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Question Number 88314 by M±th+et£s last updated on 09/Apr/20
( a,b )are complex numbers and a^2 +ab+b^2 =0 find ((a/(a+b)))^(2020) +((b/(a+b)))^(2020)
$$\left(\:{a},{b}\:\right){are}\:{complex}\:{numbers}\:{and}\:{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} =\mathrm{0} \\ $$$${find}\:\left(\frac{{a}}{{a}+{b}}\right)^{\mathrm{2020}} +\left(\frac{{b}}{{a}+{b}}\right)^{\mathrm{2020}} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 09/Apr/20
a is slution of x^2 +bx +b^2 =0 Δ=b^2 −4b^2 =−3b^2 =(i(√3)b)^2 ⇒x_1 =((−b+i(√3)b)/2) =b e^((i2π)/3) and x_2 =((−b−i(√b))/2)=be^(−((i2π)/3)) a=x_1 ⇒(a/(a+b)) =((b e^((i2π)/3) )/(be^((i2π)/3) +b)) =(e^((i2π)/3) /(1+e^((i2π)/3) )) =(1/(1+e^(−((i2π)/3)) )) =(1/(1−(1/2)−((i(√3))/2)))= =(1/((1/2)−((i(√3))/2)))=e^((iπ)/3) (b/(a+b)) =(b/(be^((i2π)/3) +b)) =(1/(e^((i2π)/3) +1)) =(1/(1−(1/2)+((i(√3))/2))) =(1/((1/2)+((i(√3))/2))) =e^(−((iπ)/3)) ⇒ ((a/(a+b)))^(2020) +((b/(a+b)))^(2020) =(e^((iπ)/3) )^(2020) +(e^(−((iπ)/3)) )^(2020) =2cos(((2020π)/3)) this value can be simplified... wr find the same value for x =x_2
$${a}\:{is}\:{slution}\:{of}\:{x}^{\mathrm{2}} \:+{bx}\:+{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Delta={b}^{\mathrm{2}} −\mathrm{4}{b}^{\mathrm{2}} \:=−\mathrm{3}{b}^{\mathrm{2}} \:=\left({i}\sqrt{\mathrm{3}}{b}\right)^{\mathrm{2}} \:\Rightarrow{x}_{\mathrm{1}} =\frac{−{b}+{i}\sqrt{\mathrm{3}}{b}}{\mathrm{2}}\:={b}\:{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \\ $$$${and}\:{x}_{\mathrm{2}} =\frac{−{b}−{i}\sqrt{{b}}}{\mathrm{2}}={be}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \\ $$$${a}={x}_{\mathrm{1}} \:\Rightarrow\frac{{a}}{{a}+{b}}\:=\frac{{b}\:{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} }{{be}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:+{b}}\:=\frac{{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} }{\mathrm{1}+{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} }\:=\frac{\mathrm{1}}{\mathrm{1}+{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} }\:=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}}=\:=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}−\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}}={e}^{\frac{{i}\pi}{\mathrm{3}}} \\ $$$$\frac{{b}}{{a}+{b}}\:=\frac{{b}}{{be}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:+{b}}\:=\frac{\mathrm{1}}{{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}+\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}}\:={e}^{−\frac{{i}\pi}{\mathrm{3}}} \:\Rightarrow \\ $$$$\left(\frac{{a}}{{a}+{b}}\right)^{\mathrm{2020}} +\left(\frac{{b}}{{a}+{b}}\right)^{\mathrm{2020}} \:=\left({e}^{\frac{{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2020}} +\left({e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2020}} \\ $$$$=\mathrm{2}{cos}\left(\frac{\mathrm{2020}\pi}{\mathrm{3}}\right)\:{this}\:{value}\:{can}\:{be}\:{simplified}… \\ $$$${wr}\:{find}\:{the}\:{same}\:{value}\:{for}\:{x}\:={x}_{\mathrm{2}} \\ $$
Answered by MJS last updated on 09/Apr/20
a^2 +ab+b^2 =0 ⇒ b=a(−(1/2)±((√3)/2)i) ⇒ (a/(a+b))=(1/2)∓((√3)/2)i ∧ (b/(a+b))=(1/2)±((√3)/2)i ⇒ ((a/(a+b)))^n +((b/(a+b)))^n =e^(−i((nπ)/3)) +e^(i((nπ)/3)) now you should be able to find the answer
$${a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{b}={a}\left(−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right) \\ $$$$\Rightarrow\:\frac{{a}}{{a}+{b}}=\frac{\mathrm{1}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\:\wedge\:\frac{{b}}{{a}+{b}}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$\Rightarrow\:\left(\frac{{a}}{{a}+{b}}\right)^{{n}} +\left(\frac{{b}}{{a}+{b}}\right)^{{n}} =\mathrm{e}^{−\mathrm{i}\frac{{n}\pi}{\mathrm{3}}} +\mathrm{e}^{\mathrm{i}\frac{{n}\pi}{\mathrm{3}}} \\ $$$$\mathrm{now}\:\mathrm{you}\:\mathrm{should}\:\mathrm{be}\:\mathrm{able}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{answer} \\ $$
Commented by M±th+et£s last updated on 09/Apr/20
thanx sir . i think ans=−1
$${thanx}\:{sir}\:.\:{i}\:{think}\:{ans}=−\mathrm{1} \\ $$
Commented by MJS last updated on 09/Apr/20
yes
$$\mathrm{yes} \\ $$
Answered by Joel578 last updated on 10/Apr/20
(a + b)^2 − ab = 0 ⇒ (a + b)^2 = ab ⇒ ((ab)/((a + b)^2 )) = 1 ⇒ ((a/(a + b)) )((b/(a + b))) = 1 ((a/(a + b))) + ((b/(a + b))) = ((a + b)/(a + b)) = 1 Let x = (a/(a + b)), y = (b/(a + b)) with xy = 1, x + y = 1 Define f(n) = x^n + y^n f(1) = x + y = 1 f(2) = x^2 + y^2 = (x + y)^2 − 2xy = −1 f(3) = x^3 + y^3 = (x + y)^3 − 3xy(x + y) = −2 f(4) = x^4 + y^4 = (x^3 + y^3 )(x + y) − xy(x^2 + y^2 ) = −1 f(5) = 1, f(6) = 2, f(7) = 1, f(8) = −1, f(9) = −2 f(n) = f(n+6) ∴ f(2020) = f(4) = −1
$$\left({a}\:+\:{b}\right)^{\mathrm{2}} \:−\:{ab}\:=\:\mathrm{0}\:\Rightarrow\:\left({a}\:+\:{b}\right)^{\mathrm{2}} \:=\:{ab}\:\Rightarrow\:\frac{{ab}}{\left({a}\:+\:{b}\right)^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\left(\frac{{a}}{{a}\:+\:{b}}\:\right)\left(\frac{{b}}{{a}\:+\:{b}}\right)\:=\:\mathrm{1} \\ $$$$ \\ $$$$\left(\frac{{a}}{{a}\:+\:{b}}\right)\:+\:\left(\frac{{b}}{{a}\:+\:{b}}\right)\:=\:\frac{{a}\:+\:{b}}{{a}\:+\:{b}}\:=\:\mathrm{1} \\ $$$$ \\ $$$$\mathrm{Let}\:{x}\:=\:\frac{{a}}{{a}\:+\:{b}},\:{y}\:=\:\frac{{b}}{{a}\:+\:{b}}\:\mathrm{with}\:{xy}\:=\:\mathrm{1},\:{x}\:+\:{y}\:\:=\:\mathrm{1} \\ $$$$\mathrm{Define}\:{f}\left({n}\right)\:=\:{x}^{{n}} \:+\:{y}^{{n}} \\ $$$${f}\left(\mathrm{1}\right)\:=\:{x}\:+\:{y}\:=\:\mathrm{1} \\ $$$${f}\left(\mathrm{2}\right)\:=\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:=\:\left({x}\:+\:{y}\right)^{\mathrm{2}} \:−\:\mathrm{2}{xy}\:=\:−\mathrm{1} \\ $$$${f}\left(\mathrm{3}\right)\:=\:{x}^{\mathrm{3}} \:+\:{y}^{\mathrm{3}} \:=\:\left({x}\:+\:{y}\right)^{\mathrm{3}} \:−\:\mathrm{3}{xy}\left({x}\:+\:{y}\right)\:=\:−\mathrm{2} \\ $$$${f}\left(\mathrm{4}\right)\:=\:{x}^{\mathrm{4}} \:+\:{y}^{\mathrm{4}} \:=\:\left({x}^{\mathrm{3}} \:+\:{y}^{\mathrm{3}} \right)\left({x}\:+\:{y}\right)\:−\:{xy}\left({x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \right)\:=\:−\mathrm{1} \\ $$$${f}\left(\mathrm{5}\right)\:=\:\mathrm{1},\:{f}\left(\mathrm{6}\right)\:=\:\mathrm{2},\:{f}\left(\mathrm{7}\right)\:=\:\mathrm{1},\:{f}\left(\mathrm{8}\right)\:=\:−\mathrm{1},\:{f}\left(\mathrm{9}\right)\:=\:−\mathrm{2} \\ $$$${f}\left({n}\right)\:=\:{f}\left({n}+\mathrm{6}\right) \\ $$$$\therefore\:{f}\left(\mathrm{2020}\right)\:=\:{f}\left(\mathrm{4}\right)\:=\:−\mathrm{1} \\ $$
Commented by Joel578 last updated on 10/Apr/20
without complex method
$$\mathrm{without}\:\mathrm{complex}\:\mathrm{method} \\ $$

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