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 5 4. MI Promo 9758/2020/PU2/P2/Q2(i) Do not use a calculator in answering this question. The roots of the equation 2 8 6iz =− − are 1z and 2z . Find 1z and 2z in cartesian form, ixy+ , showing your working. [5] Click here or scan this to view video example on how to solve such question! Answer: 1 3i−+ , 1 3i− ( ) 2 22 Let i . Then, i 8 6i. 2 i 8 6i z x y x y x y xy = + + =− −  − + =− − Comparing real parts, 22 8xy− =− . -- (1) Method 1: Comparing imaginary parts, 326xy x y=−  =− -- (2) Sub (2) into (1): ( )( ) 2 2 4 2 42 22 3 8 9 8 8 9 0 9 1 0 y y yy yy yy − − =−  − =− − − = − + = ( ) 2 2 2 9 3 1 rejected since 0y y y y y=  = =−    From (2): 1x= Therefore, the roots of the equation 2 8 6iz =− − are 1 1 3iz =− + and 2 1 3iz =− .

Teacher’s copy 6 5. VJC Prelim 9758/2021/01/Q8(a)(i) The complex number w is given by iewr = , where 0r  and 0 2  . Given that ( )1 i 3zw= −  , find z in terms of r and arg( )z in terms of  . [2] Click here or scan this to view video example on how to solve such question! Answers: 3  −+ , 2r ( )arg( ) arg 1 i 3 arg( ) 3 zw   = − + =− + 1 i 3 2 zw r =− =

Teacher’s copy 7 Standard Questions 1. RI Promo 9758/2020/Q7(a) Do not use a calculator in answering this question. One root of the equation * 2i 6i,zz z a+ = + where a is real, is 3 7i.z=− Find the value of a and the other root. [4] Answer: a = 72, 3 9i+ Given 3 7iz=− is a root, substitute it into * 2i 6izz z a+ = + , ( )( ) ( )3 7i 3 7i 2i 3 7i 6i a − + + − = + Comparing real part, 223 7 14 72 aa+ + =  = Let the other root be ixy+ . ( ) 22 2i i 72 6ix y x y+ + + = + Comparing imaginary part, 2 6 3xx=  = Comparing real part, 223 2 72yy+ − = ( )( ) 2 2 63 0 7 9 0 7 or 9 yy yy yy − − = + − =  =− = Hence the other root is 3 9i.+

Teacher’s copy 8 2. VJC Promo 9758/2020/Q10(a), (e) It is given that the complex number ( )3iw=−  − . Find the value of w . [1] Find the exact value of arg(w). [1] Without using a calculator, find the three smallest positive whole number values of n such that *nww is a real number. [3] Click here or scan this to view video example on how to solve such question! Answer: 2w = , 5arg 6w =− , 1, 7,13n= ( ) 2 23 1 2w =  + = 5arg 6w =− ( ) ( ) 55arg * arg arg * 66 * is real arg * , where 55 66 6( 1) 5 61 5 For the three smallest positive whole number values of , we take 0, 5, 10, 1, 7,13 n nn w w n w w n w w w w k k nk kn kn n kn     = + =− +  =   − + =  − + =  = − = − −  = ( )3, 1−−