Complex Number Question 2
Question 2 - 24 January - Shift 2
The value of $(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}})^{3}$ is
(1) $\frac{-1}{2}(1-i \sqrt{3})$
(2) $\frac{1}{2}(1-i \sqrt{3})$
(3) $\frac{-1}{2}(\sqrt{3}-i)$
(4) $\frac{1}{2}(\sqrt{3}+i)$
Show Answer
Answer: (3)
Solution:
Formula: Algebra of complex numbers, Demoivre’s theorem , Properties of conjugate
Let $\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}=z$
$(\frac{1+z}{1+\bar{z}})^{3}=(\frac{1+z}{1+\frac{1}{z}})^{3}=z^{3}$
$\Rightarrow(i(\cos \frac{2 \pi}{9}-i \sin \frac{2 \pi}{9}))^{3}$
$=-i(\cos \frac{2 \pi}{3}-i \sin \frac{2 \pi}{3})=-i(\frac{-1}{2}-i \frac{\sqrt{3}}{2})$
$\Rightarrow \frac{-1}{2}(\sqrt{3}-i)$.
