RI Complex numbers C5 Add Prac (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ______________________________________ Additional Practice C5 Complex Numbers Page 1 of 15 Additional Practice Questions for Chapter 5: Complex Numbers 1 Given that 2(2 3 i ) (2 3 i ) 0     , find the real numbers  and  . [6] [ 4, 13 ] 9233/2003/01/Q2 Solution: 2(2 3 i ) (2 3 i ) 0     (4 12i 9) ( 2 3i) 0     ( 5 2 ) i(3 12) 0      Comparing real and imaginary parts, 52 0   and 31 2 0 Solving, 4, 13 2 The complex number ixy is such that 2(i )ixy . Find the possible values of , xy   , giving your answers in exact form. [4] Hence find the possible values of the complex number w such that 2 iw  . [2] [ 11, 22 xy , 11, 22 xy  ; 11 11i o r i 22 22   ] 9233/2002/01/Q5 Solution: 2(i )ixy 22() i 2 ixy x y  Comparing real and imaginary parts, 22 2 2 0 (1)xy x y  and 22 4 2 2 112 1 (2) 4 1 4 1 or (N.A. since ) 22xy x y x x x x         1 2 x  (2)  x and y must have the same sign, and (1)  x = y

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ __________________________________ Additional Practice C5 Complex Numbers Page 2 of 15  When 11, 22 xy ; When 11, 22 xy  22 2 i i (i ) iww w      11 11 i i i i or i 22 22 wx y wy x       3 The complex number is given by , where a is a non-zero real number. Given that is real, find the possible values of z. [3] [ 13 iz  or 13 iz  ] TPJC Prelim 9740/2014/01/Q6bi Solution:  33 12 iza  23 13 2 i 3 2 i 2 iaa a     2311 2 6 8 iaa a   Since 3z is real, 368 0aa   223 4 0aa  Since 0a , 234 0a 2 3 4a  3 2a 312 i 2z    or 312 i 2z    13 iz  or 13 iz  4 The complex number w is such that iwa b , where a and b are non-zero real numbers. The complex conjugate of w is denoted by *w . Given that 2 * w w is purely imaginary, find the possible values of w in terms of a. [5] [ i o r i 33 aaaa ] z 12 iza 3z

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ __________________________________ Additional Practice C5 Complex Numbers Page 3 of 15 9740/2015/01/Q9(a) Solution:     232 22 32 2 3 22 32 2 3 22 ii i *i i 3i 3 i 3i 3 ab abwa b wa b a b a b aa b a b b ab aa b a b b ab            22 32 22 Given is purely imaginary, Re 0** 3 0 ww ww aa b ab       22 30aa b 0 (rejected, since 0) or 3 aaa b    Thus, i or i . 33 aawa wa  5 (a) Solve the simultaneous equations i1 iwz   , 402( 1 i ) 62 izw   . [3] (b) Two complex numbers w and z are such that * 2iwz , 2 6wz  . Find w in the form iab , where a and b are real and positive. [4] [(a) 22 iw  , 1iz (b) 22 iw ] Solution: (a) i1 iwz   ----- (1) 402( 1 i ) 62 izw   ----- (2) 2 x (1)  (2) gives 40[2i (1 i)] 2 2i 8 4i 62 iw        84 i 22 i 13 iw     and 1ii 1ii (2 2 i ) 1izw         

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ __________________________________ Additional Practice C5 Complex Numbers Page 4 of 15 (b) 2 6wz   z is real. Let z = ,   Then ** 2i 2i 2iwz w w      Then 2 2 46w    2 20    1 (reject since Re( ) 0) or 2 22 i w w       Alternative method: ** 2i 2iwz z w    ----- (1) 2 6wz  ----- (2) Substitute (1) into (2) gives 2 * 2i 6ww  Let iwa b . We have 22 (i ) 2 i 6 (6 ) i ( 2)ab a b a b  Comparing real and imaginary parts, 20 2bb   22 46 2 0aa a a   ( 2)( 1) 0 2 or 1(N.A. since is positive)aa a a     22 iw  6 The complex numbers s and w satisfy the equations 6isw and 10sw . Given that Re(s) > 0, solve the equations for s and w, giving all answers in the form ix y , where x and y are real. [4] Hence find a solution to the following equations 6uv  and 10uv . Give your answers for u and v in the form ix y , where x and y are real. [2] [ 13 i , 13 isw  ; 3iu  and 3iv ] AJC Prelim 9740/2013/01/Q11(a) Solution: 6isw  The real part of s and ware the same.

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ __________________________________ Additional Practice C5 Complex Numbers Page 5 of 15 Let is ab and iwac , where a > 0. 6bc  --- (1)    2 ii 1 0 () i 1 0 ab ac ab c a b c     2 10 (2)ab c   and since a  0, 0( 3 )bc  Solving (1) and (3), 3b and 3c Using (2), 1a Since a > 0, 13 i , 13 isw  Alternative method: Substitute 10w s into 6isw ,   2 2 10 6i 10 6i 6i 10 0 6i 36 4( 10) 13 i2 s s ss ss s         Since Re(s) > 0, 13 is and 13 iw Let ius and ivw and we would arrive at the original pair of given equations. 3iu   and 3iv Alternative method: 1i, i i 3 i i 1 and i 3 i i svwu v s s uw w             3iv  and 3iu  7 Do not use a calculator in answering this question. Given that 23 i is a root of the equation 43 2 10 48 122 143 0zzz z  , solve the equation, giving your answers in exact form. [4] [ and 32 i ] HCI Prelim 9740/2013/01/Q11(i) Solution: 23 i

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ __________________________________ Additional Practice C5 Complex Numbers Page 6 of 15 Since all the coefficients of the polynomial are real, non-real roots will occur in conjugate pairs. Given that 23 i is a root, then 23 i is also a root.     22 2 2 (2 3i) (2 3i) ( 2) 3i ( 2) 3i