x and cos x – i sin 2x are complex conjugate of each other

                             ∴ Im(z) = 0

                             ∴ cos 2x = 0 and sin 2x = 0

                   Which is not possible. Hence value of x does not exist.