Foundations, part 6
( − i + 3 i + 3 ) 2 0 2 0 + ( i + 3 i − 3 ) 2 0 1 7 + 1 = ?
Clarification
:
i
=
−
1
denotes the
imaginary unit
.
The answer is 0.
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2 solutions
How could you simplify ( ω 1 ) 2 0 2 0 + ω 2 0 1 7 + 1 into ω 2 + ω + 1 ?
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ω 2 0 2 0 1 = ω 1 = ω 3 ω 2 = ω 2
The first equality follows as ω 3 n + 1 = ω .
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I still dont get it.. Can you help me?
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@Fidel Simanjuntak – ω and ω 2 are both the complex cube roots of unity. Now, if z is a cube root of unity, then it must satisfy z 3 = 1 . It is not hard to see that further, we can write z 3 n = 1 where n ∈ N . Now, based on the property above, we see that ω 3 n + 1 = ω 3 n ⋅ ω = ω and ω 3 n + 2 = ω 3 n ⋅ ω 2 = ω 2 . Thus, we can write
ω 2 0 2 0 1 = ω 3 × 6 7 3 + 1 1 = ω 1
Further, we have
ω 1 = ω 1 ⋅ ω 2 ω 2 = ω 3 ω 2 = ω 2
Also ω 2 0 1 7 = ω 3 × 6 7 2 + 1 = ω
x = ( − i + 3 i + 3 ) 2 0 2 0 + ( i + 3 i − 3 ) 2 0 1 7 + 1 = ⎝ ⎛ 2 ( 2 3 − 2 1 i ) 2 ( 2 3 + 2 1 i ) ⎠ ⎞ 2 0 2 0 + ⎝ ⎛ 2 ( 2 3 + 2 1 i ) 2 ( − 2 3 + 2 1 i ) ⎠ ⎞ 2 0 1 7 + 1 = ( e − 6 π i e 6 π i ) 2 0 2 0 + ( e 6 π i e 6 5 π i ) 2 0 1 7 + 1 = ( e 3 π i ) 2 0 2 0 + ( e 3 2 π i ) 2 0 1 7 + 1 = e 6 7 3 3 π i + e 1 3 4 4 3 2 π i + 1 = e 3 4 π i + e 3 2 π i + 1 = − 2 1 − 2 3 i − 2 1 + 2 3 i + 1 = 0 By Euler’s formula: e θ i = cos θ + i sin θ
The roots of the quadratic equation x 2 + x + 1 = 0 are given by
ω = 2 − 1 + 3 i ω 2 = 2 − 1 − 3 i
Now, we can see that
⎩ ⎪ ⎨ ⎪ ⎧ i + 3 − i + 3 i − 3 = − 2 i ω = 2 i ω 2 = − 2 i ω 2
Thus
( − i + 3 i + 3 ) 2 0 2 0 + ( i + 3 i − 3 ) 2 0 1 7 + 1 = ( 2 i ω 2 − 2 i ω ) 2 0 2 0 + ( − 2 i ω − 2 i ω 2 ) 2 0 1 7 + 1 = ( ω − 1 ) 2 0 2 0 + ( ω ) 2 0 1 7 + 1 = ω 2 0 2 0 1 + ω 2 0 1 7 + 1 = ω 2 + ω + 1 = 0