Solution:
Given the complex number:
[tex]2\sqrt{3}\text{ -2i}[/tex]
Step 1: Express the complex number in polar form.
In polar form, we have
[tex]\begin{gathered} z=r(cos\theta+isin\theta)=rcis(\theta) \\ where \\ r\Rightarrow modulus\text{ of the complex number} \\ \theta\Rightarrow argument\text{ of the complex number} \end{gathered}[/tex]
Step 2: Evaluate the modulus of the complex number,
The modulus of the complex number is expressed as
[tex]\begin{gathered} r=\sqrt{x^2+y^2} \\ in\text{ this case,} \\ x=2\sqrt{3}\text{ ,y=-2} \\ thus, \\ r=\sqrt{(2\sqrt{3})^2+(-2)^2} \\ =\sqrt{12+4} \\ =\sqrt{16} \\ \Rightarrow r=4 \end{gathered}[/tex]
Step 3: Evaluate the argument of the complex number.
The argument of the complex number is expressed as
[tex]\begin{gathered} \theta=\tan^{-1}(\frac{y}{x}) \\ \Rightarrow\theta=\tan^{-1}(-\frac{2}{2\sqrt{3}})=\tan^{-1}(-\frac{1}{\sqrt{3}}) \\ =-\frac{\pi}{6} \\ \end{gathered}[/tex]
Thus, in polar form, the complex number becomes
[tex]z=4cis(-\frac{\pi}{6})[/tex]
To evaluate the fourth root, we use the De Moivres's theorem.
According, to the DeMoivres's threorem,
[tex]z^n=r^ncis(n\theta)[/tex]
In this case,
[tex]n=\frac{1}{4}[/tex]
Thus,
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