Question Number 8336 by Rasheed Soomro last updated on 08/Oct/16
$$\mathrm{Determine}\:\mathrm{smallest}\:\mathrm{n}\left(\neq\mathrm{0}\right),\:\mathrm{for}\:\mathrm{which} \\ $$$$\left(\omega+\mathrm{i}\right)^{\mathrm{n}} =\mathrm{1}. \\ $$
Commented by prakash jain last updated on 08/Oct/16
$${w}=\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{3}}+{i}\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}}=−\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${w}+{i}=−\frac{\mathrm{1}}{\mathrm{2}}+{i}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{1}\right) \\ $$$$\mid{w}+{i}\mid=\sqrt{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}}=\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\mathrm{Since}\:\mid{w}+{i}\mid=\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\mathrm{of}\:\left({w}+{i}\right)^{{n}} =\mathrm{1}\Rightarrow{n}=\mathrm{0} \\ $$$$\mathrm{There}\:\mathrm{is}\:\mathrm{no}\:\mathrm{value}\:\mathrm{of}\:{n}\neq\mathrm{0}\:\mathrm{for}\:\mathrm{which} \\ $$$$\left({w}+{i}\right)^{{n}} =\mathrm{1} \\ $$
Commented by Rasheed Soomro last updated on 09/Oct/16
$$\mathrm{Thanks}\:\mathrm{a}\:\mathrm{lot}! \\ $$