13 2016 - 2017 H2 Maths Complex Numbers Tutorial Solutions updated

National Junior College Mathematics Department 2017 Complex Numbers (Tutorial Solutions) Page 1 of 11 National Junior College 2016 – 2017 H2 Mathematics Complex Numbers Tutorial Solutions Basic Mastery Questions 1 (a)   3 8i 5 7i 15 21i-40i 56     71 19i (b) 7 5i 7 5i 4 3i 4 3i 4 3i 4 3i                  28 21i 20i 15 16 9 13 41 i25 25      2 (a)  2i 2 , arg 2i 2     i 22i 2e 2 cos isin 22                      (b) 1 3i 1 3 2       2i 3 2arg 1 3i 33 1 3i 2e 222 cos isin33             (c)   1 i 1 3i     1 i 1 3i    22   arg 1 i 1 3i       arg 1 i arg 1 3i 32 43 12           =    i12 1 i 1 3i 2 2e 2 2 cos isin 12 12                         (d) 1 3i1 3i 2 21 i 1 i 2       

National Junior College Mathematics Department 2017 Complex Numbers (Tutorial Solutions) Page 2 of 11 1 3iarg 1i      arg 1 3i arg 1 i 23 34 17 12           17Principalarg( ) 2 12 7 12 z     7i 121 3i 2e1i 772 cos i sin12 12                    (e) i 3 1 3 3 31 e 1 cos isin 1 i i3 3 2 2 2 2           3 3 9 3 12i32 2 4 4 4     13 3 3arg i tan2 2 3 6     ii661 e 3e 3 cos isin66      Alternative Method for 2(e):   ii i i 63 6 6 i 6 i i i6 6 6 1 e e e e e cos isin cos isin6 6 6 6 3e 2cos 2 e 3 e62                                          (f) 33e 3   i , 3arg 3e 3     i , 33e 3 cos isin 33   i (g) 100 100 ,  arg 100 π ,   i100 100 cos isin 100e     (h)   115 5 5 11 i 2 ii i i i i 666 6 6 63e 3 1 e 3e e 3e 3e 3e 3 cos isin66                                      

National Junior College Mathematics Department 2017 Complex Numbers (Tutorial Solutions) Page 3 of 11 Alternatively,       5π 5πii 66 5π 5πii 66 3e 3 e 3 5π 11π πarg 3e arg 3 arg e π 6 6 6             5πi 6 ππ3e 3 cos isin 66                  3 (a) 5 5 3 12 cos isin 2 i 3 i6 6 2 2                       (b) i 3 132e 2 cos + isin 2 cos isin 2 i 1 i 33 3 3 3 2 2                                   4 1 1 1   Practice Questions 1 3 i 4 4izw    2 4 2 5arg( ) arg( )64 zw zw    (i) 1 1 1 2zz   15arg arg( 1) arg( ) 66zz        11 cos isin2 6 6z    (ii) * * 1 1 1 1 2zz z    * * 15arg arg(1) arg( ) 0 arg( ) 6zzz       * 1 1 5 5 cos isin2 6 6z   (iii)   33 3* 3 *( ) 4 2 128 2w w w    * 3 * 3arg( ) 3arg( ) 3arg( ) 3 44w w w     

National Junior College Mathematics Department 2017 Complex Numbers (Tutorial Solutions) Page 4 of 11 *3 33( ) 128 2 cos isin 44w                 (iv) * 2 1 2 44 2 2 2 zz ww    * * 5 13arg arg( ) arg( ) arg( ) arg( ) 6 4 12 z z w z ww           * 13 11Principal arg 2 12 12 z w      * 2 11 11cos isin4 12 12 z w   (v) 232 3 2 2 (128 2) 512 2z w z w   23 5 29arg( ) 2arg( ) 3arg( ) 2 3 6 4 12z w z w                  23 29 5Principal arg( ) 2 12 12zw     23 55512 2 cos isin12 12zw   2 (i) 22 223 2 pp w w       2 2arg arg 2 arg 2arg 7 5 1702 8 8 8 8 p pww               (ii) 2 23 cos sin2 8 8 nn p n n iw                  For 2 2 n p w   to be purely imaginary,     cos 0 8 33..., , , , ,...8 2 2 2 2 21 ,82 4 2 1 , ..., 12, 4, 4,12,... n n kn k n k k                   

National Junior College Mathematics Department 2017 Complex Numbers (Tutorial Solutions) Page 5 of 11 Thus, smallest positive value of n = 4. 3 (i) 1** wwwp www    ; arg arg arg arg * 2arg 2* wp w w ww      (ii) 2i 1 i 5 0eepp    Since 5p is real, sin10 0  and cos10 0  . This implies 10θ is of a multiple of 2π. Solving sin10 0  , we have , 1,2,3,45 k k  . Thus, 2 or 55   . Alternatively, 5p is real and positive  arg( 5p ) is of a multiple of 2π. 10 2 , 2 or since 10 π2 5 5 5 kk k          4 * * 1 arg arg( ) arg( ) π bb aa b baa      5 Re Im 0 A D C B − 1 r r E

National Junior College Mathematics Department 2017 Complex Numbers (Tutorial Solutions) Page 6 of 11 2 i 2 iee     cos2 isin 2 cos 2 isin 2         2cos2 ,which is real for all    4i 2 i 2 i 2 i 2i 22 1e e e e 2e 2cos 2 cos 2 isin 2 cos 2 1 i tan 2 w                 Re( ) 1w Alternatively,           4i 4 i 4 i 4i 4 i 4 i 4i 2 2 1 e 1 e 1 e 2 1 e 1 e e 1 2 1 e 2 2cos 4 1 cos 4 isin 4 1 cos 4 1 cos 4 isin 4 sin 4 1i1 cos 4 1 cos 4 2sin 2 cos 21i 1 2cos 2 1 sin 21 i 1 itan 2 Re( ) 1cos 2 w w                                                 6 Let iw a b     22 * 2 3 4 2 i 3 4i ww w i a b a b         22 223 1 2i124 ba b a wab           7 i 2 1 2i i 2i i (1)z w z w z w         4 (2 ) * 6 (2)z i w    Substitute (1) into (2), 4(2i i) (3 i) * 6ww    Let iw x y 8i( i ) (3 i)( i ) 6 4ix y x y     

National Junior College Mathematics Department 2017 Complex Numbers (Tutorial Solutions) Page 7 of 11 8i 8 3 3i i 6 4ix y x y x y       ( 8 3 ) (8 3 )i 6 4iy x y x x y        Compare real and imaginary parts, 9 3 6 3 2 (3) 7 3 4 (4) Solving (3) & (4) 7(3 2) 3 4 18 18 11 y x y x xy y y y yx                    So 1iw  2i(1 i) i 2 iz     8 2( i ) 12i 5xy   22 2i 12i 5x y y    Comparing real and imaginary parts, 22 5 (1) 2 12 (2) xy xy      From (2), 6x y . Substitute 6x y into (1), we get 2 4 2 2 22 36 5 5 36 0 9 or 4 (N.A.because 0 as is a real number) 3, 2 3, 2 i 2 3i or 2 3i y y yy y y y yx yx xy                   2 2 2 4 12i 9 12i 5 4 4 4 12i 5 ( 2) 12i 5 2 (2 3i) 3i or 4 3i zz zz z z z                    9 (i) Since 1i is a root of the equation 322 2 0,w aw bw                    322 1 i 1 i 1 i 2 0 2 2 2i 2i 1 i 2 0 6 4 2 i 0+0i ab ab b a b                    Comparing real parts, Comparing imaginary parts,

National Junior College Mathematics Department 2017 Complex Numbers (Tutorial Solutions) Page 8 of 11 60 6 b b   4 2 0 4 264 52 ab ba a        (ii) Since the polynomial equation has real coefficients, 1i and 1i are roots to the equation.       322 5 6 2 1 i 1 i 2w w w w w w A         Comparing constants,         2 1 i 1 i 2 1 i 2 1 1 2 1 A A A A                 322 5 6 2 0 1 i 1 i 2 1 0 11 i, 1 i, . 2 w w w w w w w              Alternative solutions to parts (ii) and (iii) Since coefficients are real, if first root is 1 + i, then second root is 1  i Quadratic factor is (w  1  i)(w  1 + i)= w2  2w + 2 2w3 + aw2 + bw  2 = (w2  2w + 2)(2w  1) = (2w3  4w2 + 4w) + (w2 + 2w 