# What does (1-3i)/sqrt(1+3i) equal?

Jan 5, 2016

$\frac{1 - 3 i}{\sqrt{1 + 3 i}}$

$= \left(- 2 \sqrt{\frac{\sqrt{10} + 1}{2}} + \frac{3}{2} \sqrt{\frac{\sqrt{10} - 1}{2}}\right) - \left(2 \sqrt{\frac{\sqrt{10} - 1}{2}} + \frac{3}{2} \sqrt{\frac{\sqrt{10} + 1}{2}}\right) i$

#### Explanation:

In general the square roots of $a + b i$ are:

$\pm \left(\left(\sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} + a}{2}}\right) + \left(\frac{b}{\left\mid b \right\mid} \sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} - a}{2}}\right) i\right)$

In the case of $1 + 3 i$, both Real and imaginary parts are positive, so it's in Q1 and has a well defined principal square root:

$\sqrt{1 + 3 i}$

$= \sqrt{\frac{\sqrt{{1}^{2} + {3}^{2}} + 1}{2}} + \sqrt{\frac{\sqrt{{1}^{2} + {3}^{2}} - 1}{2}} i$

$= \sqrt{\frac{\sqrt{10} + 1}{2}} + \sqrt{\frac{\sqrt{10} - 1}{2}} i$

So:

$\frac{1 - 3 i}{\sqrt{1 + 3 i}}$

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

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

$= \frac{{\left(1 - 3 i\right)}^{2} \sqrt{1 + 3 i}}{4}$

$= \frac{1}{4} {\left(1 - 3 i\right)}^{2} \left(\sqrt{\frac{\sqrt{10} + 1}{2}} + \sqrt{\frac{\sqrt{10} - 1}{2}} i\right)$

$= \frac{1}{4} \left(- 8 - 6 i\right) \left(\sqrt{\frac{\sqrt{10} + 1}{2}} + \sqrt{\frac{\sqrt{10} - 1}{2}} i\right)$

$= - \frac{1}{2} \left(4 + 3 i\right) \left(\sqrt{\frac{\sqrt{10} + 1}{2}} + \sqrt{\frac{\sqrt{10} - 1}{2}} i\right)$

$= - \frac{1}{2} \left(\left(4 \sqrt{\frac{\sqrt{10} + 1}{2}} - 3 \sqrt{\frac{\sqrt{10} - 1}{2}}\right) + \left(4 \sqrt{\frac{\sqrt{10} - 1}{2}} + 3 \sqrt{\frac{\sqrt{10} + 1}{2}}\right) i\right)$

$= \left(- 2 \sqrt{\frac{\sqrt{10} + 1}{2}} + \frac{3}{2} \sqrt{\frac{\sqrt{10} - 1}{2}}\right) - \left(2 \sqrt{\frac{\sqrt{10} - 1}{2}} + \frac{3}{2} \sqrt{\frac{\sqrt{10} + 1}{2}}\right) i$