TS EAMCET 10 May 2024 Shift 2 Solved Paper

Given the equation:
(‌

a+bω+cω2

c+aω+bω2

)k+(‌

a+bω+cω2

b+aω2+cω

)l=2
where 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)

)k
Multiply and divide the second term by ω :
(‌

ω(a+bω+cω2)

ω(b+aω2+cω)

)l
This simplifies to:
(ω2k)+(ωl)=2
For 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.