Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 49169 by rahul 19 last updated on 04/Dec/18

Find the least positive integer n such  that (((2i)/(1+i)))^n is a +ve integer ?

$${Find}\:{the}\:{least}\:{positive}\:{integer}\:{n}\:{such} \\ $$$${that}\:\left(\frac{\mathrm{2}{i}}{\mathrm{1}+{i}}\right)^{{n}} {is}\:{a}\:+{ve}\:{integer}\:? \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Dec/18

(((2i(1−i))/2))^n   =(i+1)^n    for n=2k  =(1+2i−1)^k =(2i)^k      [k=2p  =(2i)^(2p) =(−4)^p   for p=2  (−4)^2 =16  so n=2×(2×2)=8  so n=8

$$\left(\frac{\mathrm{2}{i}\left(\mathrm{1}−{i}\right)}{\mathrm{2}}\right)^{{n}} \\ $$$$=\left({i}+\mathrm{1}\right)^{{n}} \:\:\:{for}\:{n}=\mathrm{2}{k} \\ $$$$=\left(\mathrm{1}+\mathrm{2}{i}−\mathrm{1}\right)^{{k}} =\left(\mathrm{2}{i}\right)^{{k}} \:\:\:\:\:\left[{k}=\mathrm{2}{p}\right. \\ $$$$=\left(\mathrm{2}{i}\right)^{\mathrm{2}{p}} =\left(−\mathrm{4}\right)^{{p}} \\ $$$${for}\:{p}=\mathrm{2}\:\:\left(−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{16} \\ $$$${so}\:{n}=\mathrm{2}×\left(\mathrm{2}×\mathrm{2}\right)=\mathrm{8} \\ $$$${so}\:{n}=\mathrm{8}\: \\ $$$$ \\ $$

Commented by rahul 19 last updated on 04/Dec/18

thanks sir ����

Terms of Service

Privacy Policy

Contact: info@tinkutara.com