Solve the equation z^4+81i=0?

Mar 16, 2017

There are four solutions;

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

Explanation:

Let $\omega = - 81 i$, and let ${z}^{4} = \omega$

First let us plot the point $\omega$ on the Argand diagram:

And we will put the complex number into polar form (visually):

$| \omega | = 81$
$a r g \left(\omega\right) = - \frac{\pi}{2}$

So then in polar form we have:

$\omega = 81 \left(\cos \left(- \frac{\pi}{2}\right) + i \sin \left(- \frac{\pi}{2}\right)\right)$

We now want to solve the equation for $z$ (to gain $4$ solutions):

${z}^{3} = 81 \left(\cos \left(- \frac{\pi}{2}\right) + i \sin \left(- \frac{\pi}{2}\right)\right)$

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 = 81cos(2npi-(pi)/2) + isin(2npi-(pi)/2)) \ \ \ n in ZZ

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

$z = {\left\{\cos \left(2 n \pi - \frac{\pi}{2}\right) + i \sin \left(2 n \pi - \frac{\pi}{2}\right)\right\}}^{\frac{1}{4}}$
$\setminus \setminus = {\left\{81\right\}}^{\frac{1}{4}} {\left\{\cos \left(2 n \pi - \frac{\pi}{2}\right) + i \sin \left(2 n \pi - \frac{\pi}{2}\right)\right\}}^{\frac{1}{4}}$
$\setminus \setminus = 3 \left\{\cos \left(\frac{2 n \pi - \frac{\pi}{2}}{4}\right) + i \sin \left(\frac{2 n \pi - \frac{\pi}{2}}{4}\right)\right\}$
$\setminus \setminus = 3 \left(\cos \theta + i \sin \theta\right)$

Where:

$\theta = \frac{2 n \pi - \frac{\pi}{2}}{4}$
$\setminus \setminus = \frac{1}{8} \left(4 n - 1\right) \pi$

Put:

$n = - 1 \implies \theta = \frac{- 5 \pi}{8}$
$\text{ } \therefore z = 3 \left(\cos \left(\frac{- 5 \pi}{8}\right) + i \sin \left(\frac{- 5 \pi}{8}\right)\right)$
$\text{ } \therefore z = 3 \left(- \frac{\sqrt{2 - \sqrt{2}}}{2} - \frac{\sqrt{2 + \sqrt{2}}}{2} i\right)$

$n = 0 \implies \theta = \frac{- \pi}{8}$
$\text{ } \therefore z = 3 \left(\cos \left(\frac{- \pi}{8}\right) + i \sin \left(\frac{- \pi}{8}\right)\right)$
$\text{ } \therefore z = 3 \left(\frac{\sqrt{2 + \sqrt{2}}}{2} - \frac{\sqrt{2 - \sqrt{2}}}{2} i\right)$

$n = 1 \implies \theta = \frac{3 \pi}{8}$
$\text{ } \therefore z = 3 \left(\cos \left(\frac{3 \pi}{8}\right) + i \sin \left(\frac{3 \pi}{8}\right)\right)$
$\text{ } \therefore z = 3 \left(\frac{\sqrt{2 - \sqrt{2}}}{2} + \frac{\sqrt{2 + \sqrt{2}}}{2} i\right)$

$n = 2 \implies \theta = \frac{7 \pi}{8}$
$\text{ } \therefore z = 3 \left(\cos \left(\frac{7 \pi}{8}\right) + i \sin \left(\frac{7 \pi}{8}\right)\right)$
$\text{ } \therefore z = 3 \left(- \frac{\sqrt{2 + \sqrt{2}}}{2} + \frac{\sqrt{2 - \sqrt{2}}}{2} i\right)$

After which the pattern continues.

We can plot these solutions on the Argand Diagram