# How do you find all solutions to x^3+64i=0?

##### 1 Answer
Mar 14, 2017

$x = 4 i , \pm 2 \sqrt{3} - 2 i$

#### Explanation:

Let $\omega = - 64 i$, and then ${x}^{3} = \omega$

First let us plot the point $\omega$ on the Argand diagram: And we will put the complex number into polar form (visually):
$| \omega | = 64$
$a r g \left(\omega\right) = - \frac{\pi}{2}$

So then in polar form we have:

$\omega = 64 \left(\cos \left(- \frac{\pi}{2}\right) + i \sin \left(- \frac{\pi}{2}\right)\right)$

We now want to solve the equation for $x$ (to gain $3$ solutions):

${x}^{3} = 64 \left(\cos \left(- \frac{\pi}{2}\right) + i \sin \left(- \frac{\pi}{2}\right)\right)$

Whenever dealing with complex variable equation such as this it is essential to remember that the complex exponential has a period of $2 \pi$, so we can equivalently write (incorporating the periodicity):

${x}^{3} = 64 \left(\cos \left(2 n \pi - \frac{\pi}{2}\right) + i \sin \left(2 n \pi - \frac{\pi}{2}\right)\right) \setminus \setminus \setminus n \in \mathbb{Z}$

By De Moivre's Theorem we can write this as:

$x = 64 {\left(\cos \left(2 n \pi - \frac{\pi}{2}\right) + i \sin \left(2 n \pi - \frac{\pi}{2}\right)\right)}^{\frac{1}{3}}$
$\setminus \setminus = {\left\{64\right\}}^{\frac{1}{3}} {\left\{\cos \left(2 n \pi - \frac{\pi}{2}\right) + i \sin \left(2 n \pi - \frac{\pi}{2}\right)\right\}}^{\frac{1}{3}}$
$\setminus \setminus = 4 \left\{\cos \left(\frac{2 n \pi - \frac{\pi}{2}}{3}\right) + i \sin \left(\frac{2 n \pi - \frac{\pi}{2}}{3}\right)\right\}$
$\setminus \setminus = 4 \left(\cos \theta + i \sin \theta\right) \setminus \setminus \setminus \setminus$ where $\theta = \frac{\left(4 n - 1\right) \pi}{6}$

Put:

$n = - 1 \implies \theta = \frac{- 5 \pi}{6}$
$\text{ } \therefore x = 4 \left(\cos \left(\frac{- 5 \pi}{6}\right) + i \sin \left(\frac{- 5 \pi}{6}\right)\right)$
$\text{ } \therefore x = 4 \left(- \frac{\sqrt{3}}{2} - \frac{1}{2} i\right)$
$\text{ } \therefore x = - 2 \sqrt{3} - 2 i$

$n = 0 \implies \theta = \frac{- \pi}{6}$
$\text{ } \therefore x = 4 \left(\cos \left(\frac{- \pi}{6}\right) + i \sin \left(\frac{- \pi}{6}\right)\right)$
$\text{ } \therefore x = 4 \left(\frac{\sqrt{3}}{2} - \frac{1}{2} i\right)$
$\text{ } \therefore x = 2 \sqrt{3} - 2 i$

$n = 1 \implies \theta = \frac{\pi}{2}$
$\text{ } \therefore x = 4 \left(\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right)$
$\text{ } \therefore x = 4 \left(0 + i\right)$
$\text{ } \therefore x = 4 i$

After which the pattern continues.

We can plot these solutions on the Argand Diagram 