Menu Close

Question-46110

Tinku Tara 0 Comments
Pin




Question Number 46110 by Tawa1 last updated on 21/Oct/18
Commented by math khazana by abdo last updated on 22/Oct/18
Δ=1−4=−3=(i(√3))^2 ⇒α=((1+i(√3))/2) and β=((1−i(√3))/2) ⇒α=e^(i(π/3)) and β =e^(−i(π/3)) ⇒α^(101) =e^(i((101)/3)π) = e^(i(((99 +2)/3))π) =e^(33iπ) e^(i((2π)/3)) =−j β^(107) = e^(−i((107)/3)π) =e^(−i((101)/3)π) e^(−i((6π)/3)) = e^(−i33π) e^(−i((2π)/3)) =−j^− ⇒α^(101) +β^(107) =−j−j^− =−(j+j^− ) =−(1+j+j^− ) +1=0+1 =1
$$\Delta=\mathrm{1}−\mathrm{4}=−\mathrm{3}=\left({i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \Rightarrow\alpha=\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:{and}\:\beta=\frac{\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\alpha={e}^{{i}\frac{\pi}{\mathrm{3}}} \:{and}\:\beta\:={e}^{−{i}\frac{\pi}{\mathrm{3}}} \:\Rightarrow\alpha^{\mathrm{101}} ={e}^{{i}\frac{\mathrm{101}}{\mathrm{3}}\pi} \\ $$$$=\:{e}^{{i}\left(\frac{\mathrm{99}\:+\mathrm{2}}{\mathrm{3}}\right)\pi} ={e}^{\mathrm{33}{i}\pi} \:{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:=−{j} \\ $$$$\beta^{\mathrm{107}} =\:{e}^{−{i}\frac{\mathrm{107}}{\mathrm{3}}\pi} \:={e}^{−{i}\frac{\mathrm{101}}{\mathrm{3}}\pi} \:{e}^{−{i}\frac{\mathrm{6}\pi}{\mathrm{3}}} =\:{e}^{−{i}\mathrm{33}\pi} {e}^{−{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$$=−\overset{−} {{j}}\:\Rightarrow\alpha^{\mathrm{101}} \:+\beta^{\mathrm{107}} \:=−{j}−\overset{−} {{j}}=−\left({j}+\overset{−} {{j}}\right) \\ $$$$=−\left(\mathrm{1}+{j}+\overset{−} {{j}}\right)\:+\mathrm{1}=\mathrm{0}+\mathrm{1}\:=\mathrm{1} \\ $$$$ \\ $$
Commented by Tawa1 last updated on 22/Oct/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by ajfour last updated on 21/Oct/18
x^3 +1 = (x+1)(x^2 −x+1) roots of x^3 +1=0 are −1, −ω^2 , −ω So (−ω)^(101) +(−ω^2 )^(107) = −ω^2 −ω = 1 .
$${x}^{\mathrm{3}} +\mathrm{1}\:=\:\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right) \\ $$$${roots}\:{of}\:\:\:\:\:{x}^{\mathrm{3}} +\mathrm{1}=\mathrm{0}\:\:{are} \\ $$$$−\mathrm{1},\:−\omega^{\mathrm{2}} ,\:−\omega \\ $$$${So}\:\:\:\:\left(−\omega\right)^{\mathrm{101}} +\left(−\omega^{\mathrm{2}} \right)^{\mathrm{107}} \:=\:−\omega^{\mathrm{2}} −\omega \\ $$$$\:\:\:\:\:\:=\:\mathrm{1}\:. \\ $$
Commented by Tawa1 last updated on 21/Oct/18
God bless you sir. but i don′t really understand sir. from the (− ω) and the substitution to get 1
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{but}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{really}\:\mathrm{understand}\:\mathrm{sir}.\:\:\mathrm{from}\:\mathrm{the}\:\:\left(−\:\omega\right) \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{substitution}\:\mathrm{to}\:\mathrm{get}\:\:\mathrm{1} \\ $$
Commented by $@ty@m last updated on 21/Oct/18
x^3 +1=0 ⇒(x+1)(x^2 −x+1)=0 ⇒x+1=0 or x^2 −x+1=0 ⇒x=−1 or x=−ω, −ω^2 i.e. roots of x^2 −x+1 are −ω, −ω^2 ∴ α=−ω, β=−ω^2
$${x}^{\mathrm{3}} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}+\mathrm{1}=\mathrm{0}\:{or}\:{x}^{\mathrm{2}} −{x}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}=−\mathrm{1}\:{or}\:{x}=−\omega,\:−\omega^{\mathrm{2}} \\ $$$${i}.{e}.\:{roots}\:{of}\:{x}^{\mathrm{2}} −{x}+\mathrm{1}\:{are}\:−\omega,\:−\omega^{\mathrm{2}} \\ $$$$\therefore\:\alpha=−\omega,\:\beta=−\omega^{\mathrm{2}} \\ $$
Commented by Tawa1 last updated on 21/Oct/18
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Pin

Leave a Reply

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