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Question Number 140104 by EnterUsername last updated on 04/May/21

$$\mathrm{Let}z=1+\cos \left(10\pi /9\right)+i\sin \left(10\pi /9\right).\mathrm{Then}$$$$\left(\mathrm{A}\right)\mid z\mid =2\cos \left(\frac{2\pi }{9}\right)\left(\mathrm{B}\right)\arg z=\frac{8\pi }{9}$$$$\left(\mathrm{C}\right)\mid z\mid =2\cos \left(\frac{4\pi }{9}\right)\left(\mathrm{D}\right)\arg z=\frac{5\pi }{9}$$

Commented by EnterUsername last updated on 04/May/21

$$\mathrm{One}\mathrm{or}\mathrm{more}\mathrm{answers}\mathrm{may}\mathrm{be}\mathrm{correct}.$$

Answered by MJS_new last updated on 04/May/21

$$\cos \frac{10\pi }{9}+\mathrm{i}\sin \frac{10\pi }{9}=-\cos \frac{\pi }{9}-\mathrm{i}\sin \frac{\pi }{9}$$$$\mid z\mid =\sqrt{{(1-\cos \theta )}^2+{(-\sin \theta )}^2}=\sqrt{2-2\cos \theta }=$$$$=2\mid \sin \frac{\theta }{2}\mid =2\sin \frac{\pi }{18}=2\cos \frac{4\pi }{9}$$$$\tan \left(\arg z\right)=\frac{-\sin \theta }{1-\cos \theta }=-\cot \frac{\theta }{2}=-\cot \frac{\pi }{18}\Rightarrow$$$$\Rightarrow \arg z=\frac{5\pi }{9}$$

Commented by EnterUsername last updated on 04/May/21

$$\mathrm{Thank}\mathrm{you}\mathrm{Sir}$$$$\arg z=-{\tan }^{-1}\left(\cot \frac{\pi }{18}\right)=\frac{\pi }{2}+{\cot }^{-1}\left(\cot \frac{\pi }{18}\right)=\frac{\pi }{2}+\frac{\pi }{18}=\frac{5\pi }{9}$$$$\mathrm{Understood}!$$

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