Given the following complex number:
[tex]2\sqrt{3}-2i[/tex]
We will find the fourth root by the following formula:
[tex]\begin{gathered} if,z=r(cos\theta+i*sin\theta) \\ so,\sqrt[n]{r}(cos\frac{\theta+2\pi k}{n}+i*sin\frac{\theta+2\pi k}{n}) \end{gathered}[/tex]
For the given number, we will find the fourth root
n = 4, k = 0, 1, 2, 3
First, we will convert the number from rectangular form to the polar form:
[tex]z=2\sqrt{3}-2i=4(cos330+i*sin330)[/tex]
Apply the previous formula, the four roots will be as follows:
[tex]\begin{gathered} k=0\rightarrow\sqrt[4]{4}(cos\frac{330}{4}+i*sin\frac{330}{4})=\sqrt[]{2}(cos82.5+i*sin82.5) \\ \\ k=1\rightarrow\sqrt[4]{4}(cos\frac{330+90}{4}+i*sin\frac{330+90}{4})=\sqrt{2}(cos105+i*sin105) \\ \\ k=2\rightarrow\sqrt[4]{4}(cos\frac{330+180}{4}+i*sin\frac{330+180}{4})=\sqrt{2}(cos127.5+i*sin127.5) \\ \\ k=3\rightarrow\sqrt[4]{4}(cos\frac{330+270}{4}+i*sin\frac{330+270}{4})=\sqrt{2}(cos150+i*sin150) \end{gathered}[/tex]
