Find the fifth roots of 243(cos 260° + i sin 260°).?
How can I find the fifth roots of the complex number represented in polar form as 243 (cos 260° + i sin 260°)?
4 Answers
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: zâµ = 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°
z₠= 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))
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