Question

How do you simplify `root3(1)`?

Answer 1

`1` or `1^(1/3)` =`1`

Explanation:

The cubed root of 1 is the same as raising 1 to the power of `1/3`. 1 to the power of anything is still 1.

Answer 2

Working in the reals we get `root[3]{1}=1`.

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

`root[3]{1} = 1 or -1/2 pm i sqrt{3}/2`

Explanation:

If we're working in real numbers we just note `root[3]{1} = root[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(z)=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\pi i} = 1`

Euler's True Identity shows `e^z` is periodic with period `2pi i`:

`f(z + 2pi i) = e^{z + 2 pi i} = e^z e^{2 pi i} = e^z = f(z)`

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

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

`root[3]{1} = 1^{1/3} = (e^{2\pi k i})^{1/3} = e^{i {2pi k }/3} = cos(2pi k/3) + i sin(2pi k/3)`

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

Since we have the `2pi` 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({2pi k}/3) + i sin({2pi k}/3) = cos 0 + i sin 0 = 1`

`k`=1`quad quad cos({2pi} /3) + i sin({2pi} /3) =-1/2 + i sqrt{3}/2`

`k`=-1`quad quad cos(-{2pi} /3) + i sin(-{2pi} /3) =-1/2 - i sqrt{3}/2`

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

`root[3]{1} = 1 or -1/2 pm i sqrt{3}/2`