Find all cube roots of the complex number 64(cos(219degrees)+I sin(219 degrees)). Leave answers in polar form and show all work.
To find all cube roots of a complex number in polar form, we first need to convert the complex number to polar form.
Given complex number: 64(cos(219°) + i sin(219°))
To convert to polar form:
r = |z| = √(Re(z)^2 + Im(z)^2) = √(64^2 + 0^2) = 64
θ = arg(z) = tan^(-1)(Im(z)/Re(z)) = tan^(-1)(sin(219°)/cos(219°)) ≈ -1.326 radians or -75.3 degrees
The complex number in polar form is therefore: 64(cos(-1.326) + i sin(-1.326))
Now, to find all cube roots of this complex number, we can use De Moivre's Theorem, which states that:
(z^(1/n)) = r^(1/n)*(cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)), where k = 0,1,...,n-1
For n = 3, we have:
Cube root 1: 64^(1/3)(cos((-1.326 + 2kπ)/3) + i sin((-1.326 + 2kπ)/3)), for k = 0,1,2
Therefore, the three cube roots in polar form are:
Cube root 1: 4(cos((-1.326)/3) + i sin((-1.326)/3))
≈ 4(cos(-0.442) + i sin(-0.442))
≈ 3.797(cos(107.254°) + i sin(107.254°))
Cube root 2: 4(cos((2.744)/3) + i sin((2.744)/3))
≈ 4(cos(-0.361) + i sin(-0.361))
≈ 3.797(cos(-20.746°) + i sin(-20.746°))
Cube root 3: 4(cos((5.69)/3) + i sin((5.69)/3))
≈ 4(cos(-0.558) + i sin(-0.558))
≈ 3.797(cos(-31.554°) + i sin(-31.554°))
Therefore, the three cube roots of the given complex number are:
3.797(cos(107.254°) + i sin(107.254°))
3.797(cos(-20.746°) + i sin(-20.746°))
3.797(cos(-31.554°) + i sin(-31.554°))