Find the fifth roots of 243(cos 260° + i sin 260°).?

Z = 243.[cos(260°) + i.sin(260°)]The modulus of Z is: M = 243The argument of Z is: β = 260° Now you must find z (a complex number) such as: z⁵ = ZAccording your mathematics courses, you’ve seen that:the modulus of z is m such as: m = M^(1/5)the argument of z is α such as: α = β/5For the modulus m = M^(1/5) → given that: M = 243m = 243^(1/5) → but you know that: 243 = 2 3 3 3 3 = 3^(5)m = [3^(5)]^(1/5) → you know that: [x^(a)]^(b) = x^(ab)m = 3^[5 * (1/5)]m = 3^(1)m = 3 ← this the modulus of zFor the argumentα = β/5 → given that: β = 260°α = 260/5α = 52° ← this the argument of the first root of Z, and to obtain the next root, you add (360°/5) → i.e. 72°z₁ = 3.[cos(52°) + i.sin(52°)] → then you add an angle of (360°/5) i.e. 72°z₂ = 3.[cos(124°) + i.sin(124°)] → then you add an angle of (360°/5) i.e. 72°z₃ = 3.[cos(176°) + i.sin(176°)] → then you add an angle of (360°/5) i.e. 72°z₄ = 3.[cos(228°) + i.sin(228°)] → then you add an angle of (360°/5) i.e. 72°z₅ = 3.[cos(280°) + i.sin(280°)]... Show More

243 = 3^5(cos(t) + i sin(t))^n = cos(n t) + i sin(n t)Remember that cos(t) and sin(t) are periodic to 360 degrees3^5 (cos(260 + 360 k) + i sin(260 + 360 k)Take the fifth root of that3 (cos(52 + 72 k) + i sin(52 + 72 k))3 (cos(52) + i sin(52))3 (cos(124) + i sin(124))3 (cos(196) + i sin(196))3 (cos(268) + i sin(268)) 3 * (cos(340) + i * sin(340))... Show More

3 ( cos { [ 260 + n360]/5} + i sin { [ 260 + n360]/5 } for n= 0,1,2,3,4... Show More

5 and 22... Show More