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Complex Numbers & Quadratic Equations Test - 4
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Complex Numbers & Quadratic Equations Test - 4

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  • Question 1/8
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    If  is an imaginary fifth root of unity, then 

     

    Solutions

    Since, α is the fifth root of unity 

    Hence, option D is correct.

  • Question 2/8
    1 / -0

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

    Solutions

    (x – 1)3 + 8 = 0 
    ⇒ (x – 1) = (-2) (1)1/3 
    ⇒ x – 1 = -2 or -2ω or -2ω2 
    or n = -1 or 1 – 2ω or 1 – 2ω2 
    Hence, option C is correct.

  • Question 3/8
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    Solutions

    Sum of n, nth roots of unity is zero. 

  • Question 4/8
    1 / -0

    If αis a complex constant such that  has a real root then

    Solutions

  • Question 5/8
    1 / -0

    Let z1 and z2 be two roots of the equation z2 + az + b = 0, z being complex. Further, assume that the origin, z1 and z2 form an equilateral triangle, then

    Solutions

    Given that z1 and z2 be two roots of the equation z2 + az + b = 0, where z is a complex number.
    So, we have z1 + z2 = -a, z1z2 = b (sum and product of roots)
    Since z1 and z2 form an equilateral triangle with the origin, we have

    z2 = z1 (cos 60° + isin 60°)
    = z1 (1/2 + i√3/2)
    Or, 2z2 – z1 = √3 i z1
    This gives, (2z2 -z1)2 = -3z12
    Hence, (z12 + z22) = z1z2
    So, a2 – 2b = b
    this gives a2 = 3b.

  • Question 6/8
    1 / -0

    Solutions

    Hence option B is correct.

  • Question 7/8
    1 / -0

    Let S denote the set of complex numbers z such that log1/3 (log1/2 (|z|2 + 4 |z| + 3)] < 0, then S is contained in-

    Solutions

    log1/3 [log1/2 (| z |2 + 4 | z | + 3)] < 0 
    ⇒log1/2 (| z |2 + 4 | z | + 3)] < 1 
    ⇒| z |2 + 4 | z | + 3 < 1/2 
    ⇒| z |2 + 4 | z | + 5/2 < 0 
    ⇒2 | z |2 + 8 | z | + 5 < 0 
    We can solve this quadratic equation as.

    Which is not possible

  • Question 8/8
    1 / -0

    If z1, z2, z3 are complex numbers such that  

    then |z1 + z2 + z3| is-

    Solutions

    Given: |z1| = |z2| = |z3| 

    ∴Option A is correct answer.

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