RVHS 23 J2 H2 MA CT Std Soln

River Valley High School 1 2023 JC2 H2 MA CT Markers Comments 1 The complex numbers z and w satisfy the equations: 22 * 2z ww w−= and 2122 2z w w+ = − . (i) Show that 2 2 2 0ww+ + = . [2] (ii) Hence find all possible pairs of values for z and w. [3] 1 Solution [6] Complex Numbers (i) 2 22 2 * 2 22 z ww w z w w −= =+ Substitute 221 22z w w=+  into 2122 2z w w+ = − ( ) ( ) 222 2 2 11 2 2 222 2 2 2 2 2 0 2 2 0 Shown w w w w ww ww  + + = − + + = + + = Majority were able to show the result either via the substitution or e limination method. A number of s tudents did not see that ww* = |w| 2, and hence una ble to continue or instead wrote w = x + iy and expanded. (ii) From GC, 1iw=−  When 1iw=− + , 2 2iw =− , 2 2w = and ( )2 2i 2 1 2i2z −+= = − When 1iw=− − , 2 2iw = , 2 2w = and ( )2 2i 2 1 2i2z += = + Thus, 1 2iz=− and 1iw=− + 1 2iz=+ and 1iw=− − w can be found via the GC. However some students went on to substitute w = x + i y and solve v ia comparing real and imaginary parts (which was too much work for 1 mark). Though the m ethod to find z was correct, many made careless mistakes in the computation.

River Valley High School 2 2 The complex numbers z, w and v are such that 13π 13π2 cos isin30 30z =+  , ( ) πarg 12w = and 4 i* zwv w= . (i) Find the modulus of v and show that ( ) 3arg π5v =− . [4] (ii) It is known that nv is a negative real number. Find the least positive integer value of n. [3] 2 Solution [6] Complex Numbers (i) ( ) 4 4 4 44 i* i* 1 2 4 units zwv w zw w zw w z = = = = = = 13π 13π2 cos isin30 30z =+  ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) 4 arg 4arg arg arg i arg *i* 4arg arg arg arg 13π π π42 30 12 2 7π 3π Principal argument Shown55 zw z w ww z w i w  = + − + = + − − −    = + −       −== Most candidates can work out that 4v = Some candidates are not aware that ( ) ( )arg * argww =− ; ( ) πarg 2i = ; ( )7π 3π Principal argument55 −= since 7π 3π2π55 −−= (ii) ( ) ( ) 3 πarg arg 5 n nv n v −== , ( ) ( )arg negative real number 2 1 π, where kk= +  ( )3 π 21 π5 n k− =+ ( )5 213nk −=+ Least positive value of n = 5 when 2k =− . Many candidates are wrongly state that ( )arg negative real number kπ, where k=

River Valley High School 3 Note: Setting 3 π π5 n k− = is not correct.

River Valley High School 4 3 The position vectors of distinct points A, B and Q are given by a , b and q respectively with reference to the origin. (i) Given ( ) ( )−  − =q a b a 0 , explain why A, B and Q are collinear. [2] (ii) It is given further that (1 )= + −q a b where 01  , 1| | | | 2=ba , 3 ||6AQ= a and 60AOB =  . By f