If A = Z 1 + Z 2 + Z 3 , B = Z 1 + Z 2 ω + Z 3 ω 2 , C = Z 1 + Z 2 ω 2 + Z 3 ω , where ω is a complex cube root of unity, then prove that |A| 2 | + |B| 2 + |C| 2 | = 3 (|Z 1 | 2 + |Z 2 | 2 + |Z 3 | 2 ).
Text Solution
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Sol. A + B + C = 3Z 1 + Z 2 (1 + ω + ω 2 ) + Z 3 (1 + ω 2 + ω ) = 3Z 1 as 1 + ω + ω 2 = 0
∴ Z 1 = 
; Similarly A + B ω 2 + C ω and A + B ω + C ω 2 give Z 2 = 
and Z 3 = 
|A| 2 = (Z 1 + Z 2 + Z 3 ) ( 
+ 
+ 
) ( |A| 2 = A 
)
= |Z 1 | 2 + |Z 2 | 2 + |Z 3 | 2 + 
(Z 2 + Z 3 ) + 
(Z 3 + Z 1 ) + 
(Z 1 + Z 2 ) … (1)
|B| 2 = B 
= (Z 1 + Z 2 ω + Z 3 ω 2 ) ( 
+ 
ω 2 + 
ω )
( 
= ω 2 , 
= ω )
= Z 1 
+ Z 2 
+ Z 3 
+ 
(Z 2 ω + Z 3 ω 2 ) + 
(Z 1 ω 2 + Z 3 ω ) + 
(Z 1 ω + Z 2 ω 2 )
= |Z 1 | 2 + |Z 2 | 2 + |Z 3 | 2 + 
(Z 2 ω + Z 3 ω 2 ) + 
(Z 1 ω 2 + 3 ω ) + 
(Z 1 ω + Z 2 ω 2 ) … (2)
Similarly,
|C| 2 = |Z 1 | 2 + |Z 2 | 2 + |Z 3 | 2 + 
(Z 2 ω 2
+ Z 3 ω )+ 
(Z 1 ω + Z 3 ω 2 ) + 
(Z 1 ω 2 + Z 2 ω ) …(3)
Adding (1), (2) and (3), |A| 2 + |B| 2 + |C| 2 = 3(|Z 1 | 2 + |Z 2 | 2 + |Z 3 | 2 ).
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