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)
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)