F08 - Further Complex Numbers - Assignment 1 (Solutions)

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Assignment 1 Solutions) Page 1 of 5 National Junior College 2016 – 2017 H2 Further Mathematics Topic F7: Further Complex Numbers (Assignment 1 Solutions) 1.   4 2i 3 2i 2 3 8 8 3 i 16e 16e , k z k                    i 6 22e , 2, 1,0,1. k z k         2 5i i6 2 6When 2, 2e 2e .k z                i i6 2 3When 1, 2e 2e .k z                0i i6 2 6When 0, 2e 2e .k z               2i i6 2 3When 2, 2e 2e .k z                   8 4 8 4 24 4 16 256 0 16 64 192 8 192 8 192 8 3 i z z z z z z                         *4 4 5 2i i i i4 6 3 6 3 5 2* i i i i4 6 3 6 3 8 8 3 i or 8 8 3 i 8 8 3 i 8 8 3 i 2e , 2e , 2e , 2e or 8 8 3 i 2e , 2e , 2e , 2e z z z z                                                                      www.KiasuExamPaper.com 797

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Assignment 1 Solutions) Page 2 of 5 2. (i)        12 12 12 12 12 i i43 1 11 3i 2e 2 e 2 cos 4 i sin 4 2 4096                   . (ii)         /6 3 2 2 3 3 3 3 12 126 6 i0 i 26 i i i i i 1 11 3i 1 1 3i e4096 4096 1 e , 0, 1, 2,34096 1 e , 0, 1, 2,34 1 1 1 1 1 1e , e , e , e , ,4 4 4 4 4 4 n n z z z n z n z                              (iii)    2 2 2 2 ( ) (2cos ) z re z re z r e e z r z r z r                i i i i (iv)       2 2 3 3 3 3 2 2 2 2 2 2 126 i i i i 1 3i 1 1 1 1 1 1 14096 e e e e4 4 4 4 4 4 1 1 1 2 1 14096 cos cos2 3 16 2 3 16 16 16 4 1 16 4 1 16 1 z z z z z z z z z z z z z z z z z                                                           1 4 1 4 2 3-i1 e4  3i1 e4  3-i1 e4  2 3i1 e4  π/3 www.KiasuExamPaper.com 798

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Assignment 1 Solutions) Page 3 of 5 3. (i)       2 1 1 1 1 2 2 1( ) 1 11 1 1 1 1 (1 )1 (1 ) kn n k k k n k k n k k n n z zz z z z z zz z n zz z z znz z n z z zz z                                    (ii)   i i i i i i2 2 2 e 1 1 e e e e e cos i sin2 2 cos i sin cos isin2 2 2 2 cos isin i2 2 i2isin 2 i cos isin2 22sin 2 z z                                          www.KiasuExamPaper.com 799

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Assignment 1 Solutions) Page 4 of 5 (iii)    1 2 1 2 2 2 (sin sin 2 sin ) Im ( ) Im (1 )1 (1 ) i iIm cos i sin cos i sin 1 cos i sin2 2 2 22sin 2sin2 2 iIm cos i sin2 22sin 2 n k n k k n n k z z z n z z zz z n n                                                                               2 2 2 2 cos i sin 1 cos i sin 4sin 2 cos cos sin sin sin cos2 2sin 4sin2 2 2 sin cos cos sin sin sin cos2 2 4sin 2 sin cos sin sin sin cos 4sin 2 ( 1)sin sin ( n n n n n n n n n n n n n                                                         2 1) 4sin 2    4.                   5 2 35 4 3 2 4 5 5 3 2 4 5 3 2 2 4 5 3 cos5 Re cos5 i sin 5 Re cos i sin Re[cos 5cos i sin 10 cos i sin 10 cos i sin 5cos i sin i sin ] cos 10cos sin 5cos sin cos 10cos 1 cos 5cos 1 2 cos cos 16 cos 20 cos 5cos                                                    Let cosx  where     2 1 πcos5 0 5 2 2 1 π , 0,1,3, 410 n n n          Hence  2 1 πcos , 0,1,3, 410 nx n  . www.KiasuExamPaper.com 800

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Assignment 1 Solutions) Page 5 of 5 Since 9cos cos10 10   and 3 7cos cos10 10   , 2 2 2 2 316 cos cos 0 10 10x x          . 216 20 5x x   has roots 2 2 3cos and cos .10 10   Using product of roots, 2 2 3 5 3 5cos cos cos cos10 10 16 10 10 4       . www.KiasuExamPaper.com 801