# How do you simplify (1/2+sqrt3i)/(1-sqrt2i)?

Jun 11, 2017

$\frac{\frac{1}{2} + \sqrt{3} i}{1 - \sqrt{2} i} = \left(\frac{1}{2 \sqrt{2}} - \sqrt{3}\right) + \left(\frac{1}{2} + \frac{\sqrt{3}}{\sqrt{2}}\right) i$

#### Explanation:

To simpplify ratio of two complex numbers such as $\frac{a + b i}{c + \mathrm{di}}$, just multiply numerator and denominator by the complex conjugate of denominator i.e. $c - \mathrm{di}$.

Here, we have $\frac{\frac{1}{2} + \sqrt{3} i}{1 - \sqrt{2} i}$ and hence we should multipky numerator and denominator by $1 + \sqrt{2} i$

$\therefore \frac{\frac{1}{2} + \sqrt{3} i}{1 - \sqrt{2} i} = \frac{\left(\frac{1}{2} + \sqrt{3} i\right) \left(1 + \sqrt{2} i\right)}{\left(1 - \sqrt{2} i\right) \left(1 + \sqrt{2} i\right)}$

= $\frac{\frac{1}{2} + \frac{\sqrt{2} i}{2} + \sqrt{3} i + \sqrt{6} {i}^{2}}{{1}^{2} - {\left(\sqrt{2} i\right)}^{2}}$

= $\frac{\frac{1}{2} + \frac{\sqrt{2} i}{2} + \sqrt{3} i + \sqrt{6} \left(- 1\right)}{1 - 2 {i}^{2}}$

= $\frac{\frac{1}{2} - \sqrt{6} + \left(\frac{1}{\sqrt{2}} i + \sqrt{3} i\right)}{1 - 2 \left(- 1\right)}$

= $\frac{1 - 2 \sqrt{6} + \left(\sqrt{2} + 2 \sqrt{3}\right) i}{2 \sqrt{2}}$

= $\frac{1 - 2 \sqrt{6}}{2 \sqrt{2}} + \left(\sqrt{2} + 2 \sqrt{3}\right) \frac{i}{2 \sqrt{2}}$

= $\left(\frac{1}{2 \sqrt{2}} - \sqrt{3}\right) + \left(\frac{1}{2} + \frac{\sqrt{3}}{\sqrt{2}}\right) i$