Confused About (-1)^{2/6}: Get Help Here

[joao_pimentel]

Hi guys

I was wondering after reading some other stuff on the web..

How much is (-1)^{2/6}

It is (-1)^{\frac{2}{6}}=(-1)^{\frac{1}{3}}=\sqrt[3]{-1}=-1

OR

(-1)^{\frac{2}{6}}=\sqrt[6]{(-1)^2}=\sqrt[6]{1}=1

I'm really confused with this...

Any suggestions?

Thanks

João

 

[joao_pimentel]

-1 or 1 ?

Thank you :)

 

[micromass]

First of all, exponents are classically only defined for positive real numbers (except where the exponent is an integer). If you want to talk about (-1)^{2/6}, then you first need to define what exactly it means. Such a thing can be defined with some help of complex numbers. Indeed, we define

(-1)^(2/6)=e^{\frac{2}{6}Log(-1)}

but what does Log(-1) mean?? Well, the logarithm is classically also only defined for positive real numbers. But it can also be defined for complex numbers. Sadly, it becomes multivalued. That is, the logarithm can take up more than one value. Here, we have

Log(-1)=(2k+1)\pi i

For every number k, we have a value of the logarithm. So Log(-1) is each value in the set \{...,-3\pi i, -\pi i,\pi i,3\pi i,...\}

We do define a principal value of the logarithm. This is \pi i. So we could say that Log(-1) has infinitely many values, but \pi i is the most important one (by definition).

So, now we now what Log(-1) is, we can answer what (-1)^{2/6} is. Indeed

(-1)^{2/6}=(-1)^{\frac{2}{6}Log(-1)}=(-1)^{\frac{2}{6}(2k+1)\pi i}

This is by definition equal to

\cos(\frac{2}{6}(2k+1)\pi) + i\sin(\frac{2}{6}(2k+1)\pi)

So we see that the expression (-1)^{2/6} is also multivalued. To find its values, note that our expression is

\cos(\frac{(2k+1)\pi}{3})+i\sin(\frac{(2k+1)\pi}{3})

Letting k=0,1,2 we get

\{\frac{1}{2}+i\frac{\sqrt{3}}{2},-1,\frac{1}{2}-i\frac{\sqrt{3}}{2}\}

All other values of k yield the same answer. So our expression has three answers. Note that -1 is one of the answers. However, the principal (most important) value is

\frac{1}{2}+i\frac{\sqrt{3}}{2}

Why can't the answer be 1. Well, in your OP you reasoned that

(-1)^{2/6}=((-1)^2)^{1/6}

But this law doesn't need to hold anymore. Indeed, it is in general false that a^{bc}=(a^b)^c (unless a is positive).

Sorry that this answer might be difficult to understand. But this is the example of a simple question that does not have a simple answer.

 

[Joffan]

And in distinction to your argument #2...

-1^{1/6} has six answers, none of which are -1 or 1, but when squared they map onto a set of three answers as per micromass' post.

1^{1/6} also has six answers, which include the three answers identified.

You might say that by squaring first, you've lost information about which answers are valid.

 

[joao_pimentel]

I completely understood everything :)

Thank you very much for all your attention...

I was not just realizing that I needed to go into the Complex numbers...

And also, I didn't know that (a^{b})^c=a^{bc} is just valid when a> 0

Thanks a lot :)

 

[checkitagain]

And in distinction to your argument #2...

-1^{1/6} has six answers, none of which are -1 or 1,...

(-1)^{1/6} \ \ must \ have \ been \ meant \ here.

 

[joao_pimentel]

But ({a}^b)^c=a^{bc} can't still be valid even when a<0, if b and c are integers?

I mean, are there any values of b,c\in\Re which make the above statement valid when a<0?

 

[BloodyFrozen]

(-1)^{2/6}=(-1)^{\frac{2}{6}Log(-1)}=(-1)^{\frac{2}{6}(2k+1)\pi i}[/QOUTE]

I may be wrong, but I believe this should be:

(-1)^{2/6}=\color{red}e^{\frac{2}{6}Log(-1)}=\color{red}e^{\frac{2}{6}(2k+1)\pi i}

 

[micromass]

(-1)^{2/6}=(-1)^{\frac{2}{6}Log(-1)}=(-1)^{\frac{2}{6}(2k+1)\pi i} [/QOUTE]

I may be wrong, but I believe this should be:

(-1)^{2/6}=\color{red}e^{\frac{2}{6}Log(-1)}=\color{red}e^{\frac{2}{6}(2k+1)\pi i}

Indeed! Thanks a lot!