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Complex Numbers question

2023 · 24 Jan · Shift 2 · Q34
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  5. /2023 · 24 Jan · Shift 2 · Q34

Complex Numbers question

2023 · 24 Jan · Shift 2 · Q34

JEE MainMathematicsComplex NumbersMCQ+4 / −1
The value of (1+sin⁡2π9+icos⁡2π91+sin⁡2π9−icos⁡2π9)3{\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} \right)^3}(1+sin92π​−icos92π​1+sin92π​+icos92π​​)3 is
  1. A
    −12(1−i3)- {1 \over 2}\left( {1 - i\sqrt 3 } \right)−21​(1−i3​)
  2. B
    −12(3−i)- {1 \over 2}\left( {\sqrt 3 - i} \right)−21​(3​−i)
  3. C
    12(1−i3){1 \over 2}\left( {1 - i\sqrt 3 } \right)21​(1−i3​)
  4. D
    12(3+i){1 \over 2}\left( {\sqrt 3 + i} \right)21​(3​+i)
View written solutionFree

Correct answer: D

  1. Let
θ=2π9.\theta=\frac{2\pi}{9}.θ=92π​.

We need to evaluate

(1+sin⁡θ+icos⁡θ1+sin⁡θ−icos⁡θ)3.\left(\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}\right)^3.(1+sinθ−icosθ1+sinθ+icosθ​)3.
  1. Rewrite the numerator in a useful form. Using
sin⁡θ=cos⁡(π2−θ),cos⁡θ=sin⁡(π2−θ),\sin\theta=\cos\left(\frac{\pi}{2}-\theta\right), \qquad \cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),sinθ=cos(2π​−θ),cosθ=sin(2π​−θ),

observe that

1+sin⁡θ+icos⁡θ1+\sin\theta+i\cos\theta1+sinθ+icosθ

is of the form a+iba+iba+ib with

a=1+sin⁡θ,b=cos⁡θ.a=1+\sin\theta, \qquad b=\cos\theta.a=1+sinθ,b=cosθ.

Hence

1+sin⁡θ+icos⁡θ1+sin⁡θ−icos⁡θ\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}1+sinθ−icosθ1+sinθ+icosθ​

is of the standard form

a+iba−ib=e2iϕ,where tan⁡ϕ=ba.\frac{a+ib}{a-ib}=e^{2i\phi}, \quad \text{where } \tan\phi=\frac{b}{a}.a−iba+ib​=e2iϕ,where tanϕ=ab​.

So first compute

tan⁡ϕ=cos⁡θ1+sin⁡θ.\tan\phi=\frac{\cos\theta}{1+\sin\theta}.tanϕ=1+sinθcosθ​.
  1. Simplify the tangent. Using the identity
cos⁡θ1+sin⁡θ=1−sin⁡θcos⁡θ=tan⁡(π4−θ2),\frac{\cos\theta}{1+\sin\theta}=\frac{1-\sin\theta}{\cos\theta}=\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right),1+sinθcosθ​=cosθ1−sinθ​=tan(4π​−2θ​),

we get

ϕ=π4−θ2.\phi=\frac{\pi}{4}-\frac{\theta}{2}.ϕ=4π​−2θ​.

Since θ=2π9\theta=\frac{2\pi}{9}θ=92π​,

ϕ=π4−12⋅2π9=π4−π9=5π−4π36=π36.\phi=\frac{\pi}{4}-\frac{1}{2}\cdot\frac{2\pi}{9} =\frac{\pi}{4}-\frac{\pi}{9} =\frac{5\pi-4\pi}{36} =\frac{\pi}{36}.ϕ=4π​−21​⋅92π​=4π​−9π​=365π−4π​=36π​.

Therefore,

1+sin⁡θ+icos⁡θ1+sin⁡θ−icos⁡θ=e2iϕ=eiπ/18.\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=e^{2i\phi}=e^{i\pi/18}.1+sinθ−icosθ1+sinθ+icosθ​=e2iϕ=eiπ/18.
  1. Raise to the third power:
(eiπ/18)3=eiπ/6.\left(e^{i\pi/18}\right)^3=e^{i\pi/6}.(eiπ/18)3=eiπ/6.

Now

eiπ/6=cos⁡π6+isin⁡π6=32+i2.e^{i\pi/6}=\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}=\frac{\sqrt3}{2}+\frac{i}{2}.eiπ/6=cos6π​+isin6π​=23​​+2i​.

Thus the value is

12(3+i).\frac{1}{2}(\sqrt3+i).21​(3​+i).
  1. Match with the options. This is exactly option D:
12(3+i).\boxed{\frac{1}{2}(\sqrt3+i)}.21​(3​+i)​.
  1. Compare with stored answer. The stored correct answer is B:
−12(3−i)=−3+i2,-\frac12(\sqrt3-i)=\frac{-\sqrt3+i}{2},−21​(3​−i)=2−3​+i​,

which is not equal to 3+i2\frac{\sqrt3+i}{2}23​+i​. So the stored answer appears to be incorrect.

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