Question

A b c are integers not all simultaneously equal and w is cube root of unity

Answer 1

1 + ω + ω² = 0 
=> l a + bω + cω² l 
= l a + bω + c (-1 - ω) l 
= l (a -c) + (b -c)ω l 
Putting ω = -1/2 ± (√3/2) i 
= l (a - c) - (1/2)(b - c) ± (√3/2)(b - c) i l 
= l (a - b/2 - c/2) ± (√3/2)(b - c) i l 
= √ [ a² + b²/4 + c²/4 + (3/4)(b² - 2bc + c²) ] 
= √ [ a² + b² + c² - ab - bc - ca ] 
= √ (1/2) [ (a -b)² + (b -c)² + (c - a)² ] 
a, b, c are not all equal, but any two can be equal 
=> if a = b (a - b)² = 0 and (b - c) = (c - a) = 1 
=> √ (1/2) [ (a -b)² + (b -c)² + (c - a)² ] 
= √ (1/2) [ 0 + 1 + 1 ] 
= 1 
=> minimum value of l a + bω + cω² l 
= 1. 

Answer 2

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