RVHS 23 Prelim H2MA P2 Std Soln

River Valley High School 1 2023 H2MA Prelim Paper 2 1 Solution [5] Complex Numbers P2 Q1 (i) 12 12 22 Distance between and 4 2 2(4)(2) cos - Using Cosine Rule33 28 zz zz          Many students converted the z=r form of the complex number to the cartesian form z=x+iy first, before plotting out the points on the Argand Diagram. This probably shows that the student is not conversant with the z=r form. Students must be familiar with all forms of complex numbers given by a question. The r from the z=r form must not be seen as just a value obtained by applying a formula. Students ought to know that it measures the “distance” of complex number z from the Origin. Similarly, the represents the angle measured from the Real axis. Majority of students do not know how to handle complex numbers ( suggesting that students lack preparation for linking complex numbers on argand diagrams to simple Vectors addition/subtraction concept. (ii) 1 12 2 zz az a z   Same thing, many students solved for a by converting everything to cartesian form, not knowing that they could have just made a EM BE D Eq uati on. DS MT 4 EM BE D Eq uati on. DS MT 4 E M B E D E q u at io n. D S M T 4 E M B E D E q u at io n. D S M T 4 E M B E D E q u a ti o n . D S M T 4 E M B E D E q u a ti o n . D S M T 4

River Valley High School 2 i 3 ii33 i 3 2i 3 4 2 2 2 eae e e         Either: 1z is the scaling of 2z by factor of 2 and rotating 2z 2 3  anti-clockwise about the Origin, Or: 2z is the scaling of 1z by factor of 1 2 and rotating 1z 2 3  clockwise about the Origin, the subject immediately and working with z=r form. Some students do not understand the phrase “geometrically… relationship…”, leaving this part blank. For those who understand, tried answering using phrases such as “reflection”, “symmetry”, “half of the complex number”, etc. Note that students should use words such as suggested by the solution.

River Valley High School 3 2 Solution [6] P2 Sequence (i) When 0n , 00 10 4 1 2 3 3 3u u Q              00 4 4 1 233 3 3 3 Q             81 333Q   and 11 21 4 1 1 2 3 3 3 3uu              2 4 1 1 2 4 2 3 3 3 3 3 3u              Many students made arithmetic mistakes while solving for Q and . No evidence to show that students do not understand how to start this part of the question. (ii) 1 4 1 1 2 3 3 3 3 nn nnuu               1 1 1 2 1 2 2443 3 3 3 3 3 n n n n nnuu                           1 52 33 n nnuu   Thus 1nnuu   when 52 33 n   23 35 n  