Question

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Answer 1

The complex number to find the 4th root is given to be:

$2\sqrt{3}-2i$

Step 1: Transform the complex number in the r.cisθ form

$\begin{gathered} r=\sqrt{(2\sqrt{3})^2+(-2)^2}=\sqrt{12+4}=\sqrt{16} \\ r=4 \\ and \\ \theta=\arctan(\frac{-2}{2\sqrt{3}})=-\frac{\pi}{6} \end{gathered}$

Therefore, the polar form is:

$\Rightarrow4cis(-\frac{\pi}{6})$

Step 2: According to the De Moivre's Formula, all n-th roots of a complex number

$\begin{gathered} rcis\theta \\ \Rightarrow r^{\frac{1}{n}}cis(\frac{\theta+2\pi k}{n}) \\ for \\ k=0...n-1 \end{gathered}$

Therefore, the roots are as follows.

$n=4$

k = 0

$\begin{gathered} \Rightarrow4^{\frac{1}{4}}cis(\frac{-\frac{\pi}{6}+2\pi(0)}{4}) \\ \Rightarrow\sqrt{2}cis(-\frac{\pi}{24}) \end{gathered}$

k = 1

$\begin{gathered} \Rightarrow\sqrt{2}cis(\frac{-\frac{\pi}{6}+2\pi}{4}) \\ \Rightarrow\sqrt{2}cis(\frac{11\pi}{24}) \end{gathered}$

k = 2

$\Rightarrow\sqrt{2}cis(\frac{23\pi}{24})$

k = 3

$\Rightarrow\sqrt{2}cis(\frac{35\pi}{24})$