Exercise 1.2.2

Prove (1.14) and (1.15)

Answers

Proof. We know by 1.2(B) that

z 1 − z 2 = − i 3 ( x 2 − x 3 ) .

By (1.10) :

z 1 − ω z 2 = 1 3 [ ( x 1 + ω 2 x 2 + ω x 3 ) − ( x 3 + ω 2 x 2 + ω x 1 ) ] = 1 3 ( 1 − ω ) ( x 1 − x 3 ) .

and

1 3 ( 1 − ω ) = 1 3 ( 1 − e 2 iπ ∕ 3 ) = 1 3 e iπ ∕ 3 ( e − iπ ∕ 3 − e iπ ∕ 3 ) = 1 3 ω 2 ( e iπ ∕ 3 − e − iπ ∕ 3 ) = 2 3 i sin ⁡ ( π ∕ 3 ) ω 2 = i 3 ω 2 .

So

z 1 − ω z 2 = i 3 ω 2 ( x 1 − x 3 ) .

Similarly

z 1 − ω 2 z 2 = 1 3 [ ( x 1 + ω 2 x 2 + ω x 3 ) − ( x 2 + ω 2 x 1 + ω x 3 ) ] = 1 3 ( 1 − ω 2 ) ( x 1 − x 2 ) ,

and 1 3 ( 1 − ω 2 ) = 1 3 ( 1 − ω ) ¯ = i 3 ω 2 ¯ = − i 3 ω .

Thus

z 1 − ω 2 z 2 = − i 3 ω ( x 1 − x 2 ) .

Using these three results,

D = z 1 3 − z 2 3 = ( z 1 − z 2 ) ( z 1 − ω z 2 ) ( z 1 − ω 2 z 2 ) = − i 3 i 3 ω 2 ( − i 3 ω ) ( x 1 − x 2 ) ( x 1 − x 3 ) ( x 2 − x 3 ) = − i 3 3 ( x 1 − x 2 ) ( x 1 − x 3 ) ( x 2 − x 3 ) ,

which is the formula (1.15). □

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2022-07-19 00:00
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