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

Z = 243.[cos(260°) + i.sin(260°)]

The modulus of Z is: M = 243

The argument of Z is: β = 260°

Now you must find z (a complex number) such as: z⁵ = Z

According 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/5

For the modulus

m = M^(1/5) ��� given that: M = 243

m = 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 z

For the argument

α = β/5 → given that: β = 260°

α = 260/5

α = 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°)]

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 degrees

3^5 * (cos(260 + 360 * k) + i * sin(260 + 360 * k)

Take the fifth root of that

3 * (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))

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

5 and 22