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...

- Akshay_Budhkar

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- Akshay_Budhkar

question from iit 2011

- anonymous

ohkay

- Akshay_Budhkar

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

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## More answers

- anonymous

What's iit?

- anonymous

yeah it is

- Akshay_Budhkar

IIT-indian institute of technology

- anonymous

it is the best institute in india for engineering

- anonymous

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

- anonymous

Not really an engineering question:-)

- 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

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

- anonymous

jee to get into iit

- Akshay_Budhkar

jee-joint entrance exam

- anonymous

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

- anonymous

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

- 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

well 3a = x+y+z

- Akshay_Budhkar

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

- Akshay_Budhkar

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

- Akshay_Budhkar

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

- anonymous

me too but wait a minute

- anonymous

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

- anonymous

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

- 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

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

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

- anonymous

not including (a+b+c)

- Akshay_Budhkar

-ac dude.. all signs were same no?

- anonymous

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

- Akshay_Budhkar

ya right.. that was d formula..

- anonymous

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

- 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

your right

- anonymous

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

- anonymous

x + y +z = 3a

- anonymous

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

- anonymous

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

- anonymous

yeah

- Akshay_Budhkar

right

- anonymous

so ^2 make minus sign vanish

- anonymous

it implies x^2 = |x|^2

- Akshay_Budhkar

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

- anonymous

-ve sign don't exist in squares

- anonymous

letme solve a lil

- anonymous

:D

- 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

@face give it a try

- anonymous

damn iit never let someone enter easily

- Akshay_Budhkar

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

- anonymous

well what was the options

- Akshay_Budhkar

*always

- anonymous

lol

- Akshay_Budhkar

its an integer type ques 0 to 9 all options

- anonymous

aahhhh scary

- Akshay_Budhkar

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

- anonymous

Ok , Here I go

- Akshay_Budhkar

ya go go bye. lol js kidding

- 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

Beauty

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