Follow steps (a), (b), (c) above to find all the values of the indicated roots.

$\mathbf{2}\mathbf{+}\mathbf{2}\mathbf{i}\sqrt{\mathbf{3}}$

The given expression is $2+2i\sqrt{3}.$

Complex numbers comprise real numbers and imaginary numbers; a complex can be written in the form of:

$\mathit{z}\mathbf{=}\mathit{a}\mathbf{+}\mathit{i}\mathit{b}$

Here a and b are real numbers, and i is the imaginary number which is known as iota, whose value is $\sqrt{\mathbf{-}\mathbf{1}}$.

The Complex number is in the form $2+2\sqrt{3i}$.

$x=2,y=2\sqrt{3}$

The polar coordinates of the point are in the form of $z=r{e}^{i\theta }$.

$$\begin{gathered} r=\sqrt{x^2+y^2}=\sqrt{2^2+{\left(2\sqrt{3}\right)}^3}=4 \\ \theta =\frac{\mathrm{\pi }}{6},\frac{7\mathrm{\pi }}{6} \end{gathered}$$

The equation role="math" localid="1658730779449" $z=r{e}^{1\theta }$can also be written in another form.

$$\begin{gathered} z^{\frac{1}{n}}={\left(r{e}^{1\theta }\right)}^{\frac{1}{n}} \\ {z}^{\frac{1}{n}}=r^{\frac{1}{n}}{e}^{\frac{1\theta }{n}} \\ {z}^{\frac{1}{n}=\sqrt[n]{r}}\left(\cos \frac{\theta }{n}+is\mathrm{in}\frac{\theta }{n}\right) \end{gathered}$$ ……….. (1)

When n =2, the equation becomes the $2^{nd}$root of the complex number.

$$\begin{gathered} z^{\frac{1}{2}}=r^{\frac{1}{2}}{e}^{\frac{1\theta }{2}} \\ r=4 \\ \theta =\frac{\mathrm{\pi }}{6},\frac{7\mathrm{\pi }}{6} \end{gathered}$$

$$\begin{gathered} z_0=2e^{\frac{i\mathrm{\pi }}{6}} \\ =2\left(\cos \frac{\mathrm{\pi }}{6}+i\sin \frac{\mathrm{\pi }}{6}\right) \\ =\sqrt{3}+i \\ {z}_0=2e^{\frac{7i\mathrm{\pi }}{6}} \\ =2\left(\cos \frac{7\mathrm{\pi }}{6}+i\sin \frac{7\mathrm{\pi }}{6}\right) \\ =-\sqrt{3}-i \end{gathered}$$

Hence, the value of $2+2i\sqrt{3}$is in the form of $z_0$and $z_1$:$z_0=\sqrt{3}+i,z_1=-\sqrt{3}-i$.