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                   By comparing imaginary parts, 12 2 18 0 15 s s      Since the coefficients of the equation are all real, and 1i is a root of the equation, 1i is also a root of the equation.

River Valley High School, Mathematics Department 2023 Complex Numbers T4       2 2 2 1 i 1 i 1 i 2 2 x x x xx               By long division,    4 3 2 2 26 15 18 10 2 2 4 5 x x x x x x x x         Solving 2 4 5 0xx   ,          2 4 4 4 1 5 21 44 2 2 i x         The other roots are 1i , 2i and 2i . (iii)           1 11 1 11 arg arg arg arg arg π1 4 nz n z zz n z z n        Since 1 1 nz z  is purely imaginary,     ππ1 π, where 42 11 142 1 2 4 14 n k k nk nk nk            The two smallest positive integers of n are 1 and 5. 5(i) The assumption is that a, b and c are all real. (ii) Let 32 ( (3 i))( (3 i))( 2)x ax bx c x x x         = 2( 6 10)( 2)x x x   = 32 8 22 20x x x   By comparing coefficients, we have a = –8, b = 22 and c = –20.

River Valley High School, Mathematics Department 2023 Complex Numbers T5 6   2 1 4i 15 8i        2 223 1 15 8i i 1 4i2 z          2 2 3 4 i2 z    3 4 i2 z    or 3 4 i2 z    2 2iz or 14 2iz  7(i) 2 6 36 4(1)(36)6 36 0 3 3 3i 2z z z        Thus, ii33 12 6 and 6z e z e    (ii) 4 3 7()3 6 23 3 6 6 6 55 6 cos sin 66 i i i e e e i                           (iii) i 32 6ze    i 3 2 6 n nnze  Since 2 nz  , 3 2n k  for some integer k. 6nk . ... , 12, 6, 0, 6, 12, ...n   Smallest positive integer n = 6. 8(i)   2 * * ** 2 i i 1 0 ( ) i( ) 1 0 ( i)(2 ) i(2 i) 1 0 2 2 2 i 1 ( ) kw kww w w kw w w w w k a b a b ka b abk                    Real part 2 2 212 2 1 2 kaka b b     ---(1) Im part

River Valley High School, Mathematics Department 2023 Complex Numbers T6 0  0 ie, is either purely r 0 eal or ima ginar . 0 y ab ba w k     or (ii) Hence 2 2 Since is real, 0. Using 2 and 0 From part : 2(2) 1 02 141 4 11ie, or 22 wb kb a aa ww          (i) Otherwise 22 2 Since is real, 0, ie, Using 2 and eqn becomes: 2 2 i i 1 0 141 4 11ie, or 22 w b w a k w a a a a a aa ww             9 The statement is only true if p is real. (i) Using GC, p = 5. (ii) We have )))(21())(21((534 2234 bazzizizzzzz  , = ))(52( 22 bazzzz  Comparing coefficients of similar terms, we have a = b = 1 For 012  zz , we have iz 2 3 2 1  2 33 Ra a a a  arg(q) = arg (a )3 - arg (a ) = 3 arg(a) + arg(a) = 4  Thus,  )4sin()4cos(2  iRq  6 1 q =               3 2sin3 2cos3 1 iR

River Valley High School, Mathematics Department 2023 Complex Numbers T7 Given that 03 2cos       , 4 3 23 2   or 2.36 radians 10a         5arg 5arg 0, , 2 ... 22arg 0, , , , ,...5 5 5 5 Since 0, 2arg or . 55 ww w k w               2tan or tan5533 23 tan or 3 tan55 kk kk                             12 or 55n  bi Method 1          22 22 22 2 2 1 1 cos isin 1 cos 2i cos sin isin 1 1 sin 2isin cos sin 1 1 2sin 2isin cos 2sin 2isin cos 2sin sin i cos z                                      Method 2         22 2 2 1 1 cos isin 1 cos 2 isin 2 1 cos 2 isin 2 1 1 2sin 2isin cos 2sin 2isin cos 2sin sin i cos z                             bii Method 1

River Valley High School, Mathematics Department 2023 Complex Numbers T8   2 22 1 2sin sin i cos 2sin sin cos 2sin z             Given that 0 2          2 1 1 arg 1 arg 2sin sin i cos arg 2sin arg sin i cos cos0 tan sin tan tan 2 2 2 z                                 Method 2          2 i 2i 1 2sin sin i cos 2sin i cos isin 2isin e 1 2isin e 2sin z z                                2i i arg 1 arg 2isin e arg 2isin arg e 2 z               Method 3   21 2sin sin i cos 2sin cos isin 22 2sin cos +isin 22 z                                           

River Valley High School, Mathematics Department 2023 Complex Numbers T9 21 2sinz    2arg 1 2z    11i ii iii Considering the imaginary part of 2ab , we have