Given the equation:
(‌)k+(‌)l=2where
ω is the complex cube root of unity, meaning
ω3=1 and
1+ω+ω2=0.
To solve this, we first manipulate the terms by using properties of complex numbers. Consider multiplying and dividing:
Multiply and divide the first term by
ω2 :
(‌| ω2(a+bω+cω2) |
| ω2(c+aω+bω2) |
)kMultiply and divide the second term by
ω :
(‌| ω(a+bω+cω2) |
| ω(b+aω2+cω) |
)lThis simplifies to:
(ω2k)+(ωl)=2For these powers to sum to 2 , and knowing the properties of cube roots of unity, we have specific implications for the exponents:
Since
ωn can only take values
1,ω, or
ω2, for the sum
ω2k+ωl=2 to be a real number (i.e.,
2), both
ω2k and
ωl must equal 1 .
Thus, this requires both 2 k and I to be multiples of
3(. since
ω3=1).
Hence, the expression
2k+l is also a multiple of 3 , indicating that
2k+l is always divisible by 3.