find the fifth roots of 243(cos 260° + i sin 260°)
Accepted Solution
A:
Answer:z1=3 cos (52 + i sin 52)z2 =3 cos (124 + i sin 124)z3 = 3 cos (196 + i sin 196)z4 =3 cos (268 + i sin 268)z5= 3 cos (340 + i sin 340)Step-by-step explanation:To find the fifth roots of 243 (cos 260° + i sin 260°).z ^ 1/5 = r^1/5 ( cis ( theta + 360 *k)/5) where k=0,1,2,3,4 So the first root of 243 (cos 260° + i sin 260°) is z1 = 243^1/5 ( cis ( 260 + 360 *0)/5) 3 cis ( 260/5) = 3 cis (52) = 3 cos (52 + i sin 52) The second root of 243 (cos 260° + i sin 260°) is z2 = 243^1/5 ( cis ( 260 + 360 *1)/5) 3 cis ( 620/5) = 3 cis (124) = 3 cos (124 + i sin 124) The third root of 243 (cos 260° + i sin 260°) is z3 = 243^1/5 ( cis ( 260 + 360 *2)/5) 3 cis ( 980/5) = 3 cis (196) = 3 cos (196 + i sin 196) The fourth root of 243 (cos 260° + i sin 260°) is z4 = 243^1/5 ( cis ( 260 + 360 *3)/5) 3 cis ( 1340/5) = 3 cis (268) = 3 cos (268 + i sin 268) The fifth root of 243 (cos 260° + i sin 260°) is z5 = 243^1/5 ( cis ( 260 + 360 *4)/5) 3 cis ( 1700/5) = 3 cis (340) = 3 cos (340 + i sin 340)