H2MA Remedial _ Complex Numbers_Solution (RVHS)

Teacher’s copy 1 RVHS H2 Mathematics Remedial Programme Topic: Complex Numbers Basic Mastery Questions 1. MI Promo 9758/2020/PU2/P1/Q8(a) Given that 3ia=− and 5 2ib=+ , find the following complex numbers in the form ixy+ , (i) *ab , [2] (ii) * b a . [3] Answer: (i) 13 11i− , (ii) ( )1 17 i10 + (i) ( )( ) 2 * 3 i 5 2i 15 6i 5i 2i 15 6i 5i 2 13 11i ab = − − = − − + = − − − =− (ii) ( ) 2 2 5 2i 3 i * 3 i 3 i 15 5i 6i 2i 9i 15 5i 6i 2 91 1 17 i10 b a +−= +− − + −= − − + += + =+

Teacher’s copy 2 2. ASRJC Promo 9758/2020/Q8(a) It is given that two complex numbers z and w satisfy the following equations ( ) 2 i5 + 4i 1 z = 11 18i wz w += − − + Find z and w. [4] Click here or scan this to view video example on how to solve such question! Answers: 6 2i; i 2 or 3 2i; 2i 2z w z w= + = − = + =− − i5wz+= …(1) ( ) 2 + 4i 1 = 11 18iwz − − + …(2) From [1], 5izw=− …(3) Sub [3] into [2], 2+(4i 1)(5 i) = 11 18iww − − − + 2 2 20i 4 5 i = 11 18i (i 4) (2i 6) = 0 w w w ww  + + − + − +  + + + + 2(i 4) (i 4) 4(1)(2i 6) 2w − +  + − += 2i 4 i 16 8i 8i 24 2w − −  + + − −= i 4 9 2w − −  −= i 4 3i 2w − − = i 2 or 2i 2w = − − − Substitute w into [3]: 5 (i 2)i or 5 ( 2i 2)i 6 2i or 3 2i z z = − − − − −  = + +

Teacher’s copy 3 3. MI Promo 9758/2020/PU2/P1/Q8(b) (i) Express i 6eiz  =+ in the form of ier  . [3] (ii) Given that the complex number zw has modulus 12 and argument 2 3  , find the exact modulus and argument of complex number w. [3] Answer: (i) i 33e  , (ii) 4 3, 3r == (i) i 6ei = cos isin i66 33 i22 z   =+ ++ =+ Since z is in the first quadrant, ( ) 3 1 2 3 2 1 22 i 3 arg tan tan 3 3 33 22 3 3e z r r z    − − == == =+    = = (ii) Method 1 ( ) ( ) ( ) ( ) ( ) 12 3 12 43 3 arg arg arg 2 arg33 arg 3 zw z w w w zw z w w w   = = == =+ =+ = ========================== Method 2 Re (z) Im (z)

Teacher’s copy 4 2i 3 2ii33 2i 3 i 3 2ii 33 i 3 12e 3e 12e 12e 3e 4 3e 4 3e zw w w       − =  =  = = = 4 3, 3r  = =

Teacher’s copy