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Akshay_Budhkar
 4 years ago
let w=e^{(i*pie)/3} and a,b,c,x,y,z be non zero complex numbers such that
a+b+c=x
a+bw+cw^2=y
a+bw^2+cw=z
thn the value of ([x]^2+[y]^2+[z]^2)/([a]^2+[b]^2+[c]^2) is???
where [q] is mod q...
Akshay_Budhkar
 4 years ago
let w=e^{(i*pie)/3} and a,b,c,x,y,z be non zero complex numbers such that a+b+c=x a+bw+cw^2=y a+bw^2+cw=z thn the value of ([x]^2+[y]^2+[z]^2)/([a]^2+[b]^2+[c]^2) is??? where [q] is mod q...

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Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1question from iit 2011

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1w is cube root of unity i believe.. the complex cube root of unity

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1IITindian institute of technology

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is the best institute in india for engineering

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, thks, I didn't know...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Not really an engineering question:)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah but after 12 th we have to give entrance test jee( i don't know the full form) it's from that

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1i had solved the ques using a short cut to save time

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1jeejoint entrance exam

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Mind u, maybe u have to be an engineer to solve it..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh well let me try it well i get confused jee e for entrance or engineering lol

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1just take a=b=c=1 and get x,y,z and ur done!!!!! lol!!!! but now i want the technical answer

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1that did get me 4 marks in the paper..substitution

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1right add all the three. 1+w+wsquare is 0

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1any suggestions estudier??? ishaan i cant find a way ahead of it..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0me too but wait a minute

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't like this kind of problems..:

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well 3a = x + y+z

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1dude do u remeber (a+b+c)(a+bw+cw^2)(a+bw^2+cw)=?????? its a formula.. i am not remembering

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but x^2 makes the possible minus vanish and hence x^2 + y^2+x^2 = x^2 + y^2 + z^2 \ just wait like 20 sec

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0a^2 + b^2 +c^2 ab +ac cb

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not including (a+b+c)

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1ac dude.. all signs were same no?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well including a+b+c it becomes x^3 + y^3 +z^3 3xyz x=a y=b z=c

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1ya right.. that was d formula..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lets do it x^2 + y^2 +z^2 you find y while i find x^2 and z^2

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1ok.. but i believe it is not so lenthy dude.. we get jus 3 minutes for a problem.. there must be some trick

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hey (x + y+z) ^2 = x2 +y2 + z2 +2xy + 2xz+2yz

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hey (x + y+z) ^2 = x2 +y2 + z2 +2xy + 2xz+2yz I saw that in your puzzle from yesterday...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.09a^2 = x^2 + y^2 +z^2 +2xy + 2zy +2xz

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so ^2 make minus sign vanish

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it implies x^2 = x^2

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1ofcourse.. its given just to confuse mod is not needed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ve sign don't exist in squares

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1i believe iit didnt want us to use the substitution method.... coz i have not found a way out till now after 3 months i gae the exam

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1@face give it a try

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0damn iit never let someone enter easily

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1but the substitution method is alays there!!!!! lol!! the only weak point of iit jee.. lol!!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well what was the options

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1its an integer type ques 0 to 9 all options

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1generally we mark 2 if we dun kno... 40 percent of ques hav ans 2 lol!!!

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1ya go go bye. lol js kidding

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay Here we Have \[x = a+b+c\] \[y =a+bw+cw^2\] \[z = a + bw^2+cw\] Now we have to find \[\large{\frac{x^2 + y^2 + z^2 }{a^2 + b^2 +c^2}}\] \[z^2 = z*(conj.z)\] \[conj.y = a+bw^2 +cw\] \[conj.z = a+bw+cw^2\] Okay So we have now \[\frac{(a+b+c)^2 + 2(a + bw+cw^2)(a+bw^2+cw)}{a^2+b^2+c^2}\] \[\frac{3(a^2+b^2+c^2) +2(ab+bc+ac)  2(ab+bc+ac)}{a^2+b^2+c^2}\]\[3\] Beautiful Solution : )
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