# How do you solve x^4+i=0?

Apr 4, 2017

$= c i s \left(\frac{- \pi}{8}\right) , c i s \left(\frac{3 \pi}{8}\right) , c i s \left(\frac{7 \pi}{8}\right) , c i s \left(\frac{11 \pi}{8}\right)$

#### Explanation:

${x}^{4} = - i$

$x = {\left(- i\right)}^{\frac{1}{4}}$

Using Euler's formula (or DeMoivre's theorem, if you like):

$- i = \cos \left(\frac{- \pi}{2} + 2 n \pi\right) + i \sin \left(\frac{- \pi}{2} + 2 n \pi\right) \equiv c i s \left(\frac{- \pi}{2} + 2 n \pi\right)$

$= {e}^{i \left(\frac{- \pi}{2} + 2 n \pi\right)}$

So:

${\left(- i\right)}^{\frac{1}{4}} = {e}^{i \frac{\frac{- \pi}{2} + 2 n \pi}{4}}$

$= {e}^{i \left(\frac{- \pi}{8} + \frac{n \pi}{2}\right)}$

$= c i s \left(\frac{- \pi}{8} + \frac{n \pi}{2}\right)$

Running through the $n$'s:

$= c i s \left(\frac{- \pi}{8}\right) , c i s \left(\frac{3 \pi}{8}\right) , c i s \left(\frac{7 \pi}{8}\right) , c i s \left(\frac{11 \pi}{8}\right)$

And then they start repeating.

May 21, 2017

The four roots are:

$x = \pm \left(\frac{\sqrt{2 + \sqrt{2}}}{2} - \frac{\sqrt{2 - \sqrt{2}}}{2} i\right)$

$x = \pm \left(\frac{\sqrt{2 - \sqrt{2}}}{2} + \frac{\sqrt{2 + \sqrt{2}}}{2} i\right)$

#### Explanation:

Here's an alternative non-trigonometric method...

If $a , b$ are real numbers with $b \ne 0$, then the square roots of $a + b i$ are:

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

Applying this to the square root of $- i = \textcolor{b l u e}{0} + \left(\textcolor{b l u e}{- 1}\right) i$ we find that its square roots are:

$\pm \left(\left(\sqrt{\frac{\sqrt{{\textcolor{b l u e}{0}}^{2} + {\left(\textcolor{b l u e}{- 1}\right)}^{2}} + \textcolor{b l u e}{0}}{2}}\right) - \left(\sqrt{\frac{\sqrt{{\textcolor{b l u e}{0}}^{2} + {\left(\textcolor{b l u e}{- 1}\right)}^{2}} - \textcolor{b l u e}{0}}{2}}\right) i\right)$

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

The square roots of $\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i$ are:

$\pm \left(\left(\sqrt{\frac{\sqrt{{\left(\textcolor{b l u e}{\frac{\sqrt{2}}{2}}\right)}^{2} + {\left(\textcolor{b l u e}{- \frac{\sqrt{2}}{2}}\right)}^{2}} + \textcolor{b l u e}{\frac{\sqrt{2}}{2}}}{2}}\right) - \left(\sqrt{\frac{\sqrt{{\left(\textcolor{b l u e}{\frac{\sqrt{2}}{2}}\right)}^{2} + {\left(\textcolor{b l u e}{- \frac{\sqrt{2}}{2}}\right)}^{2}} - \textcolor{b l u e}{\frac{\sqrt{2}}{2}}}{2}}\right) i\right)$

=+-((sqrt((sqrt(1/2+1/2)+sqrt(2)/2))/2)) - (sqrt((sqrt(1/2+1/2)-sqrt(2)/2)/2))i)

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

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

Rather than simplifying again, note that we can find the square roots of $- \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i$ by multiplying these by $i$ to get:

$\pm \left(\frac{\sqrt{2 - \sqrt{2}}}{2} + \frac{\sqrt{2 + \sqrt{2}}}{2} i\right)$

So the fourth roots of $- i$ (i.e. the roots of ${x}^{4} + i = 0$) are:

$\pm \left(\frac{\sqrt{2 + \sqrt{2}}}{2} - \frac{\sqrt{2 - \sqrt{2}}}{2} i\right)$

$\pm \left(\frac{\sqrt{2 - \sqrt{2}}}{2} + \frac{\sqrt{2 + \sqrt{2}}}{2} i\right)$