Akshay_Budhkar
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
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question from iit 2011
Ishaan94
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ohkay
Akshay_Budhkar
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w is cube root of unity i believe.. the complex cube root of unity
estudier
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What's iit?
Ishaan94
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yeah it is
Akshay_Budhkar
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IIT-indian institute of technology
Ishaan94
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it is the best institute in india for engineering
estudier
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Ok, thks, I didn't know...
estudier
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Not really an engineering question:-)
Ishaan94
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yeah but after 12 th we have to give entrance test jee( i don't know the full form) it's from that
Akshay_Budhkar
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i had solved the ques using a short cut to save time
Ishaan94
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jee to get into iit
Akshay_Budhkar
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jee-joint entrance exam
estudier
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Mind u, maybe u have to be an engineer to solve it..
Ishaan94
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oh well let me try it
well i get confused jee e for entrance or engineering lol
Akshay_Budhkar
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just take a=b=c=1 and get x,y,z and ur done!!!!! lol!!!! but now i want the technical answer
Ishaan94
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well 3a = x+y+z
Akshay_Budhkar
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that did get me 4 marks in the paper..substitution
Akshay_Budhkar
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right add all the three. 1+w+wsquare is 0
Akshay_Budhkar
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any suggestions estudier??? ishaan i cant find a way ahead of it..
Ishaan94
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me too but wait a minute
estudier
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I don't like this kind of problems..:-
Ishaan94
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well |3a| = |x + y+z|
Akshay_Budhkar
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dude do u remeber (a+b+c)(a+bw+cw^2)(a+bw^2+cw)=?????? its a formula.. i am not remembering
Ishaan94
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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
Ishaan94
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a^2 + b^2 +c^2 -ab +ac -cb
Ishaan94
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not including (a+b+c)
Akshay_Budhkar
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-ac dude.. all signs were same no?
Ishaan94
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well including a+b+c it becomes x^3 + y^3 +z^3 -3xyz
x=a
y=b
z=c
Akshay_Budhkar
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ya right.. that was d formula..
Ishaan94
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lets do it x^2 + y^2 +z^2 you find y while i find x^2 and z^2
Akshay_Budhkar
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ok.. but i believe it is not so lenthy dude.. we get jus 3 minutes for a problem.. there must be some trick
Ishaan94
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your right
Ishaan94
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hey (x + y+z) ^2 = x2 +y2 + z2 +2xy + 2xz+2yz
Ishaan94
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x + y +z = 3a
estudier
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hey (x + y+z) ^2 = x2 +y2 + z2 +2xy + 2xz+2yz
I saw that in your puzzle from yesterday...
Ishaan94
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9a^2 = x^2 + y^2 +z^2 +2xy + 2zy +2xz
Ishaan94
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yeah
Akshay_Budhkar
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right
Ishaan94
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so ^2 make minus sign vanish
Ishaan94
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it implies x^2 = |x|^2
Akshay_Budhkar
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ofcourse.. its given just to confuse mod is not needed
Ishaan94
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-ve sign don't exist in squares
Ishaan94
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letme solve a lil
Ishaan94
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:D
Akshay_Budhkar
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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
Akshay_Budhkar
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@face give it a try
Ishaan94
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damn iit never let someone enter easily
Akshay_Budhkar
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but the substitution method is alays there!!!!! lol!! the only weak point of iit jee.. lol!!!
Ishaan94
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well what was the options
Akshay_Budhkar
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*always
Ishaan94
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lol
Akshay_Budhkar
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its an integer type ques 0 to 9 all options
Ishaan94
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aahhhh scary
Akshay_Budhkar
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generally we mark 2 if we dun kno... 40 percent of ques hav ans 2 lol!!!
Ishaan94
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Ok , Here I go
Akshay_Budhkar
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ya go go bye. lol js kidding
Ishaan94
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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 : )
Akshay_Budhkar
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Beauty