Question Number 104333 by 175mohamed last updated on 20/Jul/20
$${if}\:{i}^{\:\mathrm{5}{n}\:+\mathrm{1}\:} =\:\mathrm{1}\:\:\:\:,\:{n}\in{Z} \\ $$$${then}\:{show}\:{that}\::\:{n}\:=\:\mathrm{3}+\mathrm{4}{p}\:\:\:,{p}\in{Z} \\ $$
Answered by OlafThorendsen last updated on 20/Jul/20
$${i}^{\mathrm{5}{n}+\mathrm{1}} \:=\:\mathrm{1}\:=\:{i}^{\mathrm{4}{k}} \\ $$$$\Leftrightarrow\:\mathrm{5}{n}+\mathrm{1}\:=\:\mathrm{4}{k} \\ $$$$\Leftrightarrow\:{n}\:=\:−\mathrm{1}+\mathrm{4}{k}−\mathrm{4}{n} \\ $$$$\Leftrightarrow\:{n}\:=\:\mathrm{3}+\mathrm{4}\left({k}−{n}−\mathrm{1}\right) \\ $$$$\Leftrightarrow\:{n}\:=\:\mathrm{3}+\mathrm{4}{p} \\ $$
Answered by mr W last updated on 21/Jul/20
$${i}=\mathrm{cos}\:\frac{\pi}{\mathrm{2}}+{i}\:\mathrm{sin}\:\frac{\pi}{\mathrm{2}} \\ $$$${i}^{\mathrm{5}{n}+\mathrm{1}} =\mathrm{cos}\:\frac{\left(\mathrm{5}{n}+\mathrm{1}\right)\pi}{\mathrm{2}}+{i}\:\mathrm{sin}\:\frac{\left(\mathrm{5}{n}+\mathrm{1}\right)\pi}{\mathrm{2}} \\ $$$$\mathrm{1}=\mathrm{cos}\:\mathrm{2}{k}\pi+{i}\:\mathrm{sin}\:\mathrm{2}{k}\pi \\ $$$$\Rightarrow\frac{\left(\mathrm{5}{n}+\mathrm{1}\right)\pi}{\mathrm{2}}=\mathrm{2}{k}\pi \\ $$$$\Rightarrow\mathrm{5}{n}+\mathrm{1}=\mathrm{4}{k} \\ $$$$\Rightarrow{n}=\mathrm{4}{k}−\mathrm{4}{n}−\mathrm{1} \\ $$$$\Rightarrow{n}=\mathrm{4}\left({k}−{n}−\mathrm{1}\right)+\mathrm{3} \\ $$$$\Rightarrow{n}=\mathrm{4}{p}+\mathrm{3} \\ $$