Ch 9 Complex Numbers Tut 9B APQ Solutions (RVHS)

Complex Numbers Tutorial 9B: Polar and Exponential Forms Solutions Additional Practice Questions 1 2 2 2z z       arg arg arg( ) 4 4 arg( ) 4 2 3 4 iz i z z                   2 2 2 2 2 wz w z w     2 5arg 6 52arg( ) arg( ) 6 3 5arg( ) 2 4 6 7 3 ( )3 2 cos sin3 3 1 3 z w z w w pv w i i                                          

2 (i) 33 32 cos sin cos sin4 4 6 6w i i             Let 1 3 3cos sin4 4w i    2 cos sin6 6w i      3 1 22 w w w 3 1 22 2(1)(1) 2  w w w     3 1 2 1 2 arg( ) arg 2 arg(2) arg( ) 3arg( ) w w w w w     30 3 4 6 4             2 cos sin4 4w i        (ii)   2 πarg arg 4 π π2 (cos i sin )4 4 nn n n n n w w nw n w n nw        When n is a muliple of 4, let n = 4k, where k is an integer. 4 4 2 (cos π i sin π) ( 1) 2 n k k k w k k    (because cos πk =1 when k even, 1 when k odd and sin πk =0 for all k.

3 a = 1 + i 3 =    3 i 2  e  1 + a + a2 + a3 + … + a9 = a a   1 1 10 = )3i1(1 21 10 3 i                 e = 3i 21 3 2 i10         e = 3i )i(21 2 3 2 110   = 3i i 3512513   3i 3i = i 3171512 //

4(a) 3 i 2 cos isin 6 6z          2 cos i sin6 6 n n n nz         * 2i Im( ) 2i 2 sin 0 6 nn n n nz z z         sin 06 n  ,6 n k k     The set of values of n is  : 6 ,n n k k    (b)    3 1 i 1 3z w     Note that OWPZ is a parallelogram, 1 3 2arg( ) tan 1 3w     2 3 6 2WOZ       OZ= 2z  OW= 1 i 3 2w     Since 2WOZ   and OZ OW , OWPZ is a square. (shown) 5arg( ) arg( ) 6 4 12z w z POZ         Also, 1 3 1arg( ) tan 3 1 z w     15 3 1 tan12 3 1        2 3 15 3 1 4 2 3tan 2 312 3 1 23 1          (deduced) Im P W i Z Re O

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6(i) (ii) or (iii) 7(a) Since is a root, Hence the equation becomes . Comparing coefficient of , . For , Hence the other 2 roots are .             44 2 3 44 3 2 4 2 3 i 3 4 i 3 6 i 3 4 i 3 i 3 18 9 i 4 3 12 3 z k k k k k k k k k                    4 3 2 real 4 3 12 3 0 4 3 3 0 0 (rej) or 3 z k k k k k        3 i 3z   3 i 3z       arg 3 i 3 4 7 15arg 3 i 3 , , ,... 4 4 4 3 7 15 , , ...4 4 4 4 3 1 7 15 , , ...4 4 4 4 Least 5 n n n n n                                 55 6 36 6nz z      iz 3i 2i 0 ik k     3 2 i 0z z   3 2 2 i ( i)( 1)z z z z az      2z ia 2 i 1 0z z   2i i 4(1)(1) i 5 i i 5 2(1) 2 2 z           i+i 5 i i 5 and 2 2   

(b)(i) (ii) Comparing real and imaginary parts, Since (iii) Subst , From consider the modulus of both sides to get This leads directly to the same equation for as above, and is solved similarly to get Then From eliminate to get Solve this quadratic to get rejecting the other root, and proceed to get Thus and Finally 1 1 1 [cos( ) isin( )] (cos isin ) w r r          500 3 40i 500 (cos i sin ) 3 40i ww rr        500 500cos 3 , sin 40r r r     23 40cos , sin500 500 r r   2 22 4 2 2 2 2 3 40 1 500 500 9 1600 250000 0 ( 100)(9 2500) 0 ( 10)( 10)(9 2500) 0 r r r r r r r r r                     , 10r r   10r 3 4cos , sin 5 5    3 4 10 i 6 8i5 5w         500 3 40i,w w   2 2500 3 40i (3 ) 40 .r r r     r 10.r  500 6 8i.3(10) 40iw    23 40cos , sin ,500 500 r r   r   2 2 216 1 cos . 1 3 500 16cos sin sin 500 4 15 50            cos 3 ,5  2 500 co 1 0.s 03r   10r 40 4 .5in 5s 00 r   i sinco 6 i.s 8w r    

8(i) By conjugate root theorem, is also a root. Hence is a factor of . By comparison of coefficients, , . Hence is also a root. (ii) Hence (iii) Hence Thus least and 1 i 3z        21 i 3 1 i 3 2 4z z z z        f z    2 3 22 4 4Az B z z z z bz       1A 1B 1z 3 2 4 1 0w bw w    2 3 1 1 14 0b w w w                    1 1 1 1, , or 1 1 i 3 1 i 3 1 i 3 1 i 31, , or 4 4 w z       i31 i 3 2z e     ,3 n n k kz            3n k 3n i3 3 2 8z e  