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Question: The value of $\left(\frac{\sin \frac{2\pi}{9}-i \cos \frac{2\pi}{9}}{\cos \frac{2\pi}{3}+i \sin \fra...

The value of (sin⁡2π9−icos⁡2π9cos⁡2π3+isin⁡2π3)3\left(\frac{\sin \frac{2\pi}{9}-i \cos \frac{2\pi}{9}}{\cos \frac{2\pi}{3}+i \sin \frac{2\pi}{3}}\right)^3 is:

A

−12(1−i3)-\frac{1}{2}(1-i\sqrt{3})

B

−12(3+i)-\frac{1}{2}(\sqrt{3}+i)

C

−12(3−i)-\frac{1}{2}(\sqrt{3}-i)

D

12(1−i3)\frac{1}{2}(1-i\sqrt{3})

Answer

−12(3+i)-\frac{1}{2}(\sqrt{3}+i)

Explanation

Solution

To solve this, we express the numerator and denominator in exponential form:

sin⁡2π9−icos⁡2π9=ei(2π9−π2),cos⁡2π3+isin⁡2π3=ei2π3\sin\frac{2\pi}{9}-i\cos\frac{2\pi}{9}=e^{i\left(\frac{2\pi}{9}-\frac{\pi}{2}\right)}, \quad \cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}=e^{i\frac{2\pi}{3}}

Then, compute the division, take the power of 3, reduce modulo 2π2\pi, and convert to rectangular form.