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    Complex Numbers

    What is Complex Numbers, Imaginary unit, Conjugate Complex Number, Modulus and Argument, Square root of Complex Numbers, Cube root of Complex Numbers:


    Imaginary number or imaginary unit = √-1 = i
    i² = -1,   i3= i² × i = -1 × i = -i,   i4 = 1,   i5 = i etc.
    i4n + 1 = i,   i4n + 2 = -1,   i4n + 3 = -i,   i4n + 1 = i etc. where, n ϵ Z     [in general i4n = 1]

    Note:-
            √-a × √-b = √a × √-1 × √b × √-1 = √ab × √(-1²) = -√ab
    But, √-a × √-b = √(-a × -b) = √ab is not valid.


    Complex Number: A complex number is a number that can be expressed in the form a + bi, where ‘a’ and ‘b’ are real numbers and ‘i’ is the imaginary unit, that satisfies the equation i² = -1. In this expression, ‘a’ is the real part and ‘b’ is the imaginary part of the complex number and ‘i’ is the positive square root of -1.

    Complex Number a + ib is denoted by Z
    a + ib = Z
    Where, a = Real part, b = imaginary part
    If b = 0, complex number is real.
    And if a = 0, b ≠ 0, complex number is purely imaginary.

    Note:-
    Two complex numbers a + ib and c + id are equal if only a = c and b = d.

    Conjugate Complex Number:
    Complex numbers a + ib and a - ib are said to be conjugate to each other,
    Denoted by Z = a + ib and Z̅ = a - ib

    Properties of conjugate complex number: If Z₁ and Z₂ be any two complex numbers, then


















    Example: 01
    Reduce (1 + i)3/(2 + 3i) to the form A = iB

    Solution:
    (1 + i)3/(2 + 3i) = (1 + 3i +3i² + i3)/2 + 3i) = (-2 + 2i)/(2 + 3i)
    = (-2 + 2i) (2 - 3i) / (2 + 3i) (2 - 3i) = (2 + 10i) / (4 + 9) = (2 + 10i)/13
    => 2/13 + i(10/13) = A + iB form
    Where, A = 2/13 and B = 10/13.

    Some Important Properties:
    If Z₁ = a + ib and Z₂ = c + id
    Then, (i) Z₁ + Z₂ = (a + c) + i(b + d)
               (ii) Z₁ + Z₂ = (a - c) + i(b - d)
               (iii) Z₁. Z₂ = (ac - bd) + (ad + bc)
               (iv) Z₁/Z₂ = a + ib)/(c + id) = (a + ib) (c - id) / (c + id) (c - id)   [Since, (a + b) (a - b) = a² - b²]


    Modulus and Argument (Amplitude):
    If Z = x + iy, then modulus of Z, |Z| = √(x² + y²)
    And = x + iy = r (cosθ + i sinθ)    [here, θ is called argument]
    |Z| = √(x² + y²) = r
    And, arg. Z or amp. Z = θ
    And, tan θ = y/x

    Note:-
    1. |1| = 1, arg.(1) = 0
    2. |-1| = 1, arg. (-1) = π
    3. |i| = 1, arg.(i) = π/2
    4. |-i| = 1, arg.(-i) = -π/2

    Example: 02
    Find mod. and, arg. of
    (A) -1 + i
    (B) -1 - √3 i

    Solution: A
    Let, (-1 + i) = r (cosθ + i sinθ)
    Then, r cosθ = -1…………..eq.(i)
               r sinθ = 1………..…..eq.(ii)

    Squaring both the eq. and adding,
          r² = 2   [since cos²θ + sin²θ = 1]
    => r = √2
    On dividing eq.(ii) by eq.(i),
         tanθ = -1
    => θ = tan -1 = 3π/4
    |-1 + i| = r = √2 and, arg.(-1 + i) = θ = 3π/4

    Solution: B
    Let, (-1 -√3 i) = r (cosθ + i sinθ)
    Then, r cosθ = -1……………eq.(i)
               r sinθ = -√3………….eq.(ii)

    Squaring both the eq. and adding,
         r² = 4   [since, cos²θ + sin²θ = 1]
    => r = 2
    On dividing eq.(ii) by eq.(i),
         tanθ = √
    =>   θ = -2π/3
    Hence, |-1 - √3 i| = r = 2 and, arg. (-1 - √3 i) = θ = -2π/3

    Note: If Z₁ and Z₂ are two complex numbers, then |Z₁ + Z₂| ≤ |Z₁| + |Z₂|


    Example: 03
    Find the modulus of (3 - 4i) (1 + 7i) / (1 + i) (12 + 5i)

    Solution:
       |(3 - 4i) (1 + 7i) / (1 + i) (12 + 5i)|
    = |(3 - 4i) (1 + 7i)| / |(1 + i) (12 + 5i)|
    = |(3 - 4i)|.| (1 + 7i)| / |(1 + i)|.|(12 + 5i)|
    = √(9 + 16). √(1 + 49) / √(1 + 1). √(144 + 25)
    = 5. 5√2 / √2. 13
    = 25/13


    Example: 04
    Find the square root of 7 - 30√-2

    Solution:
    Let, √(7 - 30√-2) = x - iy
    Then, (x - iy)² = 7 - 30√-2
    => (x² - y²) - 2ixy = 7 - 30√2i   [since, √-2 = √2i]
    => (x² - y²) = 7…….…eq.(i)
         2xy = 30√2………..…..eq.(ii)

    eq.(i)² + eq.(ii)² = (x² - y²)² + (2xy)² = [(x² - y²)² + 4x²y²] = 49 + 1800
    =>    (x²)² + (y²)² - 2x²y² + 4x²y² = (x² + y²)² = 49 + 1800
    =>    (x² + y²) = √(49 + 1800)
    =>    (x² + y²) = 43……………………eq.(iii)

    eq.(i) + eq.(iii) => 2x² = 50     => x = ± 5
    And, eq.(iii) - eq.(i) = 2y² = 36    => y = ± 3√2
    √(7 - 30√-2) = x - iy = ± (5 - 3√2)
    Hence, the two roots are, (5 - 3√2) and, -(5 - 3√2)

    Alternate Method:
    (7 - 30√-2) = 25 - 18 - 2. 5. 3√2i   [since, 25 - 18 = 7, and, 2 × 5 × 3√2 = 30√2]
                        = (5 - 3√2i)²
    The square root of (7 - 30√-2) = ± (5 - 3√2)


    Cube root of Complex Numbers:
    Cube roots of unity are, 1, -½ + i √3/2 and -½ - i √3/2 of which, 1 is real and the other are complex (conjugate to each other)
    If one of them is denoted by ω and the other is denoted by ω²
    -½ + i √3/2 = ω and, -½ - i √3/2 = ω²

    Note:-
    1. As the sum of cube roots of unity is zero, so, we have, 1 + ω + ω² = 0
    2. The product of complex cube roots being unity, we have ω. ω² = 1 i.e. ω3 = 1
    3. As a consequence, of which one complex root is the reciprocal of the other,
    I.e. ω = 1/ ω²,    ω² = 1/ ω
    As, ω3 = 1,   ω3n = 1,   ω3n + 1 = ω,    ω3n + 2 = ω² etc….


    Example: 04
    Find, [(√3 + i)/2]6 + [(i - √3)/2]6

    Solution:
    (√3 + i)/2 = (i √3 + i²)/2i = -i [(-1 + √3)/2] = -iω
    (i - √3)/2 = (i² - √3 i)/2i= -i [(-1 - √3 i)/2] = -iω²

    [(√3 + i)/2]6 + [(i - √3)/2]6  = (-iω)6 + (-iω²)6  
                                                       = i6ω6 + i6ω12

                                                       = (-1) × (1) + (-1) × (1) = -2
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    Item Reviewed: Complex Numbers Rating: 5 Reviewed By: Pranab Debnath
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