Find all the cube roots of the complex number 64(cos(219)+i sin (219)). Leave answers in polar form and show all work.
To find the cube roots of a complex number in polar form, we need to find the cube roots of the magnitude and add 120 degrees to the argument for each root.
First, let's find the magnitude of the complex number:
|64(cos(219)+i sin (219))| = 64
Now, let's find the argument of the complex number:
arg = 219 degrees
Now, let's find the cube roots of the magnitude (64) and add 120 degrees to the argument for each root:
1st cube root:
∛64 = 4
arg1 = (219+120) = 339 degrees
So, the first cube root is 4(cos(339) + i sin(339))
2nd cube root:
∛64 = 4
arg2 = (219+120+120) = 459 degrees
So, the second cube root is 4(cos(459) + i sin(459))
3rd cube root:
∛64 = 4
arg3 = (219+120+120+120) = 579 degrees
So, the third cube root is 4(cos(579) + i sin(579))
Therefore, the cube roots of the complex number 64(cos(219)+i sin (219)) in polar form are:
4(cos(339) + i sin(339)), 4(cos(459) + i sin(459)), and 4(cos(579) + i sin(579)).