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Question Number 98416 by 175 last updated on 13/Jun/20

Answered by Farruxjano last updated on 13/Jun/20

$${\mathit{i}}^{5\mathit{n}+1}=1\Rightarrow 5\mathit{n}+1=2\mathit{k}(\mathit{k}\in \mathit{Z}),\Rightarrow \mathit{n}=2\mathit{t}+1(\mathit{t}\in \mathit{Z})$$

Answered by Ramajunan last updated on 13/Jun/20

$$i^{5n+1}=1$$$$i^{5n+1}={\left(i\right)}^{4k}wherek\in Z$$$$5n+1=4k$$$$n=\frac{4k-1}{5}$$

Answered by mathmax by abdo last updated on 13/Jun/20

$${\mathrm{i}}^{5\mathrm{n}+1}=1\Rightarrow {\left({\mathrm{e}}^{\frac{\mathrm{i}\pi }{2}}\right)}^{5\mathrm{n}+1}={\mathrm{e}}^{\mathrm{i}2\mathrm{k}\pi }\Rightarrow {\mathrm{e}}^{\mathrm{i}(5\mathrm{n}+1)\frac{\pi }{2}}={\mathrm{e}}^{\mathrm{i}2\mathrm{k}\pi }\Rightarrow (5\mathrm{n}+1)\frac{\pi }{2}=2\mathrm{k}\pi \Rightarrow$$$$5\mathrm{n}+1=4\mathrm{k}\Rightarrow 5\mathrm{n}=4\mathrm{k}-1\mathrm{let}\mathrm{use}\mathrm{congruence}\mathrm{modul}5\left(\mathrm{prime}\right)\Rightarrow$$$$4\stackrel{-}{\mathrm{k}}-\stackrel{-}{1}=0\Rightarrow -\stackrel{-}{\mathrm{k}}-1=0\left[5\right]\Rightarrow \mathrm{k}=4\left[5\right]\Rightarrow \mathrm{k}=5\mathrm{p}+4\Rightarrow$$$$5\mathrm{n}=20\mathrm{p}+16-1\Rightarrow 5\mathrm{n}=20\mathrm{p}+15\Rightarrow \mathrm{n}=4\mathrm{p}+3\mathrm{so}\mathrm{all}\mathrm{number}\mathrm{at}\mathrm{form}$$$$4\mathrm{p}+3\mathrm{is}\mathrm{solution}$$

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