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Complex Numbers question

2020 · 2 Sep · Shift 1 · Q32
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Complex Numbers question

2020 · 2 Sep · Shift 1 · Q32

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(3−i){1 \over 2}\left( {\sqrt 3 - i} \right)21​(3​−i)
  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(1−i3){1 \over 2}\left( {1 - i\sqrt 3 } \right)21​(1−i3​)
View written solutionFree

Correct answer: B

  1. Let θ=2π9\theta=\frac{2\pi}{9}θ=92π​ so the given expression is (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.

  2. Rewrite the numerator and denominator using a standard half-angle identity.

Notice that 1+sin⁡θ=1+2sin⁡θ2cos⁡θ21+\sin\theta=1+2\sin\frac\theta2\cos\frac\theta21+sinθ=1+2sin2θ​cos2θ​ and cos⁡θ=cos⁡2θ2−sin⁡2θ2.\cos\theta=\cos^2\frac\theta2-\sin^2\frac\theta2.cosθ=cos22θ​−sin22θ​.

A more useful identity here is: 1+sin⁡θ=(sin⁡θ2+cos⁡θ2)2,1+\sin\theta=\left(\sin\frac\theta2+\cos\frac\theta2\right)^2,1+sinθ=(sin2θ​+cos2θ​)2, cos⁡θ=(cos⁡θ2+sin⁡θ2)(cos⁡θ2−sin⁡θ2).\cos\theta=\left(\cos\frac\theta2+\sin\frac\theta2\right)\left(\cos\frac\theta2-\sin\frac\theta2\right).cosθ=(cos2θ​+sin2θ​)(cos2θ​−sin2θ​).

Hence,

=(\sin\tfrac\theta2+\cos\tfrac\theta2)\Big[(\sin\tfrac\theta2+\cos\tfrac\theta2)+i(\cos\tfrac\theta2-\sin\tfrac\theta2)\Big].$$ Now simplify the bracket: $$\sin\frac\theta2+\cos\frac\theta2+i\left(\cos\frac\theta2-\sin\frac\theta2\right) =(1-i)\sin\frac\theta2+(1+i)\cos\frac\theta2.$$ But an even cleaner approach is to observe directly that $$1+\sin\theta+i\cos\theta=(1+\sin\theta)+i\cos\theta.$$ Its modulus is $$\sqrt{(1+\sin\theta)^2+\cos^2\theta} =\sqrt{1+2\sin\theta+\sin^2\theta+\cos^2\theta} =\sqrt{2(1+\sin\theta)}.$$ Also, $$\frac{\cos\theta}{1+\sin\theta}=\frac{1-\sin\theta}{\cos\theta}=\tan\left(\frac\pi4-\frac\theta2\right).$$ So the argument of $$1+\sin\theta+i\cos\theta$$ is $$\phi=\tan^{-1}\left(\frac{\cos\theta}{1+\sin\theta}\right)=\frac\pi4-\frac\theta2.$$ Thus, $$1+\sin\theta+i\cos\theta = r e^{i\phi},$$ $$1+\sin\theta-i\cos\theta = r e^{-i\phi},$$ where $r=\sqrt{2(1+\sin\theta)}$. Therefore, $$\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=e^{2i\phi}.$$ 3. Substitute $\phi=\frac\pi4-\frac\theta2$: $$e^{2i\phi}=e^{2i(\frac\pi4-\frac\theta2)}=e^{i(\frac\pi2-\theta)}.$$ Hence the given expression becomes $$\left(e^{i(\frac\pi2-\theta)}\right)^3=e^{i(\frac{3\pi}{2}-3\theta)}.$$ Since $\theta=\frac{2\pi}{9}$, $$3\theta=\frac{2\pi}{3}.$$ So $$\frac{3\pi}{2}-3\theta=\frac{3\pi}{2}-\frac{2\pi}{3} =\frac{9\pi-4\pi}{6}=\frac{5\pi}{6}.$$ Thus, $$\left(\frac{1+\sin\frac{2\pi}{9}+i\cos\frac{2\pi}{9}}{1+\sin\frac{2\pi}{9}-i\cos\frac{2\pi}{9}}\right)^3 =e^{i5\pi/6}.$$ 4. Convert to rectangular form: $$e^{i5\pi/6}=\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6} =-\frac{\sqrt3}{2}+\frac{i}{2}.$$ So, $$-\frac{\sqrt3}{2}+\frac{i}{2}=-\frac12(\sqrt3-i).$$ 5. Compare with the options: This matches **Option B**. Therefore, the value is $$\boxed{-\frac12(\sqrt3-i)}.$$
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