11.Complex Numbers Solutions 2023 (RVHS)

River Valley High School, Mathematics Department 2023 Complex Numbers T1 Solutions (Complex Numbers) 1(i) z k i 2 2 2 2 2( ) 2( )( ) ( ) ( 1) (2 )z k i k k i i k k i        3 3 3 2 2 3( ) 3( ) ( ) 3( )( ) ( )z k i k k i k i i      32( 3 ) (3 1)k k k i    32 2 4 0z iz z i    3 2 2[( 3 ) (3 1) ] [( 1) (2 ) ] 2[ ] 4 0k k k i i k k i k i i          3 2 2[( 3 ) 2 2 ] [(3 1) ( 1) 2 4] 0k k k k i k k          32( 3 ) (2 6) 0k k i k    2( 3) 0kk  and 22 6 0k  ( 0 or 3)kk  and 3k  Hence, 3k  (ii)  3 0z i k   1 3 2z    arg( ) 6z  Method 1: By Polar Form & Trigonometry 62 2 cos sin 66 iz e i      62 2 cos sin 66 n n in n nnz e i      nz is real sin 06 n , where 6 n kk   

River Valley High School, Mathematics Department 2023 Complex Numbers T2 6 , where n k k   Hence, 0, 6, 12, 18, ...n    Method 2: By Properties of arg(z) arg( ) arg( ) 6 n nz n z  nz is real, the point representing nz on the Argand diagram is on the x-axis. Thus, arg( ) , where 6 n nz k k     6 , where n k k   i.e. 0, 6, 12, 18, ...n    Given 100nz  . 2 nnnzz  Hence, 2 100n  But n is a multiple of 6. We then have 6 12 2 64 100 2 4096 100   The least value of n is then 12. 2 2 1 2 2 (1)iz w z iw i z iw i         4 (2 ) * 6 (2)z i w    Sub (1) into (2) 4(2 ) (3 ) * 6iw i i w    Let w x yi 8 ( ) (3 )( ) 6 4i x yi i x yi i      8 8 3 3 6 4xi y x yi xi y i       ( 8 3 ) (8 3 ) 6 4y x y x x y i i        Comparing : 9 3 6 3 2 (3) 7 3 4 (4) (3) & (4) 7(3 2) 3 4 18 18 11 y x y x xy Solving y y y yx                    So 1wi  2 (1 ) 2z i i i i    

River Valley High School, Mathematics Department 2023 Complex Numbers T3 3 (2 ) 9 16z i w i    (1) * 3z w i (2) Substitute *3w i z into equation (1) *(2 )(3 ) 9 16z i i z i     *( 2 ) ( 3 6 ) 9 16z i z i i        *( 2 ) 6 10z i z i     Let z x iy ( ) ( 2 )( ) 6 10x iy i x iy i       ( ) ( 3 ) 6 10x y i x y i       Equating real parts: 66x y x y      (3) Equating imaginary parts: 3 10xy   (4) Solving equations (3) and (4): 2x and 4y 24zi 3 (2 4 ) 2 7w i i i     4(i)          2 2 2i 2i 4 1 2 21 2i 4i 8 2 2i 2 2 i 1 z           Note that   πarg i 1 4 and   3πarg i 1 4 Since    12arg argzz  ,  1 1 i shownz   (ii)   222 1 i 1 2i i 2ix          32 2i 1 i 2i 2i 2 2ix         242 2i 4i 4x                4 3 2 1 i 6 1 i 1 i 18 1 i 10 0 4 6 2 2i 2i 18 18i 10 0 s s             