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Question Number 8301 by Rasheed Soomro last updated on 06/Oct/16

$$\mathrm{Is}\{{(\omega +\mathrm{i})}^0,{(\omega +\mathrm{i})}^1,{(\omega +\mathrm{i})}^2,....,{(\omega +\mathrm{i})}^{\mathrm{n}}\}$$$$\mathrm{cyclic}\mathrm{for}\mathrm{any}\mathrm{value}\mathrm{of}\mathrm{n}?$$$$\mathrm{Determine}\mathrm{the}\mathrm{smallest}\mathrm{such}\mathrm{n}\mathrm{if}\mathrm{it}\mathrm{exists}.$$$$\omega \mathrm{is}\mathrm{a}\mathrm{complex}\mathrm{cuberoot}\mathrm{of}\mathrm{unity}\mathrm{and}$$$$\mathrm{i}=\sqrt{-1}$$

Commented by 123456 last updated on 07/Oct/16

$${(\omega +i)}^n=\underset{i=0}{\sum ^n}\left(\begin{array}{c}n\\ i\end{array}\right){\omega }^i{\mathrm{i}}^{n-i}$$$${\omega }^i={\omega }^{i\mathrm{mod}3}$$$${\mathrm{i}}^{n-i}={\mathrm{i}}^{n-i\mathrm{mod}4}$$

Answered by prakash jain last updated on 08/Oct/16

$$w=\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}$$$$w+i=\cos \frac{2\pi }{3}+i(\sin \frac{2\pi }{3}+1)$$$$\mid w+i\mid =\sqrt{{\cos }^2\frac{2\pi }{3}+{(\sin \frac{2\pi }{3}+1)}^2}=r\ne 1$$$$w+i={re}^{i\mathrm{\varnothing }}$$$$\because r\ne 1,{(w+i)}^j={(w+i)}^k\Rightarrow j=k$$

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