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1-If-is-an-imaginary-fifth-root-of-unity-then-find-value-of-log-2-1-2-3-1-2-Find-value-of-i-3-100-i-3-100-2-100-

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Question Number 49202 by rahul 19 last updated on 04/Dec/18
1) If ω is an imaginary fifth root of unity, then find value of log _2 ∣1+ω+ω^2 +ω^3 −(1/ω)∣ ? 2) Find value of : (i+(√3))^(100) +(i−(√3))^(100) +2^(100) ?
$$\left.\mathrm{1}\right)\:{If}\:\omega\:{is}\:{an}\:{imaginary}\:{fifth}\:{root}\:{of} \\ $$$${unity},\:{then}\:{find}\:{value}\:{of}\: \\ $$$$\mathrm{log}\:_{\mathrm{2}} \:\mid\mathrm{1}+\omega+\omega^{\mathrm{2}} +\omega^{\mathrm{3}} −\frac{\mathrm{1}}{\omega}\mid\:? \\ $$$$\left.\mathrm{2}\right)\:{Find}\:{value}\:{of}\:: \\ $$$$\left({i}+\sqrt{\mathrm{3}}\right)^{\mathrm{100}} +\left({i}−\sqrt{\mathrm{3}}\right)^{\mathrm{100}} +\mathrm{2}^{\mathrm{100}} \:? \\ $$
Commented by Abdo msup. last updated on 04/Dec/18
1) w^5 =1 ⇒w^5 −1=0 ⇒(w−1)(1+w+w^2 +w^3 +w^4 )=0 but w∈iR ⇒1+w+w^2 +w^3 +w^4 =0 ⇒1+w+w^2 +w^3 =−w^4 ⇒ 1+w+w^2 +w^3 −(1/w) =−(w^4 +(1/w)) =−((w^5 +1)/w) =−(2/w) ⇒ ∣1+w+w^2 +w^3 −(1/w)∣=(2/(∣w∣)) =(2/1) =2 ⇒ log_2 ∣1+w+w^2 +w^3 −(1/w)∣=log_2 (2)=1 .
$$\left.\mathrm{1}\right)\:{w}^{\mathrm{5}} =\mathrm{1}\:\Rightarrow{w}^{\mathrm{5}} −\mathrm{1}=\mathrm{0}\:\Rightarrow\left({w}−\mathrm{1}\right)\left(\mathrm{1}+{w}+{w}^{\mathrm{2}} +{w}^{\mathrm{3}} \:+{w}^{\mathrm{4}} \right)=\mathrm{0}\:{but}\:{w}\in{iR} \\ $$$$\Rightarrow\mathrm{1}+{w}+{w}^{\mathrm{2}} \:+{w}^{\mathrm{3}} \:+{w}^{\mathrm{4}} =\mathrm{0}\:\Rightarrow\mathrm{1}+{w}+{w}^{\mathrm{2}} \:+{w}^{\mathrm{3}} \:=−{w}^{\mathrm{4}} \:\Rightarrow \\ $$$$\mathrm{1}+{w}+{w}^{\mathrm{2}} \:+{w}^{\mathrm{3}} \:−\frac{\mathrm{1}}{{w}}\:=−\left({w}^{\mathrm{4}} \:+\frac{\mathrm{1}}{{w}}\right)\:=−\frac{{w}^{\mathrm{5}} \:+\mathrm{1}}{{w}}\:=−\frac{\mathrm{2}}{{w}}\:\Rightarrow \\ $$$$\mid\mathrm{1}+{w}+{w}^{\mathrm{2}} \:+{w}^{\mathrm{3}} −\frac{\mathrm{1}}{{w}}\mid=\frac{\mathrm{2}}{\mid{w}\mid}\:=\frac{\mathrm{2}}{\mathrm{1}}\:=\mathrm{2}\:\Rightarrow \\ $$$${log}_{\mathrm{2}} \:\mid\mathrm{1}+{w}+{w}^{\mathrm{2}} \:+{w}^{\mathrm{3}} −\frac{\mathrm{1}}{{w}}\mid={log}_{\mathrm{2}} \left(\mathrm{2}\right)=\mathrm{1}\:. \\ $$
Commented by Abdo msup. last updated on 04/Dec/18
2)we have i+(√3)=2(((√3)/2)+(i/2)) =2 e^(i(π/6)) and i−(√3)=−((√3)−i)=−2 e^(−((iπ)/6)) ⇒ ((√3)+i)^(100) +(−(√3) +i)^(100) +2^(100) =2^(100) e^(i((100π)/6)) +2^(100) e^(−i((100π)/6)) +2^(100) =2^(101) cos(((100π)/6))+2^(100) but cos(((100π)/6))=cos(((50π)/3)) =cos(((48π +2π)/3))=cos(16π +((2π)/3)) =cos(((2π)/3))=−(1/2) ⇒((√3)+i)^(100) +(−(√3)+i)^(100) +2^(100) =2^(101) .(−(1/2)) +2^(100) =−2^(100) +2^(100) =0 .
$$\left.\mathrm{2}\right){we}\:{have}\:{i}+\sqrt{\mathrm{3}}=\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{{i}}{\mathrm{2}}\right)\:=\mathrm{2}\:{e}^{{i}\frac{\pi}{\mathrm{6}}} \:\:{and}\: \\ $$$${i}−\sqrt{\mathrm{3}}=−\left(\sqrt{\mathrm{3}}−{i}\right)=−\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{6}}} \Rightarrow \\ $$$$\left(\sqrt{\mathrm{3}}+{i}\right)^{\mathrm{100}} \:+\left(−\sqrt{\mathrm{3}}\:+{i}\right)^{\mathrm{100}} \:+\mathrm{2}^{\mathrm{100}} \\ $$$$=\mathrm{2}^{\mathrm{100}} \:{e}^{{i}\frac{\mathrm{100}\pi}{\mathrm{6}}} \:+\mathrm{2}^{\mathrm{100}} \:{e}^{−{i}\frac{\mathrm{100}\pi}{\mathrm{6}}} \:+\mathrm{2}^{\mathrm{100}} \\ $$$$=\mathrm{2}^{\mathrm{101}} \:{cos}\left(\frac{\mathrm{100}\pi}{\mathrm{6}}\right)+\mathrm{2}^{\mathrm{100}} \:{but} \\ $$$${cos}\left(\frac{\mathrm{100}\pi}{\mathrm{6}}\right)={cos}\left(\frac{\mathrm{50}\pi}{\mathrm{3}}\right)\:={cos}\left(\frac{\mathrm{48}\pi\:+\mathrm{2}\pi}{\mathrm{3}}\right)={cos}\left(\mathrm{16}\pi\:+\frac{\mathrm{2}\pi}{\mathrm{3}}\right) \\ $$$$={cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\left(\sqrt{\mathrm{3}}+{i}\right)^{\mathrm{100}} \:+\left(−\sqrt{\mathrm{3}}+{i}\right)^{\mathrm{100}} \:+\mathrm{2}^{\mathrm{100}} \\ $$$$=\mathrm{2}^{\mathrm{101}} \:.\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\:+\mathrm{2}^{\mathrm{100}} \:=−\mathrm{2}^{\mathrm{100}} \:+\mathrm{2}^{\mathrm{100}} \:=\mathrm{0}\:. \\ $$
Commented by rahul 19 last updated on 05/Dec/18
thank you prof Abdo ☺️
Commented by maxmathsup by imad last updated on 05/Dec/18
you are welcome
$${you}\:{are}\:{welcome} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Dec/18
2)(√3) +i =2(((√3)/2)+i(1/2))=2(cos(π/6)+isin(π/6))=2e^(i(π/6)) i−(√3) =−2(((√3)/2)−i(1/2))=−2e^(−i(π/6)) {(2e^(i(π/6)) )}^(100) +{(−2e^(−i(π/6)) )}^(100) +2^(100) =2^(100) {e^(i((100π)/6)) +e^(−i((100π)/6)) }+2^(100) =2^(100) (2cos((100π)/6))+2^(100) cos(3000^o )=cos(8×360^o +120^o )=cos(120^o ) cos(120^o )=((−1)/2) so ans is 2^(100) ×2×((−1)/2)+2^(100) =0
$$\left.\mathrm{2}\right)\sqrt{\mathrm{3}}\:+{i} \\ $$$$=\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+{i}\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}\left({cos}\frac{\pi}{\mathrm{6}}+{isin}\frac{\pi}{\mathrm{6}}\right)=\mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{6}}} \\ $$$${i}−\sqrt{\mathrm{3}}\: \\ $$$$=−\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−{i}\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{2}{e}^{−{i}\frac{\pi}{\mathrm{6}}} \\ $$$$\left\{\left(\mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{6}}} \right)\right\}^{\mathrm{100}} +\left\{\left(−\mathrm{2}{e}^{−{i}\frac{\pi}{\mathrm{6}}} \right)\right\}^{\mathrm{100}} +\mathrm{2}^{\mathrm{100}} \\ $$$$=\mathrm{2}^{\mathrm{100}} \left\{{e}^{{i}\frac{\mathrm{100}\pi}{\mathrm{6}}} +{e}^{−{i}\frac{\mathrm{100}\pi}{\mathrm{6}}} \right\}+\mathrm{2}^{\mathrm{100}} \\ $$$$=\mathrm{2}^{\mathrm{100}} \left(\mathrm{2}{cos}\frac{\mathrm{100}\pi}{\mathrm{6}}\right)+\mathrm{2}^{\mathrm{100}} \\ $$$${cos}\left(\mathrm{3000}^{{o}} \right)={cos}\left(\mathrm{8}×\mathrm{360}^{{o}} +\mathrm{120}^{{o}} \right)={cos}\left(\mathrm{120}^{{o}} \right) \\ $$$${cos}\left(\mathrm{120}^{{o}} \right)=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${so}\:{ans}\:{is} \\ $$$$\mathrm{2}^{\mathrm{100}} ×\mathrm{2}×\frac{−\mathrm{1}}{\mathrm{2}}+\mathrm{2}^{\mathrm{100}} =\mathrm{0} \\ $$$$ \\ $$
Commented by rahul 19 last updated on 05/Dec/18
thank you sir! ����
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Dec/18
1)x^5 =1 (x−1)(x^4 +x^3 +x^2 +x+1)=0 w^4 +w^3 +w^2 +w+1=0 1+w+w^2 +w^3 −(1/w) =−w^4 −(1/w) =((−w^5 −1)/w)=((−1−1)/1)=−2 so ∣1+w+w^2 +w^3 −(1/w)∣=∣−2∣=2 hence ln_2 2=1
$$\left.\mathrm{1}\right){x}^{\mathrm{5}} =\mathrm{1} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${w}^{\mathrm{4}} +{w}^{\mathrm{3}} +{w}^{\mathrm{2}} +{w}+\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{1}+{w}+{w}^{\mathrm{2}} +{w}^{\mathrm{3}} −\frac{\mathrm{1}}{{w}} \\ $$$$=−{w}^{\mathrm{4}} −\frac{\mathrm{1}}{{w}} \\ $$$$=\frac{−{w}^{\mathrm{5}} −\mathrm{1}}{{w}}=\frac{−\mathrm{1}−\mathrm{1}}{\mathrm{1}}=−\mathrm{2} \\ $$$${so}\:\mid\mathrm{1}+{w}+{w}^{\mathrm{2}} +{w}^{\mathrm{3}} −\frac{\mathrm{1}}{{w}}\mid=\mid−\mathrm{2}\mid=\mathrm{2} \\ $$$${hence}\:{ln}_{\mathrm{2}} \mathrm{2}=\mathrm{1} \\ $$$$ \\ $$

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