RI Complex numbers C5 Tut Sect A (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ___________________________ Tutorial 5: Complex Numbers Page 1 of 8 Tutorial 5 : Complex Numbers Section A (Basic Questions) 1 It is given that 12 i .z (i) Without using a calculator, find the values of 2z and 3 1 z in Cartesian form ixy , showing your working. [4] (ii) The real numbers p and q are such that 2 3 qpz z is real. Find, in terms of p, the value of q and the value of 2 3 qpz z . [3] [(i) 2 34 iz   , 3 11 1 2 i125 125z   (ii) 250qp , 19 p ] Solution: (i) 12 iz   22 12 i 14 i4 34 iz         3 34 i12 i 34 i6 i8 1 12 iz         3 11 1 1 1 2 i 1 1 1 2 11 2i i11 2i 11 2i 11 2i 125 125 125z         (ii)  2 3 11 234 i i 125 125 qpz p qz     is real 2 3Im 0 qpz z   24 0 250125pq q p     2 3 11 23 4i 250 i 125 125 34 i 2 24 i 19 qpz p pz pp pp p         

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ________________________ Tutorial 5: Complex Numbers Page 2 of 8 2 Solve the following simultaneous equations: (a) 62 iwz , 203 2iwz  ; (b) 3i 2zw , 2 i5 2 i 0zw  . [(a) 11 i22z   , 13 5 i22w (b) 2i, 2 izw   or i, 2 4izw   ] Solution: (a) 62 iwz ---------(1) 203 2iwz  ---------(2) (1)(2),  20 2 i46 2 i 2i2i 1 62 i 4 02 0 i 22 i5 z        11 i22z   1162 i i 22 13 5 i 22 w   (b) 3i 2zw --------------(1) 2 i5 2 i 0zw  --------(2) From (1), 3i - 2wz --------(3) Substitute (3) into (2),   2 2 2 i3 i 2 5 2 i 0 i3 2 i 5 2 i 0 i2 0 zz zz zz         2 ii 4 ( 1 ) ( 2 ) 2(1) i9 2 i3 i 2i or i2 z     2iw  or 24 iw 

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ________________________ Tutorial 5: Complex Numbers Page 3 of 8 3 The complex number w is such that *2 3 4 i ww w , where *w is the complex conjugate of w. Find w in the form iab , where a and b are real. [4] [ 12 iw  ] Solution:     22 22 *2 3 4 i ii 2 i 3 4 i i2 2 i 3 4 i 22 i 3 4 i ww w ab ab ab ab a b ab ab               Comparing real and imaginary parts,  22 2 2 2 2 3 and 2 4 2 42 3 21 0 10 1 ab a b b aa aa a a           12 iw  

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ________________________ Tutorial 5: Complex Numbers Page 4 of 8 4 For each of the following complex number, represent it on an Argand diagram and find its modulus and argument. (a) 3i (b) 1i3 (c) 2i( 1 i ) (d) i(1 i) Solution: (a) (a) 3i 3+i 3 1 2  1tan arg( 3 i) 663        (b) (b) 1i3 1i3 13 2    tan 3 arg( 1 i 3) 3     2 3  (c) (c) 2i( 1 i ) 1 i   1i 11 2    3tan 1 arg( 1 i) ( )44          (d) (d) i(1 i) 1 i  1i 11 2  tan 1 4  arg(1 i) 4   1 Re Im  O Re Im O Re Im 1 O Re Im 1  O

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ________________________ Tutorial 5: Complex Numbers Page 5 of 8 5 The complex number z is such that 1z  and arg z  , where 0 4  . (i) Mark a possible point A representing z on an Argand diagram . Given that   2arg 2argzz  , mark the points B and C representing 2z and 2zz respectively on the same Argand diagram corresponding to point A. [2] (ii) State the geometrical shape of OACB. [1] Solution: (i) (ii) Since OACB is a parallelogram with 4 equal sides, it is a rhombus. x y O

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ________________________ Tutorial 5: Complex Numbers Page 6 of 8 6 (i) Find the roots of the equation   2i5 i 2 6 i 0zz    , giving your answers in cartesian form iab , where ,ab   . [2] (ii) Hence find the roots of the equation   2i1 5 i 2 6 i 0ww     , giving your answers in cartesian form iab , where ,ab   . [2] (iii) Given that the roots found in part (i) are also roots of the equation () 0Pz  , where ()Pz is a polynomial of degree 4 with real coefficients, find ()Pz . [3] [(i) 13 i o r 2 i (ii) 3i o r 2w (iii) 43 2 21 48 4 0zz zz   ] Solution: (i)   2 2 i5 i 2 6 i 0 5i [ ( 5i ) ] 4 ( i ) ( 26 i ) 2i 5i 2 i 2i 5i 1i 2i 62 i 4 or 2i 2i 1 3i or 2i zz z               (ii)   2i1 5 i 2 6 i 0ww     Since iwz ,    i 1 3i or i 2i 3 i or 2 w   (iii) Since ()Pz is a polynomial of degree 4 with real coefficient, 1 3i and 2i are also the roots.             22 22 22 43 2 () 2 i 2 i 1 3 i 1 3 i 4i 1 3i 1 3i 41 9 42 1 0 21 48 4 0 Pz z z z z zz z zz zz z zz zz                       

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ________________________ Tutorial 5: Complex Numbers Page 7 of 8 7 The cubic equation 32 31 212 0,az z z b  where a and b are real numbers, has a complex root 13 i .z (i) Explain why the equation must have a real root. [2] (ii) Find the values of a and b and the real root, showing your working clearly. [5] [(ii) 25, 190ab , 19 25 ] Solution: (i) Since the coefficients of 32 31 212 0az z z b  are all real , complex roots occur in conjugate pair. Since a cubic equation has three roots, the third root must be a real root. (ii) 32 31 212 0az z z b  Given 13 i is a root, 13 i is also a root.        2 2 13 i 13 i 13 i 13 i 19 21 0 zz zz z zz               32 2 31 212 2 10az z z b z z az c     By comparing coefficients of 2 0 : 3 1 2 : 212 2 10 From GC, 19, 25 : 10 190 zc a zc a ca zbc b         The real root of the equation is 19 25 c a  . Alternative method: Since 13 i is a root of 32 31 212 0az z z b  ,   32 1 3i 31 1 3i 212 1 3i 0ab         26 18i 31 8 6i 212 1 3i 0ab          26 460 18 450 i 0ab a    Comparing real and imaginary parts: 26 460 0ab  ----------- (1) 18 450 0a -----------------(2)

Raffles Institution H2 Mathematics 2025 Year 5 ________