The value of left(frac{1+sin frac{2 pi}{9}+i | vedclass

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Question

(C) Let $z = \sin \frac{2 \pi}{9} + i \cos \frac{2 \pi}{9}$.Note that $|z|^2 = \sin^2 \frac{2 \pi}{9} + \cos^2 \frac{2 \pi}{9} = 1$,so $\bar{z} = \fra

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

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(C) Let $z = \sin \frac{2 \pi}{9} + i \cos \frac{2 \pi}{9}$.Note that $|z|^2 = \sin^2 \frac{2 \pi}{9} + \cos^2 \frac{2 \pi}{9} = 1$,so $\bar{z} = \frac{1}{z}$.The expression becomes $\left(\frac{1+z}{1+\bar{z}}\right)^3 = \left(\frac{1+z}{1+\frac{1}{z}}\right)^3 = \left(\frac{1+z}{\frac{z+1}{z}}\right)^3 = z^3$.Now,$z = i(\cos \frac{2 \pi}{9} - i \sin \frac{2 \pi}{9}) = i e^{-i \frac{2 \pi}{9}}$.Then $z^3 = (i e^{-i \frac{2 \pi}{9}})^3 = i^3 e^{-i \frac{6 \pi}{9}} = -i e^{-i \frac{2 \pi}{3}}$.$z^3 = -i (\cos \frac{2 \pi}{3} - i \sin \frac{2 \pi}{3}) = -i (-\frac{1}{2} - i \frac{\sqrt{3}}{2}) = i \frac{1}{2} - \frac{\sqrt{3}}{2} = -\frac{1}{2}(\sqrt{3}-i)$.

The value of $\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^3$ is

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