# How do you find the cube root of -125/2(1+sqrt3i)?

$5 {e}^{- \frac{2 \pi i}{9}} , \setminus 5 {e}^{- \frac{4 \pi i}{9}} , \setminus 5 {e}^{- \frac{10 \pi i}{9}}$

#### Explanation:

$- \frac{125}{2} \left(1 + \setminus \sqrt{3} i\right)$

$= - \frac{125}{2} - \frac{125 \setminus \sqrt{3}}{2} i$

$= 125 {e}^{- \frac{2 \pi}{3} i}$

Now, cubic roots of $- \frac{125}{2} \left(1 + \setminus \sqrt{3} i\right)$ are given as

${\left(125 {e}^{- \frac{2 \pi}{3} i}\right)}^{\frac{1}{3}}$

$= 5 {\left({e}^{i \left(2 k \setminus \pi - \frac{2 \pi}{3}\right)}\right)}^{\frac{1}{3}}$

$= 5 {e}^{\frac{i}{3} \left(2 k \setminus \pi - \frac{2 \pi}{3}\right)}$

$= 5 {e}^{i \left(\frac{2 k \setminus \pi}{3} - \frac{2 \pi}{9}\right)}$

Where, $k = 0 , 1 , 2$

Thus, setting $k = 0 , 1 , 2$ in above general cubic root, we get three cubic roots of given complex number

$\setminus \sqrt{- \frac{125}{2} \left(1 + \setminus \sqrt{3} i\right)} = 5 {e}^{- \frac{2 \pi i}{9}} , \setminus 5 {e}^{- \frac{4 \pi i}{9}} , \setminus 5 {e}^{- \frac{10 \pi i}{9}}$

Jun 28, 2018

There are three cube roots,

$- 5 \left(\cos \left(- \frac{5 \pi}{9}\right) + i \sin \left(- \frac{5 \pi}{9}\right)\right)$

$- 5 \left(\cos \left(\frac{\pi}{9}\right) + i \sin \left(5 \frac{\pi}{9}\right)\right)$

$- 5 \left(\cos \left(\frac{7 \pi}{9}\right) + i \sin \left(\frac{7 \pi}{9}\right)\right)$

#### Explanation:

$z = {\left(- \frac{125}{2} \left(1 + \sqrt{3} i\right)\right)}^{\frac{1}{3}}$

$= {\left(- 125\right)}^{\frac{1}{3}} {\left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right)}^{\frac{1}{3}}$

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

$= - 5 {\left(\cos \left(\frac{\pi}{3} + 2 \pi k\right) + i \sin \left(\frac{\pi}{3} + 2 \pi k\right)\right)}^{\frac{1}{3}} \quad$ integer $k$

We add the $2 \pi k$ because every complex number has three cube roots and we want to find them all.

De Moivre's theorem works when $n$ is a fraction too; it's basically Euler's Formula.

$= - 5 \left(\cos \left(\frac{\pi}{9} + \frac{2 \pi k}{3}\right) + i \sin \left(\frac{\pi}{9} + \frac{2 \pi k}{3}\right)\right)$

That's three unique values, given by any three consecutive $k .$ These aren't constructible angles so there's no nice expression combining integers, addition, subtraction, multiplication, division and square rooting. We'll just write the three possibilities, $k = - 1 , 0 , 1.$

$z = - 5 \left(\cos \left(- \frac{5 \pi}{9}\right) + i \sin \left(- \frac{5 \pi}{9}\right)\right)$ or

$z = - 5 \left(\cos \left(\frac{\pi}{9}\right) + i \sin \left(5 \frac{\pi}{9}\right)\right)$ or

$z = - 5 \left(\cos \left(\frac{7 \pi}{9}\right) + i \sin \left(\frac{7 \pi}{9}\right)\right)$

We can lose the minus sign by adding $\pi$ to the angles, but I won't bother.