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