2005.......complex

cauchy scwrtz 'equality' holds wen a=b=c, 1=w=w²...

which is not possible!

Thank you Sri

that gets me to understand the soln[1][1]

But still..........What does

"since minimum diffrnce betwn two terms is 1" mean and where does it come from in wats in name's post[7][7]

\left| a+b\omega +c\omega^{2}\right| = √(a-b/2-c/2)² + 3(c-b)²/4=√((a-b)² + (b-c)² +( c-a)²)/2

the expression is min. when 2 of the a,b,c are equal

let a=b and b-c=±1 & c-a=±1

so the answer is 1

let the iven exprsn be = Z

i.e \left|Z \right| = x

thus x^{2}= Z\bar{Z}

wihch reduces to the above one becos the conjugate ofa+b\omega +c\omega ^{2} is a+b\omega ^{2}+c\omega

so this solves it na ???

nameless are U kidding or is that a serious answer

Can U explain it if it actually is ......[1]

x=mod of that crap|:P

x^2=square of dat crap :P

so nw
itz

x^2=1/2[(a-b)^2+(b-c)^2+(c-a)^2]
nw a b c r nt equal simulatneously
so d value of dat xperssion is > or = (1+1+0)

so itz > or =1

since minimum diffrnce betwn two terms is 1

so answer is 1
:P

anyone has a solution ..

so what to do ..

ok sky go on !!

here though a,b,c are variables but 1,w,w² are not...

and also given that a≠b≠c ...

we are lucky enuf to get 0 on LHS by properties of complex no..

but that doesnt mean we can use equality here..

i dnt noe this thm had done it longggggggggggggggggggg back..........

bt iit mein ni ahi so ditch mara :P

:)

ankit......can we use cauchy schwarz for complex nos.????[7][7]

can u point out the error wats i n ur name ?

eureka aren't u intrested !!

answer is 1 had done this sum few days back :P :P

yaar then how come i got a contradiction !

chauchy - shtwartzur ytu .. some inequality with this kinda name

(a² + b² + c² )(1² + ω² + (ω²)²) ≥ (a + bω + cω²)²

= 0 ≥ (a + bω + cω²)²

but |a + bω + cω²| not -ve so it is = 0

ans is 1.............plz give soln...

i got 0 ..

asnwer is...1...........

memory test :P

ans kya koi no. hai ya a,b,c, mein hai ?

okzzzz

i am trying dude