Complex-Olympiad-stuff....

Is this all ???

I mean if we assume lzl=1 and take z=1

=>1^a+1^b+1^c+1
Also we know that a^x >0 for all x

=>1^a+1^b+1^c+1 ≠0 for any vlaue of a,b,c

I 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,Z^a + Z^b + Z^c+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)]

oops yeha mistake
But otherwise is my starting correct Nishant bhayiya??

okk sir,,,,,
thax for correcting me

looks 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.

hmm yes sir..u r rite.......

my method to prove question wrong failed :P ..but urs wont [6]

That'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!