# How do you find the cube root of 64 (cos (pi/5) + isin (pi/5)) ?

##### 1 Answer
Mar 11, 2017

$\setminus \setminus \setminus \setminus \setminus \setminus 4 \left(\cos \left(\frac{\pi}{15}\right) + i \sin \left(\frac{\pi}{15}\right)\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus 4 \left(\cos \left(\frac{11 \pi}{15}\right) + i \sin \left(\frac{11 \pi}{15}\right)\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus 4 \left(\cos \left(\frac{7 \pi}{5}\right) + i \sin \left(\frac{7 \pi}{5}\right)\right)$

#### Explanation:

Let ${z}^{3} = 64 \left(\cos \left(\frac{\pi}{5}\right) + i \sin \left(\frac{\pi}{5}\right)\right)$

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):

${z}^{3} = 64 \left(\cos \left(\frac{\pi}{5} + 2 n \pi\right) + i \sin \left(\frac{\pi}{5} + 2 n \pi\right)\right) \setminus \setminus \setminus n \in \mathbb{N}$

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

$z = {\left\{64 \left(\cos \left(\frac{\pi}{5} + 2 n \pi\right) + i \sin \left(\frac{\pi}{5} + 2 n \pi\right)\right)\right\}}^{\frac{1}{3}}$
 \ \= 64^(1/3) (cos (pi/5+2npi) + isin (pi/5+2npi))}^(1/3
$\setminus \setminus = {64}^{\frac{1}{3}} \left(\cos \left(\frac{\frac{\pi}{5} + 2 n \pi}{3}\right) + i \sin \left(\frac{\frac{\pi}{5} + 2 n \pi}{3}\right)\right)$
$\setminus \setminus = 4 \left(\cos \left(\frac{\frac{\pi}{5} + 2 n \pi}{3}\right) + i \sin \left(\frac{\frac{\pi}{5} + 2 n \pi}{3}\right)\right)$

Put:

$n = 0 \implies z = 4 \left(\cos \left(\frac{\frac{\pi}{5}}{3}\right) + i \sin \left(\frac{\frac{\pi}{5}}{3}\right)\right)$
$\text{ } = 4 \left(\cos \left(\frac{\pi}{15}\right) + i \sin \left(\frac{\pi}{15}\right)\right)$

$n = 1 \implies z = 4 \left(\cos \left(\frac{\frac{\pi}{5} + 2 \pi}{3}\right) + i \sin \left(\frac{\frac{\pi}{5} + 2 \pi}{3}\right)\right)$
$\text{ } = 4 \left(\cos \left(\frac{11 \pi}{15}\right) + i \sin \left(\frac{11 \pi}{15}\right)\right)$

$n = 2 \implies 4 \left(\cos \left(\frac{\frac{\pi}{5} + 4 \pi}{3}\right) + i \sin \left(\frac{\frac{\pi}{5} + 4 \pi}{3}\right)\right)$
$\text{ } = 4 \left(\cos \left(\frac{7 \pi}{15}\right) + i \sin \left(\frac{7 \pi}{15}\right)\right)$

After which the pattern continues

So the three roots are:

$z = 4 \left(\cos \left(\frac{\pi}{15}\right) + i \sin \left(\frac{\pi}{15}\right)\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus 4 \left(\cos \left(\frac{11 \pi}{15}\right) + i \sin \left(\frac{11 \pi}{15}\right)\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus 4 \left(\cos \left(\frac{7 \pi}{5}\right) + i \sin \left(\frac{7 \pi}{5}\right)\right)$

We can view these roots on the Argand diagram: 