Note: Enter your answer and show all the steps that you use to solve this problem in the space provided. Find all cube roots of the complex number 64(cos(219°)+isin(219°)) . Leave answers in polar form and show all work.
To find all cube roots of the complex number 64(cos(219°)+isin(219°)), we first need to express this complex number in polar form.
Let z = 64(cos(219°)+isin(219°))
We can rewrite z in polar form as:
z = 64cis(219°)
Now, we can find the cube roots of z by using De Moivre's Theorem.
The cube roots of z are given by:
∛z = ∛64(cis(219°)) = 64^(1/3) * cis(219°/3 + 2kπ/3) for k = 0,1,2
We can calculate the cube roots by evaluating ∛64 and finding the three distinct arguments:
∛64 = 4
For k = 0:
∛z_0 = 4 * cis(73°)
For k = 1:
∛z_1 = 4 * cis(73° + 2π/3) = 4 * cis(433°)
For k = 2:
∛z_2 = 4 * cis(73° + 4π/3) = 4 * cis(793°)
Therefore, the three cube roots of the complex number 64(cos(219°)+isin(219°)) in polar form are:
∛z_0 = 4 * cis(73°)
∛z_1 = 4 * cis(433°)
∛z_2 = 4 * cis(793°)