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Complex Numbers and Quadratic Equation Test - 7

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Complex Numbers and Quadratic Equation Test - 7
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  • Question 1
    2 / -0.83

    If   then |α| is equal to : 

    Solution

    α= |z/z ¯|
    z = x + iy
    z ¯= x - iy
    |α| = |z/z ¯|
    ⇒[(x2 + y2 )1/2 ]/[[x2 + y2 )1/2 ]
    ⇒1

  • Question 2
    2 / -0.83

    If z = x + iy ; x, y  ∈ R then :

  • Question 3
    2 / -0.83

    The smallest positive integer n for which  

  • Question 4
    2 / -0.83

    Solution

    z1 = (√3/2 + i/2), z2 = (√3 - i/2)
    (z1)3 = (√3/2 + i/2)2  
    = 3/4 - 1/4 + √3i/2  
    = 1/2 + √3i/2
    (z3 )3 = (z1 )2 * z1  
    = (1/2 + √3i/2) * (√3/2 + i/2)
    = √3/4 + i/4 + 3i/4 - √3/4
    = i
    (z2 )2 = (√3/2 - i/2)2  
    = 3/4 - 1/4 - √3i/2  
    = 1/2 - √3i/2
    (z2 )3 = (z2 )2 * z2
    = (1/2 - √3i/2) * (√3/2 - i/2)
    = √3/4 - i/4 - 3i/4 - √3/4  
    = -i
    (z2 )5 = (z2 )3 * (z2 )2  
    = (-i) * (1/2 - √3i/2)
    = -i/2 + √3/2
    z = (z1 )5 + (z2 )5
    = i/2 - √3/2 - i/2 + √3/2
    = 0
    So Im(z) = 0

  • Question 5
    2 / -0.83

    If the cube roots of unity are 1, ω,ω2 , then roots of the equation  (x −1)3 +8 = 0 are :

    Solution

  • Question 6
    2 / -0.83

    Arg. (x) , x  ∈ R and x  < 0 is

    Solution

    z = x + iy
    z = -x + 0i {-x >0}
    Principle arg(x) = π

  • Question 7
    2 / -0.83

    The number of solutions of the equation  Im (z2 ) = 0,|z| = 2 is

    Solution

    Let z = x + iy
    Then z2 =  x2 + 2ixy - y2
    And real(z2) = x2 - y2  = 0  …………...(1)
    And |z| = √x2 + y2
    |z| = 2
    Hence √x2 + y2 = 2  ………..(2) 
    Or x2 +  y2 = 4 …….(3)
    Solving (1) and (3) , we get x = ±√2 and  
    y = ±√2
    It has 4 solutions.

  • Question 8
    2 / -0.83

    Amp. (z −2 −3i) = π/4 then locus of z is

    Solution

     z = x+iy
    Hence ,(x −2) + i(y −3) = λ
    Since the argument is π/4
    ​⇒(x −2) = (y −3)
    ⇒x −y = −1
    ⇒y = x + 1
    ⇒x - y + 1 = 0
    Thus, slope is positive.

  • Question 9
    2 / -0.83

    If  z1 and z2   are non real complex numbers such that  |z1 | = |z2 | and Arg(z1 )+Arg(z2 )= π, then z1 =

    Solution

    z1 = |z1 (cos θ1 + i sin θ1 )
    z2 = |z2 (cos θ2 + i sin θ2 )|
    Given that |z1 | = |z2 |
    And arg(z1 ) + arg(z2 ) = pi
    θ1 + θ2 = pi
    ⇒θ1 = pi - θ2
    Now, z1 = |z2|[cos θ(pi - θ2 ) + i sin(pi - θ2 )]
    = z1 = z2 (-cos θ2 + i sin θ2 )
    = z1 = - |z2 |(cos θ2 - i sin θ2 )|
    = z1 = - |z2 |(cos θ2 - i sin θ2 )
    ⇒z1 = - z2 (bar)

  • Question 10
    2 / -0.83

    The complex numbers sinx + i cos2x and cosx –i sin2x are conjugate to each other, for

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