24Is this all ???
I mean if we assume lzl=1 and take z=1
=>1a+1b+1c+1
Also we know that ax >0 for all x
=>1a+1b+1c+1 ≠0 for any vlaue of a,b,c
'z' is a complex number satisfying
za+zb+zc+1=0.
Given a,b,c are distinct integers.....
Proe |z|=1.
From Shastra
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10 Answers
24
62no eureka.. you have misunderstood the question...
The questoin is not to prove that for all z such that |z|=1 the equality holds!!
1I am not sure whether I am correct or not , but let me try Put z = r (cos 8 + i sin 8) so, Z a = r a (cos 8 + i a sin 8) Similarily,Z b = r b (cos 8 + i b sin 8) ,Z c = r c (cos 8 + i c sin 8) So,Za + Zb + Zc+1= cos 8 ( r a + r b + r c ) + i sin 8 (a+b+c) ( r a + r b + r c ) = (r a + r b + r c ) [cos 8 + i sin 8 (a+b+c)]
1oops yeha mistake
But otherwise is my starting correct Nishant bhayiya??
341looks like some mistake, as 1+z2+z4+z5=0 will have a real root which is neither 1 or -1. So the problem statement is not true.
24hmm yes sir..u r rite.......
my method to prove question wrong failed :P ..but urs wont [6]
11That's what i too thought apparantly, but unlike u, I cud not prove it to be wrong!
What Shastra gave was a,b,c are just integers, which was erred as well, as a=b=c failed too!