# Question #db9f4

Aug 23, 2016

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

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

#### Explanation:

We will use the properties that $\sqrt{2} \cdot \sqrt{2} = 2$ and $i \cdot i = - 1$ to eliminate the imaginary and irrational components from the denominators:

$y = \frac{2}{- \sqrt{2} i}$

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

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

$= \frac{2 \sqrt{2} i}{- 2 \left(- 1\right)}$

$= \frac{2 \sqrt{2} i}{2}$

$= \sqrt{2} i$

For $y = \frac{2}{\sqrt{2} i}$, we could go through the same process, however we can also just multiply the first result by $- 1$:

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

Note that this is a simple case of more general techniques for eliminating square roots or imaginary components from a denominator. The method is the similar for both, and relies on the identity $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$.

$\frac{1}{a + b \sqrt{r}} = \frac{a - b \sqrt{r}}{\left(a + b \sqrt{r}\right) \left(a - b \sqrt{r}\right)}$

$= \frac{a - b \sqrt{r}}{{a}^{2} - {\left(b \sqrt{r}\right)}^{2}}$

$= \frac{a - b \sqrt{r}}{{a}^{2} - {b}^{2} r}$

$\frac{1}{a + b i} = \frac{a - b i}{\left(a + b i\right) \left(a - b i\right)}$

$= \frac{a - b i}{{a}^{2} - {\left(b i\right)}^{2}}$

$= \frac{a - b i}{{a}^{2} + {b}^{2}}$

We say that $x - y$ is the conjugate of $x + y$, and $a - b i$ is the complex conjugate of $a + b i$.