Question

Find the cube roots of 4−6i4−6iShow all your work.Include an explanation and diagram showing how DeMoivre's Theorem helps to solve this problem.

Answer 1

Given the following complex number

`z=4-6i`

We will find the cube root of the complex number using the following formula:

`^3√(z)=\sqrt[3]z*(cos\text{ }(\theta+2\pi k)/(3)+i*sin\text{ }(\theta+2\pi k)/(3))`

The formula is called De Moivre's theorem of the nth root

We have substituted n = 3

So, first, we will convert the given number from the rectangular form to the polar form

`\begin{gathered} |z|=√(4^2+6^2)\approx7.211 \\ \theta=tan^(-1)(-6)/(4)=303.7\degree \end{gathered}`

Substitute the magnitude and the angle and k = 0, 1, 2

So, there are 3 cubic roots of the given number as follows:

`\begin{gathered} k=0\rightarrow z_1=\sqrt[3]{7.211}(cos(303.7)/(3)+i*sin(303.7)/(3))=1.932(cos101.23+i*sin101.23) \\ \\ k=1\rightarrow z_2=\sqrt[3]{7.211}(cos(303.7+2\pi)/(3)+i*sin(303.7+2\pi)/(3))=1.932(cos221.23+i*sin221.23) \\ \\ k=2\rightarrow z_3=\sqrt[3]{7.211}(cos(303.7+4\pi)/(3)+i*sin(303.7+4\pi)/(3))=1.932(cos341.23+i*sin341.23) \end{gathered}`