Zadanie 12c, Paper 1, listopad 2022, Matematyka AA IB HL

Base n=0: w0=1+i has argument π/4=atan(1²). Assume the result at n=k and put q=k+1 and D=q²(q+1)²+1. Since w_k+1 = w_kz_k+1, the argument sum is $\arctan(q^2)+\arctan((2q+1)/D)$. Here D − q²(2q + 1) = q⁴ + 1 > 0, so the product of the tangent values is less than 1. The sum is in (0,π/2) and is the principal argument. The addition identity yields $\arctan((q^2D+2q+1)/(D-q^2(2q+1)))$. Its denominator is q⁴+1, while its numerator is q⁶ + 2q⁵ + q⁴ + q² + 2q + 1 = (q + 1)²(q⁴ + 1). Therefore the argument is atan((q+1)²)=atan((k+2)²), as required for the next index. The base and induction step establish the result for all n≥0.

R1:verify n=0. M1:state the induction hypothesis. M1:use the argument of w(k)z(k+1). A1:correct sum of arctangents. M1:apply the addition identity. A1:correct ratio with common denominators removed. A1:denominator q⁴+1 and valid branch condition. A1:numerator factorization (q+1)²(q⁴+1). A1:atan((k+2)²). R1:complete induction conclusion, requiring at least four earlier marks and a genuine induction step.