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JEE Mathematics 2023 Question with Solution

The value of (1+sin⁡2π9+icos⁡2π91+sin⁡2π9−icos⁡2π9)3\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 is:

  • A

    −12(1−i3)-\frac{1}{2}(1 - i\sqrt{3})

  • B

    12(1−i3)\frac{1}{2}(1 - i\sqrt{3})

  • C

    −12(3−i)-\frac{1}{2}(\sqrt{3} - i)

  • D

    12(3+i)\frac{1}{2}(\sqrt{3} + i)

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

(1+sin⁡2π9+icos⁡2π91+sin⁡2π9−icos⁡2π9)3\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

Find: Its value.

Let

z=sin⁡2π9+icos⁡2π9z = \sin \frac{2\pi}{9} + i \cos \frac{2\pi}{9}

Then the given expression becomes

(1+z‾1+z)3\left( \frac{1 + \overline{z}}{1 + z} \right)^3

Using the relation shown in the solution,

(1+z‾1+z)3=z3\left( \frac{1 + \overline{z}}{1 + z} \right)^3 = z^3

So we need to evaluate

z3=(sin⁡2π9+icos⁡2π9)3z^3 = \left( \sin \frac{2\pi}{9} + i \cos \frac{2\pi}{9} \right)^3

Now,

sin⁡θ+icos⁡θ=i(cos⁡θ−isin⁡θ)\sin \theta + i\cos \theta = i(\cos \theta - i\sin \theta)

Hence,

z3=(i(cos⁡2π9−isin⁡2π9))3z^3 = \left(i\left(\cos \frac{2\pi}{9} - i\sin \frac{2\pi}{9}\right)\right)^3 =−i(cos⁡2π3−isin⁡2π3)= -i\left(\cos \frac{2\pi}{3} - i\sin \frac{2\pi}{3}\right)

Substituting the standard values,

cos⁡2π3=−12,sin⁡2π3=32\cos \frac{2\pi}{3} = -\frac{1}{2}, \qquad \sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2}

Therefore,

−i(−12−i32)=−12(3−i)-i\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -\frac{1}{2}(\sqrt{3} - i)

Therefore, the value is −12(3−i)-\frac{1}{2}(\sqrt{3} - i) and the correct option is C.

Using trigonometric form

Given:

z=sin⁡2π9+icos⁡2π9z = \sin \frac{2\pi}{9} + i \cos \frac{2\pi}{9}

Find: z3z^3 as obtained from the transformed expression.

Rewrite zz in trigonometric form:

z=cos⁡(π2−2π9)+isin⁡(π2−2π9)z = \cos\left(\frac{\pi}{2} - \frac{2\pi}{9}\right) + i\sin\left(\frac{\pi}{2} - \frac{2\pi}{9}\right)

which is equivalent to

z=cos⁡5π18+isin⁡5π18z = \cos \frac{5\pi}{18} + i\sin \frac{5\pi}{18}

Hence,

z3=cos⁡5π6+isin⁡5π6z^3 = \cos \frac{5\pi}{6} + i\sin \frac{5\pi}{6}

Now,

cos⁡5π6=−32,sin⁡5π6=12\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}, \qquad \sin \frac{5\pi}{6} = \frac{1}{2}

So,

z3=−32+i2=−12(3−i)z^3 = -\frac{\sqrt{3}}{2} + \frac{i}{2} = -\frac{1}{2}(\sqrt{3} - i)

Therefore, the correct option is C.

Common mistakes

  • Treating sin⁡θ+icos⁡θ\sin \theta + i\cos \theta as cos⁡θ+isin⁡θ\cos \theta + i\sin \theta directly is incorrect because the real and imaginary parts are interchanged. Rewrite it carefully before applying powers.

  • Using De Moivre's theorem on the original fraction without first simplifying it to a cleaner complex form leads to avoidable algebraic errors. First identify the substitution used in the solution.

  • Taking incorrect values of cos⁡2π3\cos \frac{2\pi}{3} or sin⁡2π3\sin \frac{2\pi}{3} changes the sign of the final complex number. Use quadrant signs carefully.

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