Answer: 
Step-by-step explanation:
<span>use De Moivre's Theorem:
⁵√[243(cos 260° + i sin 260°)] = [243(cos 260° + i sin 260°)]^(1/5)
= 243^(1/5) (cos (260 / 5)° + i sin (260 / 5)°)
= 3 (cos 52° + i sin 52°)
z1 = 3 (cos 52° + i sin 52°) ←← so that's the first root
there are 5 roots so the angle between each root is 360/5 = 72°
then the other four roots are:
z2 = 3 (cos (52 + 72)° + i sin (52+ 72)°) = 3 (cos 124° + i sin 124°)
z3 = 3 (cos (124 + 72)° + i sin (124 + 72)°) = 3 (cos 196° + i sin 196°)
z4 = 3 (cos (196 + 72)² + i sin (196 + 72)°) = 3 (cos 268° + i sin 268°)
z5 = 3 (cos (268 + 72)° + i sin (268 + 72)°) = 3 (cos 340° + i sin 340°) </span>
The answer is NO open circle is not included in an inequality. The other person answered it incorrectly
