Zadanie 8a, Paper 2, maj 2024, TZ2, Matematyka AA IB HL

(i) The real part is 2sin²θ>0. The imaginary-to-real ratio is $2\sin\theta\cos\theta/(2\sin^2\theta)=\cot\theta=\tan(\pi/2-\theta)$. Since π/2−θ lies in (−π/2,π/2), the right-half-plane argument is exactly π/2−θ. (ii) $|z|^2=(1-\cos2\theta)^2+\sin^2(2\theta)=2-2\cos2\theta=4\sin^2\theta$. As sinθ>0, |z|=2sinθ. Alternatively factor $z=2\sin\theta(\sin\theta+i\cos\theta)$ and identify polar form.

(i) A1: use imaginary/real ratio with the correct half-plane. M1: apply double-angle identities. A1: establish the stated argument with the range justification. (ii) M1: modulus from Cartesian parts. M1: expand and use sin²+cos²=1. A1: 2−2cos2θ under the square root. A1: 2sinθ using its positive sign. A full polar-factorization proof can earn all seven via two correct double-angle substitutions, correct Cartesian form, a factorization method, and correct positive radius/angle/range conclusions.