How do you simplify root3(1)?

Apr 26, 2018

$1$ or ${1}^{\frac{1}{3}}$ =$1$

Explanation:

The cubed root of 1 is the same as raising 1 to the power of $\frac{1}{3}$. 1 to the power of anything is still 1.

Apr 26, 2018

Working in the reals we get $\sqrt[3]{1} = 1$.

Every non-zero complex number has three cube roots, so there

$\sqrt[3]{1} = 1 \mathmr{and} - \frac{1}{2} \pm i \frac{\sqrt{3}}{2}$

Explanation:

If we're working in real numbers we just note $\sqrt[3]{1} = \sqrt[3]{{1}^{3}} = 1$. I'm going to assume this is about complex numbers.

One of the odd things we find out when we delve into complex numbers is that the function $f \left(z\right) = {e}^{z}$ is periodic. Exponential growth is sort of the opposite of periodic, so this is a surprise.

The key fact is Euler's Identity squared. I call it Euler's True Identity.

${e}^{2 \setminus \pi i} = 1$

Euler's True Identity shows ${e}^{z}$ is periodic with period $2 \pi i$:

$f \left(z + 2 \pi i\right) = {e}^{z + 2 \pi i} = {e}^{z} {e}^{2 \pi i} = {e}^{z} = f \left(z\right)$

We can raise Euler's True Identity to any integer power $k$:

${e}^{2 \setminus \pi k i} = 1$

What's all this got to do with the cube root of one? It's the key. It tells there are a countably infinite number of ways of writing one. Some of them have different cube roots than others. It's why non-integer exponents give rise to multiple values.

That's all a big windup. Usually I just start these by writing:

${e}^{2 \pi k i} = 1 \quad$ for integer $k$

$\sqrt[3]{1} = {1}^{\frac{1}{3}} = {\left({e}^{2 \setminus \pi k i}\right)}^{\frac{1}{3}} = {e}^{i \frac{2 \pi k}{3}} = \cos \left(2 \pi \frac{k}{3}\right) + i \sin \left(2 \pi \frac{k}{3}\right)$

The last step is of course Euler's Formula ${e}^{i \theta} = \cos \theta + i \sin \theta .$

Since we have the $2 \pi$ periodicity of the trig functions (which follows from the periodicity of the exponential and Euler's Formula) we only have unique values for three consecutive $k$s. Let's evaluate this for $k = 0 , 1 , - 1$:

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

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

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

So we get three values for the cube root of one:

$\sqrt[3]{1} = 1 \mathmr{and} - \frac{1}{2} \pm i \frac{\sqrt{3}}{2}$