Question

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...

Answer 1

question from iit 2011

Answer 2

ohkay

Answer 3

w is cube root of unity i believe.. the complex cube root of unity

Answer 4

What's iit?

Answer 5

yeah it is

Answer 6

IIT-indian institute of technology

Answer 7

it is the best institute in india for engineering

Answer 8

Ok, thks, I didn't know...

Answer 9

Not really an engineering question:-)

Answer 10

yeah but after 12 th we have to give entrance test jee( i don't know the full form) it's from that

Answer 11

i had solved the ques using a short cut to save time

Answer 12

jee to get into iit

Answer 13

jee-joint entrance exam

Answer 14

Mind u, maybe u have to be an engineer to solve it..

Answer 15

oh well let me try it
well i get confused jee e for entrance or engineering lol

Answer 16

just take a=b=c=1 and get x,y,z and ur done!!!!! lol!!!! but now i want the technical answer

Answer 17

well 3a = x+y+z

Answer 18

that did get me 4 marks in the paper..substitution

Answer 19

right add all the three. 1+w+wsquare is 0

Answer 20

any suggestions estudier??? ishaan i cant find a way ahead of it..

Answer 21

me too but wait a minute

Answer 22

I don't like this kind of problems..:-

Answer 23

well |3a| = |x + y+z|

Answer 24

dude do u remeber (a+b+c)(a+bw+cw^2)(a+bw^2+cw)=?????? its a formula.. i am not remembering

Answer 25

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

Answer 26

a^2 + b^2 +c^2 -ab +ac -cb

Answer 27

not including (a+b+c)

Answer 28

-ac dude.. all signs were same no?

Answer 29

well including a+b+c it becomes x^3 + y^3 +z^3 -3xyz
x=a
y=b
z=c

Answer 30

ya right.. that was d formula..

Answer 31

lets do it x^2 + y^2 +z^2 you find y while i find x^2 and z^2

Answer 32

ok.. but i believe it is not so lenthy dude.. we get jus 3 minutes for a problem.. there must be some trick

Answer 33

your right

Answer 34

hey (x + y+z) ^2 = x2 +y2 + z2 +2xy + 2xz+2yz

Answer 35

x + y +z = 3a

Answer 36

hey (x + y+z) ^2 = x2 +y2 + z2 +2xy + 2xz+2yz

I saw that in your puzzle from yesterday...

Answer 37

9a^2 = x^2 + y^2 +z^2 +2xy + 2zy +2xz

Answer 38

yeah

Answer 39

right

Answer 40

so ^2 make minus sign vanish

Answer 41

it implies x^2 = |x|^2

Answer 42

ofcourse.. its given just to confuse mod is not needed

Answer 43

-ve sign don't exist in squares

Answer 44

letme solve a lil

Answer 45

:D

Answer 46

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

Answer 47

@face give it a try

Answer 48

damn iit never let someone enter easily

Answer 49

but the substitution method is alays there!!!!! lol!! the only weak point of iit jee.. lol!!!

Answer 50

well what was the options

Answer 51

*always

Answer 52

lol

Answer 53

its an integer type ques 0 to 9 all options

Answer 54

aahhhh scary

Answer 55

generally we mark 2 if we dun kno... 40 percent of ques hav ans 2 lol!!!

Answer 56

Ok , Here I go

Answer 57

ya go go bye. lol js kidding

Answer 58

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 : )

Answer 59

Beauty