How do you simplify #root3(1)#?

2 Answers
Apr 26, 2018

Answer:

#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.

Apr 26, 2018

Answer:

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#