# What is the fourth root of 1296?

Apr 5, 2018

$6 , 6 i , - 6 , - 6 i$

#### Explanation:

[SIDE NOTE: De Moivre's Formula:

if $z$ is a complex number in the form $r \left(\cos \left(x\right) + i \cdot \sin \left(x\right)\right)$, then ${z}^{n} = {r}^{n} \left(\cos \left(n \cdot x\right) + i \cdot \sin \left(n \cdot x\right)\right)$]

for this problem, use the extension of De Moivre's Formula (Roots of Complex Numbers):

if $z$ is a complex number in the form $r \left(\cos \left(x\right) + i \cdot \sin \left(x\right)\right)$, then the n'th roots of z (${z}^{\frac{1}{n}}$) are:

${r}^{\frac{1}{n}} \left(\cos \left(\frac{x + 2 \pi k}{n}\right) + i \cdot \sin \left(\frac{x + 2 \pi k}{n}\right)\right)$, where k is any integer

note that $1296 = 1296 + 0 i = 1296 \left(\cos \left(0\right) + i \cdot \sin \left(0\right)\right) = {6}^{4} \left(\cos \left(0\right) + i \cdot \sin \left(0\right)\right)$

if you let the complex number $z$ equal 1296, then $r = {6}^{4}$ and $x = 0$

to find the 4th roots, let $n = 4$

so the roots are: ${\left({6}^{4}\right)}^{\frac{1}{4}} \left(\cos \left(\frac{0 + 2 \pi k}{4}\right) + i \cdot \sin \left(\frac{0 + 2 \pi k}{4}\right)\right)$
$= 6 \left(\cos \left(\frac{\pi k}{2}\right) + i \cdot \sin \left(\frac{\pi k}{2}\right)\right)$, where k is any integer

as long as you choose 4 consecutive values for k, you'll get all the solutions. i would choose: $k = 0 , 1 , 2 , 3$

$k = 0 \rightarrow 6 \left(\cos \left(\frac{\pi \left(0\right)}{2}\right) + i \cdot \sin \left(\frac{\pi \left(0\right)}{2}\right)\right) = 6 \left(1\right) = 6$
$k = 1 \rightarrow 6 \left(\cos \left(\frac{\pi \left(1\right)}{2}\right) + i \cdot \sin \left(\frac{\pi \left(1\right)}{2}\right)\right) = 6 \left(i\right) = 6 i$
$k = 2 \rightarrow 6 \left(\cos \left(\frac{\pi \left(2\right)}{2}\right) + i \cdot \sin \left(\frac{\pi \left(2\right)}{2}\right)\right) = 6 \left(- 1\right) = - 6$
$k = 3 \rightarrow 6 \left(\cos \left(\frac{\pi \left(3\right)}{2}\right) + i \cdot \sin \left(\frac{\pi \left(3\right)}{2}\right)\right) = 6 \left(- i\right) = - 6 i$

you can check by raising the roots to the 4th power and using ${i}^{2} = - 1$ to simplify