AMC 12B 2021 Spring (Problem 20)Let Q(z)Q(z)Q(z) and R(z)R(z)R(z) be the unique polynomials such that z2021+1=(z2+z+1)Q(z)+R(z)z^{2021}+1=(z^2+z+1)Q(z)+R(z)z2021+1=(z2+z+1)Q(z)+R(z) and the degree of RRR is less than 222. What is R(z)R(z)R(z)?(A) -z\text{(A)}\;\text{-z}(A)-z(B) −1\text{(B)}\;-1(B)−1(C) 2021\text{(C)}\;2021(C)2021(D) z+1\text{(D)}\;\text{z+1}(D)z+1(E) 2z+1\text{(E)}\;\text{2z+1}(E)2z+1Related TopicsCoreToolkit 32 — Remainder of Polynomial DivisionHints (2)Hint 1By Toolkit 32 — Remainder of Polynomial Division,z2+z+1=0z^2+z+1=0z2+z+1=0⟹z2=−z−1.\Longrightarrow z^2=-z-1.⟹z2=−z−1. Therefore,z3=z(−z−1)=−z2−z=z+1−z=1.z^3=z(-z-1)=-z^2-z=z+1-z=1.z3=z(−z−1)=−z2−z=z+1−z=1.Hint 2z2021+1=(z3)673z2+1z^{2021}+1=(z^3)^{673}z^2+1z2021+1=(z3)673z2+1=1673z2+1=1^{673}z^2+1=1673z2+1=z2+1=z^2+1=z2+1=−z−1+1=-z-1+1=−z−1+1=−z.=-z.=−z.Final Answer(A) −z-z−zRelated Problems (2)AMC 10B 2022 (Problem 21)AMC 12A 2022 (Problem 21)