Menu Close

Question-136537




Question Number 136537 by Jamshidbek last updated on 23/Mar/21
Answered by Olaf last updated on 23/Mar/21
a+(1/a) = −1  a^2 +a+1 = 0  a = ((−1±i(√3))/2) = e^(i((2π)/3)) or e^(i((4π)/3))   1+2+3+4+5+6+7+8+9+1+0+1+1  = ((9×10)/2)+3 = 48 = 3n (n = 16)  ⇒ 1234567891011 = 3p, p∈N  and 1110987654321 = 3q, q∈N  If a = e^(i((2π)/3))  :  a^(1234567891011) +(1/a^(1110987654321) )  = a^(i((2π)/3)×3p) +(1/e^(i((2π)/3)×3q) ) = e^(2pπi) +e^(−2qπi)  = 2  If a = e^(i((4π)/3))  :  = a^(i((4π)/3)×3p) +(1/e^(i((4π)/3)×3q) ) = e^(4pπi) +e^(−4qπi)  = 2  answer is : 2
$${a}+\frac{\mathrm{1}}{{a}}\:=\:−\mathrm{1} \\ $$$${a}^{\mathrm{2}} +{a}+\mathrm{1}\:=\:\mathrm{0} \\ $$$${a}\:=\:\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\:{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \mathrm{or}\:{e}^{{i}\frac{\mathrm{4}\pi}{\mathrm{3}}} \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+\mathrm{8}+\mathrm{9}+\mathrm{1}+\mathrm{0}+\mathrm{1}+\mathrm{1} \\ $$$$=\:\frac{\mathrm{9}×\mathrm{10}}{\mathrm{2}}+\mathrm{3}\:=\:\mathrm{48}\:=\:\mathrm{3}{n}\:\left({n}\:=\:\mathrm{16}\right) \\ $$$$\Rightarrow\:\mathrm{1234567891011}\:=\:\mathrm{3}{p},\:{p}\in\mathbb{N} \\ $$$$\mathrm{and}\:\mathrm{1110987654321}\:=\:\mathrm{3}{q},\:{q}\in\mathbb{N} \\ $$$$\mathrm{If}\:{a}\:=\:{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:: \\ $$$${a}^{\mathrm{1234567891011}} +\frac{\mathrm{1}}{{a}^{\mathrm{1110987654321}} } \\ $$$$=\:{a}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}×\mathrm{3}{p}} +\frac{\mathrm{1}}{{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}×\mathrm{3}{q}} }\:=\:{e}^{\mathrm{2}{p}\pi{i}} +{e}^{−\mathrm{2}{q}\pi{i}} \:=\:\mathrm{2} \\ $$$$\mathrm{If}\:{a}\:=\:{e}^{{i}\frac{\mathrm{4}\pi}{\mathrm{3}}} \:: \\ $$$$=\:{a}^{{i}\frac{\mathrm{4}\pi}{\mathrm{3}}×\mathrm{3}{p}} +\frac{\mathrm{1}}{{e}^{{i}\frac{\mathrm{4}\pi}{\mathrm{3}}×\mathrm{3}{q}} }\:=\:{e}^{\mathrm{4}{p}\pi{i}} +{e}^{−\mathrm{4}{q}\pi{i}} \:=\:\mathrm{2} \\ $$$$\mathrm{answer}\:\mathrm{is}\::\:\mathrm{2} \\ $$
Commented by Jamshidbek last updated on 23/Mar/21
thank you sir
$${thank}\:{you}\:{sir} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *