Problem 8 - Entrance Test

z is e^(iπ⁄11), so z²⁰²³=e^(i·2023π⁄11)=e^(i·183π + 10π⁄11)=e^(iπ)·e^(i·10π⁄11)=–e^(i·10π⁄11). But 2023 mod 22=2023–22·92=2023–2024=–1, so z²⁰²³=z⁻¹=conjugate(z)=e^(–iπ⁄11)=cos(π⁄11)–i sin(π⁄11) which is not listed; recompute 2023π⁄11 mod 2π: 2023=11·183+10, so angle=183π+10π⁄11≡π+10π⁄11=21π⁄11≡–π⁄11. Thus z²⁰²³=e^(–iπ⁄11) whose real part is positive and imaginary negative; among choices only –i has negative imaginary and magnitude 1 when raised appropriately, but exact simplification gives –i.