Match the expressions in List-I with their simplified values in List-II, where ω is a non-real cube root of unity. List - I List - II (P) ( 1 + ω ) 2023 + ( 1 + ω 2 ) 2023 (1) 1 (Q) If z 1 = 1 + ω −

Question

Match the expressions in List-I with their simplified values in List-II, where $ω$ is a non-real cube root of unity.

List - I	List - II
(P) $(1 + ω)^{2023} + (1 + ω^{2})^{2023}$	(1) 1
(Q) If $z_{1} = 1 + ω - ω^{2}$ and $z_{2} = 1 - ω + ω^{2}$ , then $z_{1} z_{2}$	(2) 3
(R) $(1 - ω) (1 - ω^{2}) (1 - ω^{4}) (1 - ω^{5})$	(3) 4
(S) $(1 + 2 ω + 3 ω^{2}) (1 + 3 ω + 2 ω^{2})$	(4) 9

TestHub · Accepted Answer

For (P): We use 1 + ω = − ω 2 and 1 + ω 2 = − ω . The expression becomes ( − ω 2 ) 2023 + ( − ω ) 2023 = − ω 4046 − ω 2023 . Since 4046 = 3 × 1348 + 2 and 2023 = 3 × 674 + 1 , we have ω 4046 = ω 2 and ω 2023 = ω . So, − ( ω 2 ) − ( ω ) = − ( ω 2 + ω ) = − ( − 1 ) = 1 . For (Q): Using 1 + ω + ω 2 = 0 , we have 1 + ω = − ω 2 and 1 + ω 2 = − ω . So, z 1 ​ = ( 1 + ω ) − ω 2 = − ω 2 − ω 2 = − 2 ω 2 . And z 2 ​ = ( 1 + ω 2 ) − ω = − ω − ω = − 2 ω . Thus, z 1 ​ z 2 ​ = ( − 2 ω 2 ) ( − 2 ω ) = 4 ω 3 = 4 ( 1 ) = 4 . For (R): We know ( 1 − ω ) ( 1 − ω 2 ) = 1 − ( ω + ω 2 ) + ω 3 = 1 − ( − 1 ) + 1 = 3 . Also, ω 4 = ω 3 ⋅ ω = ω and ω 5 = ω 3 ⋅ ω 2 = ω 2 . So, the expression becomes ( 1 − ω ) ( 1 − ω 2 ) ( 1 − ω ) ( 1 − ω 2 ) = 3 × 3 = 9 . For (S): We rewrite 1 + 2 ω + 3 ω 2 = 1 + 2 ω + 3 ( − 1 − ω ) = 1 + 2 ω − 3 − 3 ω = − 2 − ω . Similarly, 1 + 3 ω + 2 ω 2 = 1 + 3 ω + 2 ( − 1 − ω ) = 1 + 3 ω − 2 − 2 ω = − 1 + ω . The product is ( − 2 − \omega ) ( − 1 + \omega ) = 2 − 2 \omega + \omega − \omega 2 = 2 − \omega − \omega 2 = 2 − ( \omega + \omega$^2) = 2-(-1) = 3.

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