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Complex Numbers question

2020 · 7 Jan · Shift 2 · Q32
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Complex Numbers question

2020 · 7 Jan · Shift 2 · Q32

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If 3+isin⁡θ4−icos⁡θ{{3 + i\sin \theta } \over {4 - i\cos \theta }}4−icosθ3+isinθ​, θ∈\theta \inθ∈ [0, 2 θ\thetaθ], is a real number, then an argument of sin θ\thetaθ + icos θ\thetaθ is :
  1. A
    π−tan⁡−1(34)\pi - {\tan ^{ - 1}}\left( {{3 \over 4}} \right)π−tan−1(43​)
  2. B
    −tan⁡−1(34)- {\tan ^{ - 1}}\left( {{3 \over 4}} \right)−tan−1(43​)
  3. C
    tan⁡−1(43){\tan ^{ - 1}}\left( {{4 \over 3}} \right)tan−1(34​)
  4. D
    π−tan⁡−1(43)\pi - {\tan ^{ - 1}}\left( {{4 \over 3}} \right)π−tan−1(34​)
View written solutionFree

Correct answer: D

  1. Given condition

We are given that

3+isin⁡θ4−icos⁡θ\frac{3+i\sin\theta}{4-i\cos\theta}4−icosθ3+isinθ​

is a real number.

For a quotient of complex numbers to be real, its imaginary part must be zero.

  1. Make the denominator real

Multiply numerator and denominator by the conjugate of the denominator:

3+isin⁡θ4−icos⁡θ⋅4+icos⁡θ4+icos⁡θ\frac{3+i\sin\theta}{4-i\cos\theta}\cdot \frac{4+i\cos\theta}{4+i\cos\theta}4−icosθ3+isinθ​⋅4+icosθ4+icosθ​

Now compute the numerator:

(3+isin⁡θ)(4+icos⁡θ)(3+i\sin\theta)(4+i\cos\theta)(3+isinθ)(4+icosθ) =12+3icos⁡θ+4isin⁡θ+i2sin⁡θcos⁡θ=12+3i\cos\theta+4i\sin\theta+i^2\sin\theta\cos\theta=12+3icosθ+4isinθ+i2sinθcosθ =12−sin⁡θcos⁡θ+i(3cos⁡θ+4sin⁡θ)=12-\sin\theta\cos\theta+i(3\cos\theta+4\sin\theta)=12−sinθcosθ+i(3cosθ+4sinθ)

Denominator:

(4−icos⁡θ)(4+icos⁡θ)=16+cos⁡2θ(4-i\cos\theta)(4+i\cos\theta)=16+\cos^2\theta(4−icosθ)(4+icosθ)=16+cos2θ

which is real.

So the whole expression is real iff the imaginary part of the numerator is zero:

3cos⁡θ+4sin⁡θ=03\cos\theta+4\sin\theta=03cosθ+4sinθ=0
  1. Find relation between sin⁡θ\sin\thetasinθ and cos⁡θ\cos\thetacosθ

From

3cos⁡θ+4sin⁡θ=03\cos\theta+4\sin\theta=03cosθ+4sinθ=0

we get

4sin⁡θ=−3cos⁡θ4\sin\theta=-3\cos\theta4sinθ=−3cosθ tan⁡θ=−34\tan\theta=-\frac{3}{4}tanθ=−43​
  1. Argument of sin⁡θ+icos⁡θ\sin\theta+i\cos\thetasinθ+icosθ

Let

z=sin⁡θ+icos⁡θz=\sin\theta+i\cos\thetaz=sinθ+icosθ

Then

  • real part =sin⁡θ=\sin\theta=sinθ
  • imaginary part =cos⁡θ=\cos\theta=cosθ

From

tan⁡θ=−34\tan\theta=-\frac{3}{4}tanθ=−43​

we can take a reference triangle with

∣sin⁡θ∣=35,∣cos⁡θ∣=45|\sin\theta|=\frac{3}{5},\qquad |\cos\theta|=\frac{4}{5}∣sinθ∣=53​,∣cosθ∣=54​

Since tan⁡θ<0\tan\theta<0tanθ<0, sin⁡θ\sin\thetasinθ and cos⁡θ\cos\thetacosθ have opposite signs.

Thus for z=sin⁡θ+icos⁡θz=\sin\theta+i\cos\thetaz=sinθ+icosθ, one suitable case is

sin⁡θ=−35,cos⁡θ=45\sin\theta=-\frac{3}{5},\qquad \cos\theta=\frac{4}{5}sinθ=−53​,cosθ=54​

so

z=−35+i45z=-\frac35+i\frac45z=−53​+i54​

This lies in the second quadrant.

Hence its argument is

arg⁡z=π−tan⁡−1(4535)\arg z=\pi-\tan^{-1}\left(\frac{\frac45}{\frac35}\right)argz=π−tan−1(53​54​​) =π−tan⁡−1(43)=\pi-\tan^{-1}\left(\frac43\right)=π−tan−1(34​)

So an argument is

π−tan⁡−1(43)\boxed{\pi-\tan^{-1}\left(\frac43\right)}π−tan−1(34​)​
  1. Check options
  • A: π−tan⁡−1(3/4)\pi-\tan^{-1}(3/4)π−tan−1(3/4) ❌
  • B: −tan⁡−1(3/4)-\tan^{-1}(3/4)−tan−1(3/4) ❌
  • C: tan⁡−1(4/3)\tan^{-1}(4/3)tan−1(4/3) ❌
  • D: π−tan⁡−1(4/3)\pi-\tan^{-1}(4/3)π−tan−1(4/3) ✅

Therefore, the correct option is D.

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