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Question-184084

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Question Number 184084 by HeferH last updated on 02/Jan/23
Answered by Rasheed.Sindhi last updated on 03/Jan/23
w^3 =1⇒w=1,ω,ω^2 Let w_1 =ω & w_2 =ω^2 x=a+b y=a^2 ω+b^2 ω^2 z=aω^2 +bω ((x^2 −z^2 +y+2ab)/x^2 ) x^2 =a^2 +2ab+b^2 z^2 =a^2 ω^4 +2abω^2 ω+b^2 ω^2 =a^2 ω^3 ω+2abω^3 +b^2 ω^2 =a^2 ω+2ab+b^2 ω^2 y=a^2 ω+b^2 ω^2 ((x^2 −z^2 +y+2ab)/x^2 )=1+((−z^2 +y+2ab)/x^2 ) =1+((−(a^2 ω+2ab+b^2 ω^2 )+(a^2 ω+b^2 ω^2 )+2ab)/(a^2 +2ab+b^2 )) =1+((−a^2 ω−2ab−b^2 ω^2 +a^2 ω+b^2 ω^2 +2ab)/(a^2 +2ab+b^2 )) =1+0=1
$${w}^{\mathrm{3}} =\mathrm{1}\Rightarrow{w}=\mathrm{1},\omega,\omega^{\mathrm{2}} \\ $$$${Let}\:{w}_{\mathrm{1}} =\omega\:\&\:{w}_{\mathrm{2}} =\omega^{\mathrm{2}} \\ $$$${x}={a}+{b} \\ $$$${y}={a}^{\mathrm{2}} \omega+{b}^{\mathrm{2}} \omega^{\mathrm{2}} \\ $$$${z}={a}\omega^{\mathrm{2}} +{b}\omega \\ $$$$\frac{{x}^{\mathrm{2}} −{z}^{\mathrm{2}} +{y}+\mathrm{2}{ab}}{{x}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} \\ $$$${z}^{\mathrm{2}} ={a}^{\mathrm{2}} \omega^{\mathrm{4}} +\mathrm{2}{ab}\omega^{\mathrm{2}} \omega+{b}^{\mathrm{2}} \omega^{\mathrm{2}} \\ $$$$\:\:\:\:\:={a}^{\mathrm{2}} \omega^{\mathrm{3}} \omega+\mathrm{2}{ab}\omega^{\mathrm{3}} +{b}^{\mathrm{2}} \omega^{\mathrm{2}} \\ $$$$\:\:\:\:\:={a}^{\mathrm{2}} \omega+\mathrm{2}{ab}+{b}^{\mathrm{2}} \omega^{\mathrm{2}} \\ $$$${y}={a}^{\mathrm{2}} \omega+{b}^{\mathrm{2}} \omega^{\mathrm{2}} \\ $$$$ \\ $$$$\frac{{x}^{\mathrm{2}} −{z}^{\mathrm{2}} +{y}+\mathrm{2}{ab}}{{x}^{\mathrm{2}} }=\mathrm{1}+\frac{−{z}^{\mathrm{2}} +{y}+\mathrm{2}{ab}}{{x}^{\mathrm{2}} } \\ $$$$=\mathrm{1}+\frac{−\left({a}^{\mathrm{2}} \omega+\mathrm{2}{ab}+{b}^{\mathrm{2}} \omega^{\mathrm{2}} \right)+\left({a}^{\mathrm{2}} \omega+{b}^{\mathrm{2}} \omega^{\mathrm{2}} \right)+\mathrm{2}{ab}}{{a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} } \\ $$$$=\mathrm{1}+\frac{−\cancel{{a}^{\mathrm{2}} \omega}−\cancel{\mathrm{2}{ab}}−\cancel{{b}^{\mathrm{2}} \omega^{\mathrm{2}} }+\cancel{{a}^{\mathrm{2}} \omega}+\cancel{{b}^{\mathrm{2}} \omega^{\mathrm{2}} }+\cancel{\mathrm{2}{ab}}}{{a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} } \\ $$$$=\mathrm{1}+\mathrm{0}=\mathrm{1} \\ $$
Commented by HeferH last updated on 03/Jan/23
thank you sir, but why we say w^2 and w are solutions of w^3 =1, it is a property of complex numbers?
$${thank}\:{you}\:{sir},\:{but}\:{why}\:{we}\:{say}\: \\ $$$$\:{w}^{\mathrm{2}} \:{and}\:{w}\:{are}\:{solutions}\:{of}\:{w}^{\mathrm{3}} =\mathrm{1},\:{it}\:{is}\:{a}\: \\ $$$$\:{property}\:{of}\:{complex}\:{numbers}?\: \\ $$$$\: \\ $$
Commented by Rasheed.Sindhi last updated on 03/Jan/23
Solving x^3 =1: x^3 −1=0 (x−1)(x^2 +x+1)=0 x=1 , x=((−1±i(√3))/2) You can verify easily that ((−1+i(√3))/2) & ((−1−i(√3))/2) are square of one another. So if any one of the two is denoted by ω then the other is equal to ω^2 .
$${Solving}\:{x}^{\mathrm{3}} =\mathrm{1}: \\ $$$${x}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{1}\:,\:{x}=\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${You}\:{can}\:{verify}\:{easily}\:{that} \\ $$$$\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\&\:\frac{−\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:{are}\:{square}\:{of} \\ $$$${one}\:{another}.\:{So}\:{if}\:{any}\:{one}\:{of}\:{the} \\ $$$${two}\:{is}\:{denoted}\:{by}\:\omega\:{then}\:{the}\:{other} \\ $$$${is}\:{equal}\:{to}\:\omega^{\mathrm{2}} . \\ $$
Commented by HeferH last updated on 03/Jan/23
Oh, I didnt realize that until now, thank you very much :)
$${Oh},\:{I}\:{didnt}\:{realize}\:{that}\:{until}\:{now},\:{thank}\: \\ $$$$\left.{you}\:{very}\:{much}\:\::\right) \\ $$
Commented by Rasheed.Sindhi last updated on 03/Jan/23
Anyway sir, your knowledge of math is more than of mine! BTW !RAEY WEN YPPAH
$${Anyway}\:\:{sir},\:{your}\:{knowledge}\:{of} \\ $$$${math}\:{is}\:{more}\:{than}\:{of}\:{mine}! \\ $$$$\mathcal{BTW}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:!\mathrm{RAEY}\:\mathrm{WEN}\:\mathrm{YPPAH} \\ $$
Commented by HeferH last updated on 03/Jan/23
Happy new year to you as well! :)
$$\left.{Happy}\:{new}\:{year}\:{to}\:{you}\:{as}\:{well}!\::\right) \\ $$

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