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sin-pi-9-i-sin-3pi-18-9-

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Question Number 68632 by aliesam last updated on 14/Sep/19
(sin(π/9) + i sin((3π)/(18)))^(−9)
$$\left({sin}\frac{\pi}{\mathrm{9}}\:+\:{i}\:{sin}\frac{\mathrm{3}\pi}{\mathrm{18}}\right)^{−\mathrm{9}} \\ $$

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