Question

(2-i sqrt(3)) / 2+i sqrt(3))Solve in standard complex form, i = imaginary number

Answer 1

We want to solve the following equation

`\frac{2-i\sqrt[]{3}}{2+i\sqrt[]{3}}`

Where i is the imaginary number.

Using the following property

`(a+bi)(a-bi)=a^2+b^2`

We can rewrite our equation by multiplying both numerator and denominator by the denominator complex conjugate.

`\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})}`

Expanding those products, we have

`\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^{}} \\ =(1)/(7)-\frac{4\sqrt[]{3}}{7^{}}i \end{gathered}`

And this is the solution for our problem.

`(1)/(7)-\frac{4\sqrt[]{3}}{7^{}}i`