ASRJC_FM_2023_Prelim_P2 (Solutions)

1 1 2 *' kz z   2 *arg ' arg kz z          2*arg arg 0 arg arg( ) kz z z      Therefore the line PP’ passes through the origin O O,P,P’ are collinear and O,Q,Q’ are collinear. Let .OPQ  Area of triangle OP’Q’ is 4 times the Area of triangle OPQ. Then       22 ** 22 3 * * 4 3 4 4 2 11 ' ' sin 4 sin22 4 2 3 2 .3 since and 2 .323 2 .3 23 OP OQ OP OQ kk zw kk z z w wzw k k k           2 (a) 1 2 3 4 5 61, 1, 2, 3, 5, 8F F F F F F      1 2 3 4 5 61, 3, 4, 7, 11, 18L L L L L L      5 4 6 5 4 6 11 and 3 8 11 (verified) L F F L F F         (b) To prove: 11m n m n m nF F F F F   ,  2 fixed ,m m n  . For n = 1, LHS = 1mF  . RHS = 1 1 2 1 1m m m m mF F F F F F F      = LHS, implying the statement holds for n = 1. For n = 2, LHS = 2mF  .

2 RHS =  1 2 3 1 1 2m m m m m m mF F F F F F F F F        1 2 LHS, mm m FF F     implying the statement holds for n = 2. Assume statement holds for 1 and for some , 2n k n k k k     . That is, we have 1 1 1m k m k m kF F F F F    and 11m k m k m kF F F F F   . For n = k + 1, 11m k m k m kF F F         1 1 1 1 1 1 1 1 1 2 by (IH)m k m k m k m k m k k m k k m k m k F F F F F F F F F F F F F F F F F F                    implying that statement holds for n = k + 1. So 11m n m n m nF F F F F   for a fixed m  2 and for all n  . (c) Let ,mn   2 1 1 11 n n n n n n n n nn F F F F F F F F FL      3 (a) 11 1 1 1 2 2 2 2 2 1 1 1 1 11 cos sin cos sin 2 sin 2 2 2 2 2 i i i iie e e e e i i ie                                                2 21 4 sin . 2 iiee   (b)     2 3 2 2 2 2 2 2 21 .....1 2 3 1 4 sin 2 14 sin 2 i i i ni ni n i n in n n n nC iS e e e e e e e                                  

3 Equating real parts,   2 14 cos sin 2 n nCn  Equating imaginary parts,   2 14 sin sin 2 n nSn  2 6 631 , 0, 1, 2,3. kk ii w w e e k      