# How do you simplify (2-sqrt2i)/(3+sqrt6i)?

Dec 27, 2016

The answer is $= \frac{6 - \sqrt{12}}{15} - \frac{2 \sqrt{6} + 3 \sqrt{2}}{15} i$

#### Explanation:

Let a complex number be $z = {z}_{1} / {z}_{2}$

To simplify, we multiply the numerator and the denominator by the conjugate of the denominator

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

If $w = a + i b$, the conjugate is $\overline{w} = a - i b$

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

Here,

$z = \frac{2 - \sqrt{2} i}{3 + \sqrt{6} i}$

$z = \frac{\left(2 - \sqrt{2} i\right) \left(3 - \sqrt{6} i\right)}{\left(3 + \sqrt{6} i\right) \left(3 - \sqrt{6} i\right)}$

$= \frac{6 - 2 \sqrt{6} i - 3 \sqrt{2} i - \sqrt{12}}{9 + 6}$

$= \frac{6 - \sqrt{12}}{15} - \frac{2 \sqrt{6} + 3 \sqrt{2}}{15} i$