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Question Number 104333 by 175mohamed last updated on 20/Jul/20

if i^( 5n +1 ) = 1 , n∈Z then show that : n = 3+4p ,p∈Z

$${if}\:{i}^{\:\mathrm{5}{n}\:+\mathrm{1}\:} =\:\mathrm{1}\:\:\:\:,\:{n}\in{Z} \\ $$$${then}\:{show}\:{that}\::\:{n}\:=\:\mathrm{3}+\mathrm{4}{p}\:\:\:,{p}\in{Z} \\ $$

Answered by OlafThorendsen last updated on 20/Jul/20

i^(5n+1) = 1 = i^(4k) ⇔ 5n+1 = 4k ⇔ n = −1+4k−4n ⇔ n = 3+4(k−n−1) ⇔ n = 3+4p

$${i}^{\mathrm{5}{n}+\mathrm{1}} \:=\:\mathrm{1}\:=\:{i}^{\mathrm{4}{k}} \\ $$$$\Leftrightarrow\:\mathrm{5}{n}+\mathrm{1}\:=\:\mathrm{4}{k} \\ $$$$\Leftrightarrow\:{n}\:=\:−\mathrm{1}+\mathrm{4}{k}−\mathrm{4}{n} \\ $$$$\Leftrightarrow\:{n}\:=\:\mathrm{3}+\mathrm{4}\left({k}−{n}−\mathrm{1}\right) \\ $$$$\Leftrightarrow\:{n}\:=\:\mathrm{3}+\mathrm{4}{p} \\ $$

Answered by mr W last updated on 21/Jul/20

i=cos (π/2)+i sin (π/2) i^(5n+1) =cos (((5n+1)π)/2)+i sin (((5n+1)π)/2) 1=cos 2kπ+i sin 2kπ ⇒(((5n+1)π)/2)=2kπ ⇒5n+1=4k ⇒n=4k−4n−1 ⇒n=4(k−n−1)+3 ⇒n=4p+3

$${i}=\mathrm{cos}\:\frac{\pi}{\mathrm{2}}+{i}\:\mathrm{sin}\:\frac{\pi}{\mathrm{2}} \\ $$$${i}^{\mathrm{5}{n}+\mathrm{1}} =\mathrm{cos}\:\frac{\left(\mathrm{5}{n}+\mathrm{1}\right)\pi}{\mathrm{2}}+{i}\:\mathrm{sin}\:\frac{\left(\mathrm{5}{n}+\mathrm{1}\right)\pi}{\mathrm{2}} \\ $$$$\mathrm{1}=\mathrm{cos}\:\mathrm{2}{k}\pi+{i}\:\mathrm{sin}\:\mathrm{2}{k}\pi \\ $$$$\Rightarrow\frac{\left(\mathrm{5}{n}+\mathrm{1}\right)\pi}{\mathrm{2}}=\mathrm{2}{k}\pi \\ $$$$\Rightarrow\mathrm{5}{n}+\mathrm{1}=\mathrm{4}{k} \\ $$$$\Rightarrow{n}=\mathrm{4}{k}−\mathrm{4}{n}−\mathrm{1} \\ $$$$\Rightarrow{n}=\mathrm{4}\left({k}−{n}−\mathrm{1}\right)+\mathrm{3} \\ $$$$\Rightarrow{n}=\mathrm{4}{p}+\mathrm{3} \\ $$

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