# How do you simplify (sqrt3-i)div(2-2sqrt3i)?

Dec 4, 2016

The answer is $= \frac{\sqrt{3}}{4} + \frac{i}{4}$

#### Explanation:

When you have a fraction of complex numbers,

$w = {z}_{1} / {z}_{2}$

Multiply numerator and denominator by the conjugate of the denominator

$w = \frac{{z}_{1} \cdot {\overline{z}}_{2}}{{z}_{2} \cdot {\overline{z}}_{2}}$

If the complex number is $z = a + i b$

The conjugate is $\overline{z} = a - i b$

and ${i}^{2} = - 1$

$w = \frac{\sqrt{3} - i}{2 - 2 \sqrt{3} i}$

$= \frac{\left(\sqrt{3} - i\right) \left(2 + 2 \sqrt{3} i\right)}{\left(2 - 2 \sqrt{3} i\right) \left(2 + 2 \sqrt{3} i\right)}$

=(2sqrt3+2*3i-2i-2sqrt3i^2)/(4-4*3i^2

$= \frac{4 \sqrt{3} + 4 i}{16}$

$= \frac{\sqrt{3}}{4} + \frac{i}{4}$