# How do you find all solutions to x^4+(1+i)=0?

Sep 28, 2016

$x = \sqrt[8]{2} {e}^{i \left(\frac{5}{16} \pi + k \frac{\pi}{2}\right)}$ for $k = 0 , \pm 1 , \pm 2 , \cdots$

#### Explanation:

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

Using de Moivre's identity

${e}^{i \phi} = \cos \phi + i \sin \phi$ we have

$- \left(1 + i\right) = - \sqrt{2} \left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = - \sqrt{2} \left(\cos {\phi}_{0} + i \sin {\phi}_{0}\right)$

with ${\phi}_{0} = \frac{\pi}{4} + 2 k \pi , k = 0 , \pm 1 , \pm 2 , \cdots$

but

$- \sqrt{2} \left(\cos {\phi}_{0} + i \sin {\phi}_{0}\right) = \sqrt{2} {e}^{i \pi} {e}^{i \left(\frac{\pi}{4} + 2 k \pi\right)} = \sqrt{2} {e}^{i \left(\frac{5}{4} \pi + 2 k \pi\right)}$

(we used ${e}^{i \pi} + 1 = 0$ after Euler
https://en.wikipedia.org/wiki/Leonhard_Euler)

${x}^{4} = \sqrt{2} {e}^{i \left(\frac{5}{4} \pi + 2 k \pi\right)}$

finally

$x = \sqrt[8]{2} {e}^{i \left(\frac{5}{16} \pi + k \frac{\pi}{2}\right)}$

for $k = 0 , \pm 1 , \pm 2 , \cdots$