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Akshay_Budhkar

  • 3 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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  1. Akshay_Budhkar
    • 3 years ago
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    question from iit 2011

  2. Ishaan94
    • 3 years ago
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    ohkay

  3. Akshay_Budhkar
    • 3 years ago
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    w is cube root of unity i believe.. the complex cube root of unity

  4. estudier
    • 3 years ago
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    What's iit?

  5. Ishaan94
    • 3 years ago
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    yeah it is

  6. Akshay_Budhkar
    • 3 years ago
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    IIT-indian institute of technology

  7. Ishaan94
    • 3 years ago
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    it is the best institute in india for engineering

  8. estudier
    • 3 years ago
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    Ok, thks, I didn't know...

  9. estudier
    • 3 years ago
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    Not really an engineering question:-)

  10. Ishaan94
    • 3 years ago
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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

  11. Akshay_Budhkar
    • 3 years ago
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    i had solved the ques using a short cut to save time

  12. Ishaan94
    • 3 years ago
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    jee to get into iit

  13. Akshay_Budhkar
    • 3 years ago
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    jee-joint entrance exam

  14. estudier
    • 3 years ago
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    Mind u, maybe u have to be an engineer to solve it..

  15. Ishaan94
    • 3 years ago
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    oh well let me try it well i get confused jee e for entrance or engineering lol

  16. Akshay_Budhkar
    • 3 years ago
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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

  17. Ishaan94
    • 3 years ago
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    well 3a = x+y+z

  18. Akshay_Budhkar
    • 3 years ago
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    that did get me 4 marks in the paper..substitution

  19. Akshay_Budhkar
    • 3 years ago
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    right add all the three. 1+w+wsquare is 0

  20. Akshay_Budhkar
    • 3 years ago
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    any suggestions estudier??? ishaan i cant find a way ahead of it..

  21. Ishaan94
    • 3 years ago
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    me too but wait a minute

  22. estudier
    • 3 years ago
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    I don't like this kind of problems..:-

  23. Ishaan94
    • 3 years ago
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    well |3a| = |x + y+z|

  24. Akshay_Budhkar
    • 3 years ago
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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

  25. Ishaan94
    • 3 years ago
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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

  26. Ishaan94
    • 3 years ago
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    a^2 + b^2 +c^2 -ab +ac -cb

  27. Ishaan94
    • 3 years ago
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    not including (a+b+c)

  28. Akshay_Budhkar
    • 3 years ago
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    -ac dude.. all signs were same no?

  29. Ishaan94
    • 3 years ago
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    well including a+b+c it becomes x^3 + y^3 +z^3 -3xyz x=a y=b z=c

  30. Akshay_Budhkar
    • 3 years ago
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    ya right.. that was d formula..

  31. Ishaan94
    • 3 years ago
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    lets do it x^2 + y^2 +z^2 you find y while i find x^2 and z^2

  32. Akshay_Budhkar
    • 3 years ago
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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

  33. Ishaan94
    • 3 years ago
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    your right

  34. Ishaan94
    • 3 years ago
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    hey (x + y+z) ^2 = x2 +y2 + z2 +2xy + 2xz+2yz

  35. Ishaan94
    • 3 years ago
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    x + y +z = 3a

  36. estudier
    • 3 years ago
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    hey (x + y+z) ^2 = x2 +y2 + z2 +2xy + 2xz+2yz I saw that in your puzzle from yesterday...

  37. Ishaan94
    • 3 years ago
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    9a^2 = x^2 + y^2 +z^2 +2xy + 2zy +2xz

  38. Ishaan94
    • 3 years ago
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    yeah

  39. Akshay_Budhkar
    • 3 years ago
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    right

  40. Ishaan94
    • 3 years ago
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    so ^2 make minus sign vanish

  41. Ishaan94
    • 3 years ago
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    it implies x^2 = |x|^2

  42. Akshay_Budhkar
    • 3 years ago
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    ofcourse.. its given just to confuse mod is not needed

  43. Ishaan94
    • 3 years ago
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    -ve sign don't exist in squares

  44. Ishaan94
    • 3 years ago
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    letme solve a lil

  45. Ishaan94
    • 3 years ago
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    :D

  46. Akshay_Budhkar
    • 3 years ago
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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

  47. Akshay_Budhkar
    • 3 years ago
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    @face give it a try

  48. Ishaan94
    • 3 years ago
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    damn iit never let someone enter easily

  49. Akshay_Budhkar
    • 3 years ago
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    but the substitution method is alays there!!!!! lol!! the only weak point of iit jee.. lol!!!

  50. Ishaan94
    • 3 years ago
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    well what was the options

  51. Akshay_Budhkar
    • 3 years ago
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    *always

  52. Ishaan94
    • 3 years ago
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    lol

  53. Akshay_Budhkar
    • 3 years ago
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    its an integer type ques 0 to 9 all options

  54. Ishaan94
    • 3 years ago
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    aahhhh scary

  55. Akshay_Budhkar
    • 3 years ago
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    generally we mark 2 if we dun kno... 40 percent of ques hav ans 2 lol!!!

  56. Ishaan94
    • 3 years ago
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    Ok , Here I go

  57. Akshay_Budhkar
    • 3 years ago
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    ya go go bye. lol js kidding

  58. Ishaan94
    • 3 years ago
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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 : )

  59. Akshay_Budhkar
    • 3 years ago
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    Beauty

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