F08 - Further Complex Numbers - Tutorial Set 2 (Solutions)

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Tutorial Set 2) Solutions Page 1 of 14 National Junior College 2016 – 2017 H2 Further Mathematics Topic F8: Further Complex Numbers (Tutorial Set 2) Solutions Basic Mastery Questions 1 (a) 3 0 3z z    Locus of z is a circle with centre at (0,0) and radius 3 units. (b) 1 2 ( 1) 2z z      Locus of z is a circle with centre at( 1,0) and radius 2 units (c) 4 2i 6z           2i 2i 6 2i 2i 6 2 2i 6 2i 3 z z z z                  Locus of z is a circle centered at (0, 2) and radius 3 units. (d) 3 4i 3 4iz        2 23 4i 3 ( 4) 3 4i 5 z z             Locus of z is a circle centered at ( 3, 4)  and radius 5 units. (e) ( 2) ( 1 3i)z z      Locus of z is a perpendicular bisector of the line joining ( 2,0) and ( 1,3) . Im Re -3 3 -3 3 O Im Re -2 2 -3 1 O Im Re -5 1 -3 3 O O A Im Re -9 1 2 -8 5 O Im Re www.KiasuExamPaper.com 733

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Tutorial Set 2) Solutions Page 2 of 14 (f) i iz z   i ( i)z z     Locus of z is a perpendicular bisector of line joining (0,1) and (0, 1) . (g) 2 1 2i 1 2z z    1 12 i ( 2)2 2 1 12 i 2 2 2 1 1 i2 2 z z z z z z                                    Locus of z is a perpendicular bisector of the line joining 1 ,12     and 1 ,02      . (h)   2 5 1 150z  2 2 5 1 150 1 30 1 30 or 1 30 (Rejected radius 0) ( 1) 30 z z z z z               Locus of z is a circle centered at( 1,0) with radius 30 units. (i) arg( 4) 2z   Locus of z is a half-line at (4,0) that makes an angle of 2  with the horizontal in the positive real axis direction. (j) arg 3 i 3z          1 arg 3 i2 3 2arg 3 i 3 z z             Locus of z is a half-line at ( 3,1) that makes an angle of 2 3  with the horizontal in the positive real axis direction. Im Re O 1 Re Im O (0,1) Locus of (0,-1) (-3,1) 1 , 12     Im Re O z = -1 Re Im O 1 -3 Re Im (4,0) O www.KiasuExamPaper.com 734

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Tutorial Set 2) Solutions Page 3 of 14 (k) 3arg(1 i ) 4z         3arg( i)( i) 4 3arg( i) arg ( i) 4 3arg ( i)2 4 5 3arg ( i) 4 4 z z z z                         Locus of z is a half-line at (0,1) that makes an angle of 3 4  with the horizontal in the positive real axis direction. (l) 1arg 1 i 6 z         arg 1 arg 1 i 6z       arg 1 4 6 5arg( 1)( 1) 12 5arg( 1) arg( 1) 12 17 7arg( 1) 12 12 z z z z                           Locus of z is a half line at (1,0) that makes an angle of 7 12  with the horizontal in the positive real axis direction. (m) * 6zz z  2 6 6 z z z z       Locus of z is a perpendicular bisector of the line joining  0,0 and  6,0 . Im O (1,0) Re Im Re O (0,-1) Re Im O (6,0) x = 3 www.KiasuExamPaper.com 735

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Tutorial Set 2) Solutions Page 4 of 14 (n) 2 i 12i z z             2 2 22 2 2 2 2 2 2 2 2 2 i 2i Let i 2 i2 i i 2i 2 i 2 1 i 2 2 2 1 2 4 4 4 1 4 4 3 3 3 1 z z z x y x y x y x y x y x y x y x y y x y y x y x y                                      Locus of z is a circle with centre at(0,0) and radius 1 unit. 2 (a) Im( ) 2z  (b) 22 36 36 6z z z     (c) 1 2 1 15i 1z      2 211 (2) 15 12 11 2 4 2 1 1 22 2 z z z                 Re Im O (-1,0) (0,-1) (1,0) (0,1) -2 2 Re Im 1 O Im 6 6 -6 -6 Re Re Im O O www.KiasuExamPaper.com 736

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Tutorial Set 2) Solutions Page 5 of 14 (d) 2 1 3iz z     ( 2) 1 3iz z       (e) 2arg(2 2)2 3z    2arg(2)( 1)2 3 2arg(2) arg( 1)2 3 2arg( 1)2 3 z z z                    (f)   2 arg 1 i 3z        * 2 arg 1 i 3 arg (1 i) 6 1 2Re( ) 1 1Re( ) 2 z z z z z z                 (g) 1z z   arg4 4z    (h) Im( ) 3z     1arg 1 i tan 24 z      where 1tan 2  tan 2  Im Re 3 O -1 (-2,0) Im Re 1 O Re O Im -1 1 2 z = -1 + i (1,1) (-1,3) Im O (1,0) Re Re Im O z = 1 6  www.KiasuExamPaper.com 737

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Tutorial Set 2) Solutions Page 6 of 14 (i) i 2z   1 1 iz z    (j)  arg i4 2z     0 arg 2i 4z    3 (i) z a (ii) 2z a a  Re Im Locus of P A O Re Im O (0,1) (1,−1) (1,0) Re Im O (0,2) (0,1) Im Re Locus of P O www.KiasuExamPaper.com 738

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Tutorial Set 2) Solutions Page 7 of 14 (iii) z a z  (iv) arg( ) arg( )z a a  Practice Questions 1 (i)  1 1 πarg , 2 3z z   while  2 2 3πarg , 2 4z z    . Hence 1 2 π π2 cos i sin ,3 3 3π 3π2 cos isin 4 4 z z                      (ii) *πi* 3 1 3πi2 4 * -13π13π ii 1212 11πi 12 2e 2e 2e 2e 2e 11π 11π2 cos sin12 12 z z i                                                     (iv) Since the circle 1 2z z  is symmetrical about the line  Re 1z  and passes through  0, 0 , it will also passes through  2, 0 where it meets the positive real axis. Re Im O Locus of P Re Im 0 A Locus of P × (1 , √3) 2 Re (-1 , -1) ○ Im O (iii) www.KiasuExamPaper.com 739

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Tutorial Set 2) Solutions Page 8 of 14 2   arg 2 3i 3z     2 iz k   is a circle centered at (2, 1) with radius k units. (i) 2 3 6       sin 4 14sin 46 2 2 ( ) k k k shown           (ii) From the Argand diagram, observed that 2 < k < 4 such that the two loci intersect at two points. 3 (i) Let iz x y    22 2 2 2 2 2 2 2 i 2 i 2 4 4 4 4 0 1 ( ) z z x y x y x y x y x y x x y x x shown                        (ii)   1 2 1 2 1cos 3 1.23096 Let and represent complex numbers and respectively. arg( ) 1.91 arg( ) 1.91 P Q z z z z               (iii) z a b  is a circle centered at a with radius b units, passing through P and Q. PQ is the diameter of the circle such that b is least. 2 2Least 3 1 8 2 2b    1a  (Note: a is real.) Re Im 0 (2,-3) (2,1) A C k Re Im 0 P Q (3,0) (-3,0) −1 (0,-3) (0,3) www.KiasuExamPaper.com 740

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Tutorial Set 2) Solutions Page 9 of 14 4 (i) (ii) Maximum value of 2 22 5 3 29 3z      Minimum value of 2 22 5 3 29 3z      (iii) The line joining the centre of the circle to the point 6+i makes right angle with the