Question

Find the cube roots of 125 i. Graph each cube root as a vector the complex plane. Choose the correct cube roots below O A. 5(cos 90° + i sin 90°), 5(cos 210° + i sin 210°), 5(cos 330° + i sin 330%) OB. 5(cos 30° + i sin 30°), 5(cos 150° + i sin 150°), 5(cos 270° + i sin 270º) O C. 125(cos 0° + i sin 0°), 125(cos 120° + i sin 120°), 125(cos 240° + i sin 240°) OD. 125(cos 60° + i sin 60), 125(cos 180° + i sin 180°), 125(cos 300° + i sin 300°) Choose the correct graph of the cube roots of 125 i below. OA. ов. Oc. OD. 126 -125

Answer 1

Solution: 125i Since cos(?)=0 125 (i) = 125 cos (5) +i Sin ( and sin (E)=1 > 125i Then LOS () +isin () = Now [125(cos(¥)+isin (¥)] Since 3 = (125) 3 [cos()+isin() (E )+i sin ()] [cos(e) +isincez cos (axima 9) +isin ( 267 +2) K = 0,1,2,--, N-1 (53) * [urlemos. )disin (4:1) where K=0,1,2 4KTT + TT AKIT + TT 5 ti sin 3.2 where K=0,1,2

For K=1, (1) becomes 4 KTT + TT 4KIT + TT 6. ) +isin :)] where K=0, 1, 2 (! There fore the cube roots of 125i are given below: For K=0, ( be comes = 5 5[cos (10:76 +1) +i sin (4:07*")] [ws () +i sin sin (5) 5 [cos 1980 1809 = Since π 1800 (180 tisin = 5 [Cos (30°) + i sin (300)] = Therefore the first Cube root is Zo = 5 5 COS(309) + i sin (300)] (2) 5 [cas (4473")is (3) S[cos ( 5 ) + i sin (9)] [cos (5X16°) +isin (5X180°)] since TT = 180° = 5

=5[cos (5x30) + + i sin(5x300)] Cos (1509) + i sin (1509) = 5 5[o There fore Second Cure root is Z, = 5 COS(1509) +i sîn (1500)] For K=2, (1) becomes 4.2 T +T tisin 6. = 5 5[cas (12.1+T) *) [cos (97) + i sin ( PT (93.906) +i sin (9x1900)] Since IT = 180° = 5 cos E 5 cos (9x300) + i sin (9x300) ] - 5 cos (2709) + i sin (270) cos(20) (270"] Therefore the third cube root is cos (260°) +isin (2700)] Z = : 5 :2 (4)

Therefore the cubroots of are Z from (2), (3), (4), we get 125i Zo = 5[cos(309) +isin (300] Z= 5 [605(1509) + i sin (1509 Z2 = 5 [cos (270") +i sin (2700)] Zz Answer Conclusion There fore option (B) is the correct option. (Answer)