# How do you simplify w = [ ( 1 + i ) / ( 2 - i ) ] ^ 0.25?

Jun 11, 2016

${\left(\frac{1 + i}{2 - i}\right)}^{0.25}$

$= \sqrt[4]{\frac{1}{5} + \frac{3}{5} i}$

$= \frac{1}{2} \sqrt{\frac{\left(2 + \sqrt{\sqrt{10} + 1}\right) \sqrt{10}}{5}} + \frac{1}{2} \sqrt{\frac{\left(2 - \sqrt{\sqrt{10} + 1}\right) \sqrt{10}}{5}} i$

#### Explanation:

First simplify $\frac{1 + i}{2 - i}$ by multiplying both numerator and denominator by the Complex conjugate $2 + i$ of the denominator:

$\frac{1 + i}{2 - i} = \frac{\left(1 + i\right) \left(2 + i\right)}{\left(2 - i\right) \left(2 + i\right)} = \frac{2 + 3 i + {i}^{2}}{{2}^{2} - {i}^{2}} = \frac{1 + 3 i}{5} = \frac{1}{5} + \frac{3}{5} i$

Then note that:

${\left(\frac{1 + i}{2 - i}\right)}^{0.25} = \sqrt[4]{\frac{1}{5} + \frac{3}{5} i} = \sqrt{\sqrt{\frac{1}{5} + \frac{3}{5} i}}$

Then (see https://socratic.org/s/avemtNYy) since $a , b > 0$ we have:

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

So, with $a = \frac{1}{5}$ and $b = \frac{3}{5}$, we find:

$\sqrt{\frac{1}{5} + \frac{3}{5} i}$

$= \left(\sqrt{\frac{\sqrt{{\left(\frac{1}{5}\right)}^{2} + {\left(\frac{3}{5}\right)}^{2}} + \frac{1}{5}}{2}}\right) + \left(\sqrt{\frac{\sqrt{{\left(\frac{1}{5}\right)}^{2} + {\left(\frac{3}{5}\right)}^{2}} - \frac{1}{5}}{2}}\right) i$

$= \left(\sqrt{\frac{\sqrt{\frac{10}{25}} + \frac{1}{5}}{2}}\right) + \left(\sqrt{\frac{\sqrt{\frac{10}{25}} - \frac{1}{5}}{2}}\right) i$

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

Then with $a = \left(\sqrt{\frac{\sqrt{10} + 1}{10}}\right)$ and $b = \left(\sqrt{\frac{\sqrt{10} - 1}{10}}\right)$ we have:

${a}^{2} + {b}^{2}$

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

$= \left(\frac{\sqrt{10} + 1}{10}\right) + \left(\frac{\sqrt{10} - 1}{10}\right) = \frac{2 \sqrt{10}}{10} = \frac{\sqrt{10}}{5}$

$\frac{\sqrt{{a}^{2} + {b}^{2}} + a}{2} = \frac{\frac{\sqrt{10}}{5} + \left(\sqrt{\frac{\sqrt{10} + 1}{10}}\right)}{2}$

$= \frac{\frac{\sqrt{10}}{5} + \left(\sqrt{\frac{10 \left(\sqrt{10} + 1\right)}{100}}\right)}{2}$

$= \frac{2 \sqrt{10} + \sqrt{10} \sqrt{\sqrt{10} + 1}}{20}$

$= \frac{\left(2 + \sqrt{\sqrt{10} + 1}\right) \sqrt{10}}{20}$

Similarly:

$\frac{\sqrt{{a}^{2} + {b}^{2}} - a}{2} = \frac{\left(2 - \sqrt{\sqrt{10} + 1}\right) \sqrt{10}}{20}$

So:

${\left(\frac{1 + i}{2 - i}\right)}^{0.25}$

$= \sqrt{\frac{\left(2 + \sqrt{\sqrt{10} + 1}\right) \sqrt{10}}{20}} + \sqrt{\frac{\left(2 - \sqrt{\sqrt{10} + 1}\right) \sqrt{10}}{20}} i$

$= \frac{1}{2} \sqrt{\frac{\left(2 + \sqrt{\sqrt{10} + 1}\right) \sqrt{10}}{5}} + \frac{1}{2} \sqrt{\frac{\left(2 - \sqrt{\sqrt{10} + 1}\right) \sqrt{10}}{5}} i$