# Imaninary numbers, i=1 ?

1. Jul 11, 2007

### LHarriger

Imaninary numbers, i=1 ??!!

Ok, this is driving me crazy:
$i^{2}=-1$
no problem here, but
$i^{2}=\left(\sqrt{-1}\right)^{2}=\sqrt{(-1)^{2}}=\sqrt{1}=1$
oops......
I know that the error has to be in this step:
$\left(\sqrt{-1}\right)^{2}=\sqrt{(-1)^{2}}$
(works fine for positive numbers)
and I am pretty sure it has something to do with principle roots.
However, I just can't figure out a clean arguement for what exactly is wrong.

2. Jul 11, 2007

The square root is ambiguous, ie, $\sqrt{1}=\pm 1[/tex], and you need to be explicit about this ambiguity when using it or you'll make mistakes like that one. Last edited: Jul 11, 2007 3. Jul 11, 2007 ### morphism Actually, [itex]\sqrt{1}=1[/tex]. However the square roots of 1 are [itex]\pm 1[/tex]. The problem is that exponentiation for complex numbers differs than that for real numbers. In fact, if a and b are complex numbers, then we define: $$a^b = e^{b (\log |a| + i\arg(a))}$$, where [itex]\arg(a)$ is chosen to lie in $(-\pi, \pi]$. (Read up on branch cuts.)

From this it follows that $(a^b)^c = (a^c)^b$ isn't necessarily true.

Last edited: Jul 11, 2007
4. Jul 11, 2007

### LHarriger

Thank you Morphism!!!
I recall learning complex exponentiation and branch cuts from my undergraduate complex analysis that I took 3 years ago. I was just too rusty to be able to put it too work on my own to solve this problem.

5. Jul 11, 2007

### StatusX

That notation isn't standard, and couldn't have been what the OP was using since he wrote $$\sqrt{-1}$$.

Again, this is just one of many possible conventions. The important point is that for a rational number p/q (in lowest terms) and a complex number z, there are q complex numbers which all have equal claim to the name zp/q. And for irrational numbers, there are infinitely many. Complex exponentiaion is simply not a single valued function.

And $(a^b)^c = (a^c)^b$ is still true, in the sense that for any $a^b$ and $a^c$ (remember, they're multivalued) there are choices of $(a^b)^c$ and $(a^c)^b$ so that equality holds, and so that both are equal to some $a^{bc}$. The problem only comes up when you artificially try to make the function single valued.

Last edited: Jul 11, 2007
6. Jul 12, 2007

### matt grime

sqrt(-1) is a standard symbol: it means a primitive square root of -1, it does not mean both primitive square roots of -1, and of course there is no algebraic distinction between either.

Technically single valued function is a pleonasm - a function is single valued by definition. There is, perhaps, an inconsistent usage of the word function, here. However it would be better to stay away from these topics lest it confuse the reader. Incidentally, in my experience there are not q complex numbers with an equal claim to being z^{1/q}: that symbol is normally fixed at meaning the principal branch of the q'th root. This is not an artificial definition.

Last edited: Jul 12, 2007
7. Jul 12, 2007

### StatusX

I'm not arguing that it's not convenient for many purposes to use a principal branch for multivalued functions, I'm just saying that the issue in the OPs question is more closely concerned with the multi-valuedness of the functions, not with the properties of a specific branch.

8. Jul 13, 2007

### robert Ihnot

L Harringer: I know that the error has to be in this step:
$\left(\sqrt{-1}\right)^{2}=\sqrt{(-1)^{2}}$

(works fine for positive numbers)
and I am pretty sure it has something to do with principle roots.
However, I just can't figure out a clean arguement for what exactly is wrong.

I heard one reason when taking algebra: the symbol "i" was invented to prevent such a mistake as:

$$\sqrt{-1}\sqrt{-1} =\sqrt{1}$$

I can't vouch for that, but take it for what its worth.

Last edited: Jul 13, 2007
9. Jul 13, 2007

### ObsessiveMathsFreak

That's where the error is. The product rule for surds does not work if both surds are negative square roots.

10. Jul 18, 2007