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Complex Numbers and Quadratic Equation Test - 5

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Complex Numbers and Quadratic Equation Test - 5
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  • Question 1
    2 / -0.83

    The complex numbers z = x + iy which satisfy the equation  = 1 lie on

    Solution

    ∣z −5i ∣/∣z+5i| = 1
     ∣z −5i ∣=∣z+5i| (Using definition ∣z −z1 ∣=∣z −z2 ∣given Perpendicular bisector of z1 and z2.)
    ⇒Perpendicular bisector of points (0,5) and (0,−5). which lies on x-axis.

  • Question 2
    2 / -0.83

    The inequality  | z  − 4 | < | z  −2 | represents the region given by

    Solution

     Let, z = x + iy
    Putting in the given inequality,
    |(x −4)−iy| <|(x −2)+iy|
    ⇒[(x −4)2 + y2 ]11/2 <[(x −2)2 + y2 ]1/2
    squaring both sides,
    ⇒(x −4)2 + y2 <(x −2)2 + y2
    ⇒−8x+16 <−4x+4
    ⇒4

  • Question 3
    2 / -0.83

    Solution

    x = (√3+i)/2  
    x3 = 1/8(√3+i)3
    Apply formula (a+b)3 = a3 + b3 + 3a2 b + 3ab2
    = (3 √3 + i3 + 3*3*i + 3*√3*i2 )/8  
    = (3 √3 - i + 9i - 3 √3)/8  
    = 8i/8
    = i

  • Question 4
    2 / -0.83

  • Question 5
    2 / -0.83

    If (√3 + i)10 = a + ib: a, b  ∈ R, then a and b are respectively :

    Solution

    (√3 + i)10 = a + ib
    Z = √3 + i = rcos θ+ i rsin θ
    ⇒√3 = rcos θ i = rsin θ
    ⇒(√3)2 + (1)2 = r2 cos2 θ+ r2 sin2 θ
    ⇒4 = r2
    ⇒r = 2
    tan = 1/√3   
    ⇒tan π/6
    Therefore, Z = √3 + i = 2(cos π/6 + i sin π/6)
    (Z)10 = √3 + i = (2cos π/6 + 2i sin π/6)10
    = 210 (cos π/6 + i sin π/6)10
    210 (cos 10 π/6 + i sin 10 π/6)
    = 210 (cos(2 π- π/3) + i sin(2 π- π/3))
    = 210 (cos π/3 - i sin π/3)
    = 210 (1/2 - i √3/2)
    29 (1 - i √3)
    a = 29 = 512
    b = - 29 (√3) = -512 √3

  • Question 6
    2 / -0.83

    Let  x,y ∈R, hen x + iy is a non real complex number if

  • Question 7
    2 / -0.83

    Let  x,y ∈R, then x + iy is a purely imaginary number if

  • Question 8
    2 / -0.83

    Multiplicative inverse of the non zero complex number x + iy (x,y ∈R,)

    Solution

    Let u be multiplicative inverse
    zu = 1
    u = 1/z
    u = 1/(x+iy)
    Rationalise it  
    [1/(x+iy)]*[(x-iy)/(x-iy)]
    = (x-iy)/(x2 +y2 )
    u = x/(x2 +y2 ) +i(-y)/(x2 +y2 )

  • Question 9
    2 / -0.83

    The locus of z which satisfied the inequality log0.3 |z –1| >log0.3 |z –i| is given by

    Solution

    =

  • Question 10
    2 / -0.83

    The inequality  | z  − 6 | < | z  − 2 | represents the region given by

    Solution

    Let, z = x + iy

    Putting in the given inequality,

    ∣(x −4) −iy ∣<∣(x −2) + iy ∣

    squaring both sides,

    ⇒(x −4)2   + y2   <(x −2)2   + y2
    ⇒−8x + 16 <−4x + 4
    ⇒4x >12
    ⇒x >3
    ⇒Re(z) >3

  • Question 11
    2 / -0.83

  • Question 12
    2 / -0.83

    Distance of the representative of the number 1 + I from the origin (in Argand ’s diagram) is

  • Question 13
    2 / -0.83

    If  ω is a cube root of unity , then  (1+ω)(1+ω2 )(1+ω4 )(1+ω8 )...... upto 2n factors is

  • Question 14
    2 / -0.83

    If points corresponding to the complex numbers z1 , z2 , z3  and z4  are the vertices of a rhombus, taken in order, then for a non-zero real number k

    Solution

    AC = z3 = z1 ei π
    = z1 (cos π+ i sin π)
    = z3 = z1 (-1 + i(0))
    = z3 = -z1
    AC = z1 - z3
    BC = z2 - z4
    (z1 - z3 )/(z2 - z4 ) = k
    (z1 - z3 ) = ei π/2(z2 - z4)
    (z1 - z3 ) k(cos π/2 + sin π/2) (z2 - z4 )
    z1 - z3 = ki(z2 - z4 )
    z1 - z3 = ik(z2 - z4 )
     

  • Question 15
    2 / -0.83

    If k , l, m , n are four consecutive integers, then  is equal to :

  • Question 16
    2 / -0.83

    i2 +i4 +i6 +........... up to 2k + 1 terms, for all k belongs to natural numbers N.

    Solution

  • Question 17
    2 / -0.83

    If  i2 =−1, then sum  i+i2 +i3 +....... to  1000  terms is equal to

    Solution

  • Question 18
    2 / -0.83

    If |z1 | = 4, |z2 | = 4, then |z1  + z2  + 3 + 4i| is less than

    Solution

  • Question 19
    2 / -0.83

    If ω is a non real cube root of unity and  (1+ω)9   = a+b ω;a,b ∈R, then a and b are respectively the numbers :

    Solution

    Since w is the cube root of unity.
    (1+w)9 =A+Bw
    ⇒(−w2 )9 =A+Bw
    ⇒−(w3 )6 = A+Bw
    ⇒−1=A+Bw
    ∴A= -1 &B=0

  • Question 20
    2 / -0.83

    If | z | = 1, then   z  ≠-1 is

  • Question 21
    2 / -0.83

    In z = a + b ι, if i is replaced by −ι, then another complex number obtained is said to b

  • Question 22
    2 / -0.83

    If  = 2n  α, then  αis equal to :

  • Question 23
    2 / -0.83

    If (x + iy) (3 - 4i) = (5 + 12i), then  

    Solution




  • Question 24
    2 / -0.83

    The locus of the equation  

    Solution


    This is Equation of a Circle.

  • Question 25
    2 / -0.83

    Solution

    √-8 * √-18 * √-4
    = 2 √2(√-1) ×3 √2(√-1) ×2(√-1)
    = 2 √2(i) ×3 √2(i) ×2(i)
    = (2 √2 * 3 √2 * 2) *(i * i * i)
    = (24)(-i2 * i)
    = 24(-i)
    = -24i

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