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Question Number 20933 by Tinkutara last updated on 08/Sep/17

$$\mathrm{If}z\mathrm{is}\mathrm{a}\mathrm{complex}\mathrm{number}\mathrm{satisfying}$$$$z+z^{-1}=1,\mathrm{then}{z}^n+z^{-n},n\in N,\mathrm{has}$$$$\mathrm{the}\mathrm{value}$$$$\left(1\right)2{(-1)}^n,\mathrm{when}n\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}3$$$$\left(2\right){(-1)}^{n-1},\mathrm{when}n\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{multiple}\mathrm{of}$$$$3$$$$\left(3\right){(-1)}^{n+1},\mathrm{when}n\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}3$$$$\left(4\right)0\mathrm{when}n\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{multiple}\mathrm{of}3$$

Answered by $@ty@m last updated on 09/Sep/17

$$Given$$$$z+\frac{1}{z}=1$$$$\Rightarrow z^2-z+1=0$$$$\Rightarrow z=\frac{1\pm i\sqrt{3}}{2}$$$$\Rightarrow z=\cos \frac{\pi }{3}\pm i\sin \frac{\pi }{3}$$$$\therefore z^n+z^{-n}=\cos \frac{n\pi }{3}+i\sin \frac{n\pi }{3}+\cos \frac{n\pi }{3}-i\sin \frac{n\pi }{3}$$$$=2cos\frac{n\pi }{3}$$$$\mathrm{Case}\left(\mathrm{i}\right)n=1$$$$2cos\frac{n\pi }{3}=2\cos \frac{\pi }{3}=2\times \frac{1}{2}=1$$$$\mathrm{Case}\left(\mathrm{ii}\right)n=2$$$$2cos\frac{n\pi }{3}=2\cos \frac{2\pi }{3}=2\times \frac{-1}{2}=-1$$$$\mathrm{Case}\left(\mathrm{iii}\right)n=3$$$$2cos\frac{n\pi }{3}=2\cos \pi =2\times (-1)=-2$$$$andsoon....$$$$\Rightarrow \left(1\right)\mathrm{\&}\left(2\right)arecorrectanswers.$$

Commented by Tinkutara last updated on 09/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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