Akshay_Budhkar
  • 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...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Akshay_Budhkar
  • Akshay_Budhkar
question from iit 2011
anonymous
  • anonymous
ohkay
Akshay_Budhkar
  • Akshay_Budhkar
w is cube root of unity i believe.. the complex cube root of unity

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

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