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Complex Numbers question

2019 · 9 Jan · Shift 1 · Q31
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Complex Numbers question

2019 · 9 Jan · Shift 1 · Q31

JEE MainMathematicsComplex NumbersMCQ+4 / −1
Let A = {θ∈(−π2,π):3+2isin⁡θ1−2isin⁡θis purely imaginary}\left\{ {\theta \in \left( { - {\pi \over 2},\pi } \right):{{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}is\,purely\,imaginary} \right\}{θ∈(−2π​,π):1−2isinθ3+2isinθ​ispurelyimaginary} . Then the sum of the elements in A is :
  1. A
    5π6{5\pi \over 6}65π​
  2. B
    π\piπ
  3. C
    3π4{3\pi \over 4}43π​
  4. D
    2π3{{2\pi } \over 3}32π​
View written solutionFree

Correct answer: D

  1. We need to find all
θ∈(−π2,π)\theta \in \left(-\frac{\pi}{2},\pi\right)θ∈(−2π​,π)

such that

3+2isin⁡θ1−2isin⁡θ\frac{3+2i\sin\theta}{1-2i\sin\theta}1−2isinθ3+2isinθ​

is purely imaginary.

  1. Let
s=sin⁡θ.s=\sin\theta.s=sinθ.

Then the expression becomes

3+2is1−2is.\frac{3+2is}{1-2is}.1−2is3+2is​.

To check when this is purely imaginary, rationalize the denominator:

3+2is1−2is⋅1+2is1+2is=(3+2is)(1+2is)1+4s2.\frac{3+2is}{1-2is}\cdot \frac{1+2is}{1+2is} =\frac{(3+2is)(1+2is)}{1+4s^2}.1−2is3+2is​⋅1+2is1+2is​=1+4s2(3+2is)(1+2is)​.
  1. Expand the numerator:
(3+2is)(1+2is)=3+6is+2is+4i2s2.(3+2is)(1+2is)=3+6is+2is+4i^2s^2.(3+2is)(1+2is)=3+6is+2is+4i2s2.

Since i2=−1i^2=-1i2=−1,

(3+2is)(1+2is)=3+8is−4s2.(3+2is)(1+2is)=3+8is-4s^2.(3+2is)(1+2is)=3+8is−4s2.

So

3+2is1−2is=3−4s21+4s2+i8s1+4s2.\frac{3+2is}{1-2is}=\frac{3-4s^2}{1+4s^2}+i\frac{8s}{1+4s^2}.1−2is3+2is​=1+4s23−4s2​+i1+4s28s​.
  1. For this complex number to be purely imaginary, its real part must be zero:
3−4s21+4s2=0.\frac{3-4s^2}{1+4s^2}=0.1+4s23−4s2​=0.

Since 1+4s2>01+4s^2>01+4s2>0, we need

3−4s2=0⇒s2=34.3-4s^2=0 \quad\Rightarrow\quad s^2=\frac34.3−4s2=0⇒s2=43​.

Thus

sinθ=±32.sin\theta=\pm \frac{\sqrt3}{2}.sinθ=±23​​.
  1. Now find all such θ\thetaθ in the interval
(−π2,π).\left(-\frac{\pi}{2},\pi\right).(−2π​,π).
  • If sin⁡θ=32\sin\theta=\frac{\sqrt3}{2}sinθ=23​​, then θ=π3, 2π3.\theta=\frac{\pi}{3},\ \frac{2\pi}{3}.θ=3π​, 32π​.
  • If sin⁡θ=−32\sin\theta=-\frac{\sqrt3}{2}sinθ=−23​​, then in the given interval, θ=−π3.\theta=-\frac{\pi}{3}.θ=−3π​. (The other solution 4π3\frac{4\pi}{3}34π​ is not in the interval.)

So

A={−π3,π3,2π3}.A=\left\{-\frac{\pi}{3},\frac{\pi}{3},\frac{2\pi}{3}\right\}.A={−3π​,3π​,32π​}.
  1. Sum of the elements:
−π3+π3+2π3=2π3.-\frac{\pi}{3}+\frac{\pi}{3}+\frac{2\pi}{3}=\frac{2\pi}{3}.−3π​+3π​+32π​=32π​.
  1. Therefore the correct option is
2π3.\boxed{\frac{2\pi}{3}}.32π​​.

So, option D is correct.

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