Question

Find all cube roots of the complex number. Leave answers in trigonometric form.

Answer 1

We want to find the cube root of

`4-4i√(3)`

This can be written as

`\sqrt[3]{(4-i.4.√(3))}`

Step 2

Step 3

Step 4

Step 5

So, 3 degree root from input complex number has exactly 3 values:

These become

`\begin{gathered} \alpha_0=2.(cos(-(\pi)/(9))+i.sin(-(\pi)/(9))) \\ \alpha_1=2.(cos(-(5.\pi)/(9))+i.sin((5.\pi)/(9))) \\ \alpha_2=2.(cos(-(11.\pi)/(9))+i.sin((11.\pi)/(9))) \end{gathered}`

Hence the answer is

`\begin{gathered} \alpha_0=2.(cos(-(\pi)/(9))+i.sin(-(\pi)/(9))) \\ \alpha_1=2.(cos(-(5.\pi)/(9))+i.sin((5.\pi)/(9))) \\ \alpha_2=2.(cos(-(11.\pi)/(9))+i.sin((11.\pi)/(9))) \end{gathered}`