We want to solve the following equation
[tex]\frac{2-i\sqrt[]{3}}{2+i\sqrt[]{3}}[/tex]Where i is the imaginary number.
Using the following property
[tex](a+bi)(a-bi)=a^2+b^2[/tex]We can rewrite our equation by multiplying both numerator and denominator by the denominator complex conjugate.
[tex]\frac{2-i\sqrt[]{3}}{2+i\sqrt[]{3}}=\frac{(2-i\sqrt[]{3})\cdot(2-i\sqrt[]{3})}{(2+i\sqrt[]{3)}\cdot(2-i\sqrt[]{3})}[/tex]Expanding those products, we have
[tex]\begin{gathered} \frac{(2-i\sqrt[]{3})\cdot(2-i\sqrt[]{3})}{(2+i\sqrt[]{3)}\cdot(2-i\sqrt[]{3})}=\frac{(2-i\sqrt[]{3})\cdot(2-i\sqrt[]{3})}{2^2+(\sqrt[]{3})^2} \\ =\frac{4-2i\sqrt[]{3}-2i\sqrt[]{3}-3}{4+3^{}} \\ =\frac{1-4i\sqrt[]{3}}{7^{}} \\ =\frac{1}{7}-\frac{4\sqrt[]{3}}{7^{}}i \end{gathered}[/tex]And this is the solution for our problem.
[tex]\frac{1}{7}-\frac{4\sqrt[]{3}}{7^{}}i[/tex]