# A new imaginary number? $x^c = -x$

Being young, I don't have much experience with imaginary numbers outside of the basic usages of $i$. As I was sitting in my high school math class doing logs, I had an idea of something that would allow solving for logs with negative bases or with negative arguments. Taking a similar idea to i, this number, when used as an exponent, would result in a negative number. In other words, as the title says, $x^c = -x$.

For starters, I just quickly mentally ran through simple examples of how it could be used in these cases, (ex. $\log(-100) = 2c$) but I soon began to wonder if it could have other applications and or just if it could be anything else.

I'm also not 100% sure what to tag it, so I kinda guessed.

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The realm of complex number frees us, if we become a bit more careful. For example, if you are familiar with the natural logarithm, you may find that we can extend logarithm to be defined for all complex numbers except for zero. Then $\log_{10} (-100) = 2 + i \pi / \ln 10$, wheere $i$ is the imaginary unit. In fact, complex logarithm plays a very imprtant role in many theories. – Sangchul Lee Nov 24 '11 at 2:09
Clever thinking! But these questions most probably end up with a new no. Here is another mind-bender for you. What is $\displaystyle i^i$? – user13838 Nov 24 '11 at 4:07
Another nice question (see Henning Makholm's answer below): Is $i=1$, $i<1$, or $i>1$? :) – Joel Reyes Noche Nov 24 '11 at 4:11
@Joel i < 1 or i >1 doesn't make sense, as i is orthogonal to the real number line, however |i| = 1. – Chris Cudmore Nov 24 '11 at 14:45
I want to thank everyone for all the great answers. I feel like I've gained a lot of insight into all this thanks to all of you. – Warren L. Nov 25 '11 at 1:41

Robert's answer tells you correctly how to find, given an $x$, a complex $c$ such that $x^c=-x$.

However, it is worth pointing out that you cannot find a single $c$ such $x^c=-x$ will hold for every $x$. There's no existing number that achieves this, and if you try to extend the number system by simply postulating that $c$ exists (which I think was the point of the question), you run into this: $$0 = \frac08 = \frac{-4+4}{8} = \frac{4^c+4}8 = \frac{2^c2^c+4}8 = \frac{(-2)(-2)+4}8 = \frac{4+4}8 = 1$$ So if you want your $c$ to exist, but don't want to collapse everything into $0=1$, then you have to discard the exponentiation rule $(ab)^c=a^cb^c$. And without that rule, the result isn't really worthy of being called exponentiation in the first place.

I think you have been fooled by a common introduction to complex numbers, which goes like this: "There is no real number whose square is $-1$, but assume we had such a number anyway and call it $i$. Then we can calculate with the assumption $i^2=-1$, and, ta-dah! we have invented a new kind of numbers". This presentation, however, sweeps an important point under the carpet, namely that we had better be sure that assuming $i^2=-1$ doesn't allow us to prove falsehoods such as $0=1$ about the real numbers that we already know and love -- because if it does that means that we haven't invented a new kind of numbers but simply deceived ourselves.

For $i^2=-1$ it turns out that it's not too difficult to prove that it doesn't lead to nonsense. But even when this proof is shown, something is still swept under the carpet here, namely that if we assume $i^2=-1$ and the familiar rule that every number is either 0 or positive or negative (and the usual rules for positive and negative numbers), then we can still derive nonsense. So what's really true is that we can get a $\sqrt{-1}$ if and only if we're willing to drop what we know about positive and negative numbers (or "less than"/"greater than") when we're working with complex numbers.

Extending the concept of number is almost always connected to a loss of some rules that used to hold before the extension. How far we can push the extension depends on how many rules we're willing to let go of. So you can have your $c$, but the cost you must pay is the rules for how exponentiation works. And if you pay that cost, having $c$ becomes pretty useless anyway, so the whole expedition turns out to be pointless.

(But of course we couldn't have known that before we went and checked. The idea was good enough; it just didn't work. Perhaps your next one will.)

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+1, but I wouldn't say "it just didn't work". It works, it's "just fruitless". – Thomas Eding Dec 17 '12 at 4:58
Hm. Seems to me you only lose that rule about exponents half the time--if you preserve parity of the number of c exponents it should work, I think? Note that introducing an extra factor of 1^c would always suffice. You still end up with x^2c == x^2, of course, which seems peculiar. It's sort of a hypernegative unit, that constructs the inverse of two hyperoperations down, like -1 does one hyperoperation down. Unfortunately, multiplying by c doesn't give you the predecessor operation... – camccann Dec 28 '12 at 17:35

$x^c = e^{c \log x} = -x = e^{\pi i + \log(x)}$ if $c \log x = \log(x) + (2 n + 1) \pi i$ for some integer $n$. Thus the answer is $c = 1 + \frac{(2n + 1) \pi i}{\log x}$ (assuming, of course, $x \ne 1$).

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The point is that $c$ is already a complex number. – Carl Mummert Nov 24 '11 at 3:40
@Warren-l Robert's point is that your number $c$ does exist, and his answer proceeds to show exactly what it is. He is also using the $\log$ base $e$, whereas I see from your question you are familiar with $\log$ base $10$, however the idea is the same either way. I think your question was an excellent one! – process91 Nov 24 '11 at 3:45
What if x itself is negative? – DeadMG Nov 24 '11 at 15:48

Your idea is not wrong. From what I assume (and Henning seems to think this, too), your introduction to imaginary and complex numbers has been something like this:

The equation $x^2 + 1 = 0$ has no solution in the Real numbers. So, lets add a new number $i$ that is a solution to this equation and thus "extend" the set of Real numbers to the Complex numbers where every one of them will have this form (proof omitted): $a + bi$ and (again proof omitted) this new set $C$ does not drive us into deriving nonsense statements and has all the properties needed to call it a field.

This new set of Complex numbers does not have all features that the Reals have (as Henning points) like the concept of negatives and positives but it does have a lot of new features that the Reals do not have (like the fact that any polynomial equation with degree $n$ has exactly $n$ solutions, and not just for $n=2$ but for any positive $n$) and many others that you will find in any book about Complex Analysis.

So, your idea again, is quite smart and is essentially what many people have used previously in what are called algrebraic (field) extensions. Off course, it doesn't work as you expected and depending on how one interpets your equation either (as Robert explained) "your equation already has solutions in the Complex numbers" or (as Henning explained) "it leads to nonsense statements". Perhaps you can find some other equation that has no solution and thus extend them.

Or you can start with something simpler like the Rationals (the field $Q$) and extend it with a solution to the equation $x^2 - 2 = 0$. This is called if I am not wrong $Q(\sqrt{2})$ and you could try to find (and prove) the properties of this set. Is it a field (etc)?

Or the Rationals extended by a solution to $x^2 + 2 = 0$, which would be called $Q(\sqrt{-2})$.

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Log of negative numbers can be expressed with usual complex numbers: $\ln(-1)=\pi i$.

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This doesn't really answer the question. Robert Israel's answer explains in depth what I think you were trying to get at. – Zev Chonoles Nov 24 '11 at 8:14
The asker was trying to create a "new imaginary number" which makes log of negative numbers definable. All I'm saying here is that log of negative numbers is already defined in the realm of "old complex numbers". – kdewrwe Nov 24 '11 at 12:40