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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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question from iit 2011
w is cube root of unity i believe.. the complex cube root of unity

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Other answers:

What's iit?
yeah it is
IIT-indian institute of technology
it is the best institute in india for engineering
Ok, thks, I didn't know...
Not really an engineering question:-)
yeah but after 12 th we have to give entrance test jee( i don't know the full form) it's from that
i had solved the ques using a short cut to save time
jee to get into iit
jee-joint entrance exam
Mind u, maybe u have to be an engineer to solve it..
oh well let me try it well i get confused jee e for entrance or engineering lol
just take a=b=c=1 and get x,y,z and ur done!!!!! lol!!!! but now i want the technical answer
well 3a = x+y+z
that did get me 4 marks in the paper..substitution
right add all the three. 1+w+wsquare is 0
any suggestions estudier??? ishaan i cant find a way ahead of it..
me too but wait a minute
I don't like this kind of problems..:-
well |3a| = |x + y+z|
dude do u remeber (a+b+c)(a+bw+cw^2)(a+bw^2+cw)=?????? its a formula.. i am not remembering
but 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
a^2 + b^2 +c^2 -ab +ac -cb
not including (a+b+c)
-ac dude.. all signs were same no?
well including a+b+c it becomes x^3 + y^3 +z^3 -3xyz x=a y=b z=c
ya right.. that was d formula..
lets do it x^2 + y^2 +z^2 you find y while i find x^2 and z^2
ok.. but i believe it is not so lenthy dude.. we get jus 3 minutes for a problem.. there must be some trick
your right
hey (x + y+z) ^2 = x2 +y2 + z2 +2xy + 2xz+2yz
x + y +z = 3a
hey (x + y+z) ^2 = x2 +y2 + z2 +2xy + 2xz+2yz I saw that in your puzzle from yesterday...
9a^2 = x^2 + y^2 +z^2 +2xy + 2zy +2xz
so ^2 make minus sign vanish
it implies x^2 = |x|^2
ofcourse.. its given just to confuse mod is not needed
-ve sign don't exist in squares
letme solve a lil
i 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
@face give it a try
damn iit never let someone enter easily
but the substitution method is alays there!!!!! lol!! the only weak point of iit jee.. lol!!!
well what was the options
its an integer type ques 0 to 9 all options
aahhhh scary
generally we mark 2 if we dun kno... 40 percent of ques hav ans 2 lol!!!
Ok , Here I go
ya go go bye. lol js kidding
Okay 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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