# How do you find all solutions to x^3-(1-i)=0?

Dec 31, 2016

Roots are:

${2}^{\frac{1}{6}} \exp \left(- \frac{i \pi}{12}\right)$

${2}^{\frac{1}{6}} \exp \left(\frac{7 i \pi}{12}\right)$

${2}^{\frac{1}{6}} \exp \left(\frac{5 i \pi}{4}\right)$

#### Explanation:

you are looking for:

$x = {\left(1 - i\right)}^{\frac{1}{3}}$

as a rule, this complex number will have 3 cube roots, same absolute value but spaced equally about the Argand diagram.

using $c i s \left(\theta\right) \equiv {e}^{i \theta}$ in the notation, we can express this in polar coordinates as follows:

$1 - i = \setminus \sqrt{2} \setminus c i s \left(- \frac{\pi}{4} \textcolor{red}{+ 2 n \pi}\right)$ where $n \in Z$

So

${\left(1 - i\right)}^{\frac{1}{3}} = {\left(\setminus \sqrt{2} \setminus c i s \left(- \frac{\pi}{4} + 2 n \pi\right)\right)}^{\frac{1}{3}}$

$= {2}^{\frac{1}{6}} \setminus c i s \left(- \frac{\pi}{12} + \frac{2 n \pi}{3}\right)$

working through values of n

$n = 0 \to {2}^{\frac{1}{6}} c i s \left(- \frac{\pi}{12}\right) = {2}^{\frac{1}{6}} \exp \left(- \frac{i \pi}{12}\right)$

$n = 1 \to {2}^{\frac{1}{6}} c i s \left(\frac{7 \pi}{12}\right) = {2}^{\frac{1}{6}} \exp \left(\frac{7 i \pi}{12}\right)$

$n = 2 \to {2}^{\frac{1}{6}} c i s \left(\frac{5 \pi}{4}\right) = {2}^{\frac{1}{6}} \exp \left(\frac{5 i \pi}{4}\right)$

and then the roots start repeating as follows
$n = 3 \to {2}^{\frac{1}{6}} c i s \left(- \frac{\pi}{12} + 2 \pi\right) = {2}^{\frac{1}{6}} c i s \left(- \frac{\pi}{12}\right)$