Question Number 8301 by Rasheed Soomro last updated on 06/Oct/16
$$\mathrm{Is}\:\:\left\{\:\left(\omega+\mathrm{i}\right)^{\mathrm{0}} ,\:\left(\omega+\mathrm{i}\right)^{\mathrm{1}} ,\:\left(\omega+\mathrm{i}\right)^{\mathrm{2}} ,\:....,\:\left(\omega+\mathrm{i}\right)^{\mathrm{n}} \:\right\} \\ $$$$\mathrm{cyclic}\:\mathrm{for}\:\mathrm{any}\:\mathrm{value}\:\mathrm{of}\:\mathrm{n}? \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{such}\:\mathrm{n}\:\mathrm{if}\:\mathrm{it}\:\mathrm{exists}. \\ $$$$\omega\:\mathrm{is}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{cuberoot}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{and} \\ $$$$\mathrm{i}=\sqrt{−\mathrm{1}} \\ $$
Commented by 123456 last updated on 07/Oct/16
$$\left(\omega+{i}\right)^{{n}} =\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{i}}\end{pmatrix}\omega^{{i}} \mathrm{i}^{{n}−{i}} \\ $$$$\omega^{{i}} =\omega^{{i}\:\mathrm{mod3}} \\ $$$$\mathrm{i}^{{n}−{i}} =\mathrm{i}^{{n}−{i}\:\mathrm{mod}\:\mathrm{4}} \\ $$
Answered by prakash jain last updated on 08/Oct/16
$${w}=\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{3}}+{i}\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$${w}+{i}=\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{3}}+{i}\left(\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{1}\right) \\ $$$$\mid{w}+{i}\mid=\sqrt{\mathrm{cos}^{\mathrm{2}} \frac{\mathrm{2}\pi}{\mathrm{3}}+\left(\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} }={r}\neq\mathrm{1} \\ $$$${w}+{i}={re}^{{i}\emptyset} \\ $$$$\because{r}\neq\mathrm{1},\:\left({w}+{i}\right)^{{j}} =\left({w}+{i}\right)^{{k}} \Rightarrow{j}={k} \\ $$