# Why isn't x^(ab) = (x^a)^b ?

1. Nov 8, 2007

### Zorodius

(-1)^2 = 1.

1 ^ (1/3) = 1.

so why is (-1)^(2/3) a horrible complex number?

2. Nov 8, 2007

### ObsessiveMathsFreak

The power rule only works if the base number is positive. For a negative number, you have to first transform it into a complex form which has positive bases if you will.

$$-1=e^{i \pi}$$
$$(-1)^{\frac{2}{3}}=\left(e^{i \pi}\right)^{\frac{2}{3}}=e^{\frac{2 i \pi}{3}}=-\frac{1}{2}+i \frac{\sqrt{3}}{2}$$

3. Nov 8, 2007

### jostpuur

I remember fighting with these calculation rules myself in high school age. I think the confusion stems from the fact, that we learn these calculation rules when we are little kids and don't know about proving, and once we have learned more about proving things, we still tend to believe in the old calculation rules we have got used to.

The answer to your "why"-question is ridiculously simple, anyway. $z^{ab}=(z^a)^b$ doesn't work for arbitrary $z\in\mathbb{C}$ (and in particular for negative reals you mentioned), because there exists counter examples!

As long as you don't have a proof for the claim, that it should work for arbitrary z, there is no contradiction or paradox.

4. Nov 8, 2007

### D H

Staff Emeritus
That's only one of the roots. There are two others:

$$-1 = e^{(1+2n)\pi i}$$

$$(-1)^{\frac{2}{3}} = e^{\frac 2 3 (1+2n)\pi i} = e^{\frac 2 3 \pi i}, 1, e^{\frac 4 3 \pi i}$$

The power rule still does work in a sense, as one of the solutions to (-1)^(2/3) is one.

5. Nov 8, 2007

### George Jones

Staff Emeritus
This is somewhat subtle stuff.

If $x = (-1)^{\frac{2}{3}}$, then

$$\begin{equation*} \begin{split} x^3 &= (-1)^2 \\ 0 &= 1 - x^3 \\ 0 &= \left( 1 - x\right) \left(x^2 + x + 1). \end{split} \end{equation*}$$

The last equation has three roots, so there are three values for $(-1)^{\frac{2}{3}}$:

$$1;$$

$$-\frac{1}{2}+i \frac{\sqrt{3}}{2};$$

$$-\frac{1}{2}-i \frac{\sqrt{3}}{2}.$$

What is the definition $w^z$ when $w$ and $z$ are general numbers (possibly irrational and/or complex)?

$$w^z = e^{wlnz}$$.

When you think about it, this definition makes sense. However, this defrinition does depend on the choice of value of $lnz$.

In the case in question,

$$(-1)^{\frac{2}{3}} = e^{\frac{2}{3}ln(-1)$$, which depends on a specification of what $ln(-1)$ equals.

$$ln(-1) = ln \left(e^{i\pi +i2n\pi} \right) = i\pi +i2n\pi.$$

Even though there are an infinite number of possibilities for the value of $ln(-1)$, a little playing around shows that these values only give the three values listed above for $(-1)^{\frac{2}{3}}$.

After posting, I see that what I typed is just an elaboration D H's post.

Last edited: Nov 8, 2007
6. Nov 8, 2007

### uart

But it has a perfectly good real solution too, 1.

$$(-1)^{2/3} = (e^{i (2n+1) \pi})^{2/3} = e^{i 2 (2n+1) \pi / 3}$$

Just put n=1 and you've got the real solution predicted.

Edit: It looks like a few others got in before my to say much the same thing. Seems that I don't refresh my browser often enough.

Last edited: Nov 8, 2007
7. Nov 8, 2007

### jostpuur

One source of confusion, or at least it was for me, is how complex numbers are treated differently as reals.

If somebody tries to say that $\sqrt{1}=\pm 1$, usually everybody is ready to say that this is wrong, because square roots are supposed to be positive.

But if somebody says $\sqrt{-1} =\pm i$, people are less likely to complain, because in the context of complex numbers folks always start to think about solutions to some polynomial equations.

I think this is confusing, because real numbers are also complex numbers, so there should not be this difference.

I like the principal branch of the logarithm (and other definitions that use this logarithm), because it is unique, and agrees with the old definitions when restricted to the real line.

Zorodius, btw did you get that "horrible complex number" out of a calculator or something like that? I think my Texas Instrument at least chooses the principal branch out of the A^B. That means
$$(-1)^{2/3} := e^{(2/3)\cdot\textrm{Log}(-1)}$$
and
$$\textrm{Log}(-1) = i\pi$$
here.

8. Nov 8, 2007

### D H

Staff Emeritus
The standard meaning for $\sqrt x$ is indeed the positive square root of $x$ for positive $x$. People should rightfully complain if you use this notation to include the negative root. By extension, it is better to say $\sqrt{-1}=i$ rather than $\pm i$. Further extending the convention, the square root symbol means the principal root for any complex number. Some people will rightfully complain that this extension is carrying things a bit too far.

If your intent is to allow the negative root it is better to say something like $1^{1/2}$, $(-1)^{1/2}$, or $i^{1/2}$. By convention, this notation denotes all roots, not just the principal root. For example, the nth roots of unity are simply $1^{1/n}$.

9. Nov 8, 2007

### mathwonk

the problem with saying sqrt(-1) = i and not -i, is that there is no such thing as i. i.e. there is no way to choose a particular square root of -1. there are two of them and no way to prefer one over the other. so there is no way to say which number the letter i represents.

it is different for reals as you can take sqrt(1) as the one that itself has a real square root, i.e. the positive one.

put more abstractly, the complex numbers have a non trivial automorphism which interchanges i and -i, but the reals have none interchanging 1 and -1.

Last edited: Nov 8, 2007
10. Nov 8, 2007

### sennyk

I thought it was notation

$$\sqrt(-1) = i$$

I always thought that was for notation purposes only; math people don't like to write, so using a letter was a less cumbersome way to represent that expression. Thus the letter i was more of a MACRO for that expression.

By not extending that notation, one would have to accept the following as 100% correct.

$$x^2=-1 => x=\sqrt(-1)$$

I don't know any of my professors that would accept that as 100% correct.

11. Nov 8, 2007

### rbj

no.

yes.

yes! emphatically! take a look at the Wikipedia article on the Imaginary unit, particularly:

http://en.wikipedia.org/wiki/Imaginary_unit#i_and_.E2.88.92i

$$x^{a b} = \left( x^a \right)^b$$

is always valid (or can always be made valid), but you need worry about which roots of whatever number you're dealing with. x1/N has N possible values (all different, if x is not zero, and all equally valid) if N is an integer.

Last edited: Nov 8, 2007
12. Nov 8, 2007

### jostpuur

I thought that this is standard convention too:

$$a^b = e^{b\cdot\textrm{Log}(a)}$$

$$\textrm{Log}(a) = \textrm{log}(|a|) + i\textrm{Arg}(a)$$

and

$$-\pi < \textrm{Arg}(a) \leq \pi$$

But a^b has a special meaning if b=1/n for some n?

Good point, but I think the problem itself is a bit artificial. $(0,1)\in\mathbb{R}^2$ and $(0,-1)\in\mathbb{R}^2$ are distinguishable, so once i has been defined to be something, then i and -i are distinguishable too.

Am I right to guess that the point of views preferred in algebra and in complex analysis are different?

Last edited: Nov 8, 2007