Course Information x Eb Elementary Linear Age x) e Math 1B0 Summe x INFINITE WAR
ID: 3039093 • Letter: C
Question
Course Information x Eb Elementary Linear Age x) e Math 1B0 Summe x INFINITE WARFARE x C Chegg study I Guicec sc x C Secure www child 2.php?assign Your Mark: 50a) 5(a) 5(b) 50b) Problem #6: Which of the following is a solution to the equation (1- V3i)? 1/3 [cos(23 t/18) ist (23rt 18) (A) 2 sin(8nt/9)I (C) 2 cos(10nui9) i sin(10rt/9 F) 2. cos (13Ti18) i sin 131t/l 18)1 sin 19t 18)1 (E) 2 cos (7r/9) i sin 71t/9 (112 9) i sin cos(17at 18) (G) 2 i sin(17w 18) COS prohlem afi: Selec v Just Save Your work has been saved inauk La Admin Paue) Sub Problem 46 for Grading Problem #6 Attempt 21 Attempt 2 Attempt 3 Your Answer r Mark 65 Problem #7: v3 Find the value of 65 Problem 7: if your answer is a bi, hen enter a, b in the answer box Just Save l Your work has been saved Lnack La Admir Pdue) Sub Problem 47 for Grading Problem #7 Attempt 21 Attempt 2 Attempt 3 Your Answer r Mark Type here to search 1042 PM 2017-U6-02Explanation / Answer
1) z^3 = ( 1 - sqrt 3 i )
z = r ( cos theta + i sin theta )
z^3 = r^3 ( cos 3 theta + i sin 3 theta ) = 1 - sqrt 3 i
modulus | 1 - sqrt 3 i | = 2
Applying Demories theorem
zk = 2^1/3[ cos(2pik -pi/3) + i*sin(2pik - pi/3) ]^1/3
= 2^1/3[cos(2pik/3 - pi/9) +i*sin(2pik/3 - pi/9) ]
k = 0 ; z1 = 2^1/3[(cos(-pi/9) + isin(-pi/9) ]
k= 1; z2 = 2^1/3[cos(2p/3i - pi/9) +isin(2pi/3 -pi/9) ] = 2^1/3 [ cos ( 5pi/9 ) + i sin (5pi/9 ) ]
k= 2 ; z3 = 2^1/3[cos(4pi/3 - pi/9) + i*sin(4pi/3 - pi/9)] = 2^1/3 [ cos ( 7pi/9) + i sin ( 7 pi/9 ) ]
so one of the solutions is z = 2^1/3 [ cos ( 7pi/9) + i sin ( 7 pi/9 ) ]
option E is correct