Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Complex Numbers question

2020 · 4 Sep · Shift 2 · Q31
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Complex Numbers
  5. /2020 · 4 Sep · Shift 2 · Q31

Complex Numbers question

2020 · 4 Sep · Shift 2 · Q31

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If a and b are real numbers such that (2+α)4=a+bα{\left( {2 + \alpha } \right)^4} = a + b\alpha(2+α)4=a+bα where α=−1+i32\alpha = {{ - 1 + i\sqrt 3 } \over 2}α=2−1+i3​​ then a + b is equal to :
  1. A
    33
  2. B
    9
  3. C
    24
  4. D
    57
View written solutionFree

Correct answer: B

  1. First, note that
α=−1+i32\alpha=\frac{-1+i\sqrt{3}}{2}α=2−1+i3​​

This is a cube root of unity, so it satisfies

α3=1,1+α+α2=0.\alpha^3=1,\qquad 1+\alpha+\alpha^2=0.α3=1,1+α+α2=0.

Hence,

α2=−1−α.\alpha^2=-1-\alpha.α2=−1−α.
  1. We need to compute
(2+α)4(2+\alpha)^4(2+α)4

and express it in the form

a+bα.a+b\alpha.a+bα.
  1. First find (2+α)2(2+\alpha)^2(2+α)2:
(2+α)2=4+4α+α2.(2+\alpha)^2=4+4\alpha+\alpha^2.(2+α)2=4+4α+α2.

Using α2=−1−α\alpha^2=-1-\alphaα2=−1−α,

(2+α)2=4+4α−1−α=3+3α.(2+\alpha)^2=4+4\alpha-1-\alpha=3+3\alpha.(2+α)2=4+4α−1−α=3+3α.

So,

(2+α)2=3(1+α).(2+\alpha)^2=3(1+\alpha).(2+α)2=3(1+α).
  1. Now square again:
(2+α)4=(3+3α)2=9(1+α)2.(2+\alpha)^4=(3+3\alpha)^2=9(1+\alpha)^2.(2+α)4=(3+3α)2=9(1+α)2.

Compute (1+α)2(1+\alpha)^2(1+α)2:

(1+α)2=1+2α+α2.(1+\alpha)^2=1+2\alpha+\alpha^2.(1+α)2=1+2α+α2.

Again using α2=−1−α\alpha^2=-1-\alphaα2=−1−α,

(1+α)2=1+2α−1−α=α.(1+\alpha)^2=1+2\alpha-1-\alpha=\alpha.(1+α)2=1+2α−1−α=α.

Therefore,

(2+α)4=9α.(2+\alpha)^4=9\alpha.(2+α)4=9α.
  1. Comparing with
a+bα,a+b\alpha,a+bα,

we get

a=0,b=9.a=0,\qquad b=9.a=0,b=9.

Thus,

a+b=0+9=9.a+b=0+9=9.a+b=0+9=9.
  1. Checking options:
  • A: 333333 ✗
  • B: 999 ✓
  • C: 242424 ✗
  • D: 575757 ✗

Therefore, the correct answer is Option B.

PreviousNext

More from Complex Numbers

  • If the four complex numbers z,z,z−2Reolimits(z) and z−2Re(z) represent the vertices of a square of side 4 units in the Argand plane, then ∣z∣ is equal to :2020 · MCQ
  • The value of (1−i−1+i3​​)30 is :2020 · MCQ
  • The region represented by {z = x + iy ∈ C : |z| – Re(z) ≤ 1} is also given by the inequality : {z = x + iy ∈ C : |z| – Re(z) ≤ 1}2020 · MCQ
  • Let z = x + iy be a non-zero complex number such that z2=i∣z∣2, where i = −1​ , then z lies on the :2020 · MCQ
  • If Reolimits(2z+iz−1​)=1, where z = x + iy, then the point (x, y) lies on a :2020 · MCQ
  • If 4−icosθ3+isinθ​, θ∈ [0, 2 θ], is a real number, then an argument of sin θ + icos θ is :2020 · MCQ
  • If the equation, x2 + bx + 45 = 0 (b ∈ R) has conjugate complex roots and they satisfy |z +1| = 2 10​ , then :2020 · MCQ
  • Let z be complex number such that ​z+2iz−i​​=1 and |z| =25​. Then the value of |z + 3i| is :2020 · MCQ