13_2016_-_2017_H2_Maths_Complex_Numbers_Assignment_Short_answers

National Junior College Mathematics Department 2017 Complex Numbers (Assignment Solutions) Page 1 of 3 National Junior College 2016 – 2017 H2 Mathematics Complex Numbers Assignment Solutions Qn Solution 1   ii 331 i 3 2e 2 e 2 cos isin 33 n nn nn nn                                 Given that  1 3 i n  is real,    sin 0 3 sin 03 , since 03 3 3 1 3 n n n k k n n k k               Thus, n is an arithmetic progression whose first term is 3 and common difference 3. 2(a) i 2 i 2 (1)z w z w       *2 1 2i i (2)wz   Substituting (1) into (2),    *2 1 2i 2 i iww    . Let w = a + ib. Then                 2 i 1 2i 2 i i 2 2i 2 i 4i 2 i 2 2 2i 4i i 2 i i 2 2 2 i 2 4 2 i 2 i 4 4 i a b w a b w w a b a b a b a b a b b b a                                  Comparing real parts, 2 0 2bb    Comparing imaginary parts, 4 4 1 8 4 1 5 ba a a          2 and 5 5 2ib a w      Substituting 5 2iw into (1),  i 5 2i 2 5i 2 2 5i. z       After considering the polar form, what can you say about n? w = 5 +2i z= 5i

National Junior College Mathematics Department 2017 Complex Numbers (Assignment Solutions) Page 2 of 3 2(b) 2 2 2 2 2 1 i, 1 3 i, i 1 ( 1) 2, 1 ( 3) 2, 1 11arg( ) , arg( )43 p q r k p q r k pq                        2 3 2 2 3 11 22 21 1 22 prs q k      2 2 12 14 3 k k k         22arg( ) 2arg( ) arg( ) 3arg( )33s p r q      2 1 1 5arg( ) 2 3 3 4 3 6r                    If 13, then arg( ) 6kr  If 53, then arg( ) 6kr   3k  Alternative Solution              2 3 2 23 1 i i 1 3i 1 2i i i 1 3 3i 3 3i 3i 2 2 i 8 11 i44 ks k k k             

National Junior College Mathematics Department 2017 Complex Numbers (Assignment Solutions) Page 3 of 3 22 2 2 2 1 2 1 1 1 i4 4 2 11 4 4 2 11 16 16 4 14 3 3 or 3 2Since arg( ) , Re(s) Im(s) 0.3 Hence 0 3 s k k k k k k s kk                                   Qn Solution 4(ii) k= −√5 or √5 4(ii)           4 3 2 2 4 3 2 2 2 3 4 3 2 2 2 4 10 20 25 5i 5i 5 4 10 20 25 5 5 By comparison of coefficient of , 4 4 10 20 25 5 4 5 5i, 2 i z z z z z z z bz z z z z z z bz z b z z z z z z z z                               4(iii) (a)         4 3 2 4 32 4i 10 20i 25 0 i 4 i 10 i 20 i 25 0 i.e. i i 5, 1 2i w w w w w w w w wz wz w                  4(iii) (b)   4 3 2 4 3 2 25 20 10 4 1 0 1 1 1 14 10 20 25 0 1i.e. 1 i1 , 2 i55 v v v v v v v v zv v z v                                     k= −√3