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We know that negative number to the power of any integers or some fractions will always have a solution. Is it possible for us to solve $(-2)^\frac 13$ or $(-2)^e$, by modifying/extending our current number system?

abiessu
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Kutu kupret
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  • Does "our current number system" include the complex numbers? – abiessu May 08 '15 at 18:15
  • I mean in order to make (-2)^e has one absolute definite value – Kutu kupret May 08 '15 at 18:17
  • By which you mean "$(-2)^e$ has no more than a single defined value", correct? – abiessu May 08 '15 at 18:19
  • @abiessu YESS.... – Kutu kupret May 08 '15 at 18:19
  • As far as I understand it, you have the option of choosing to ignore the alternate values that something like $(-2)^\frac 13$ can take on (which has a real value at $-\sqrt [3]{2}$), but doing so can make other things that produced the need for this value to be limited in their results and/or produce incorrect or unexpected results. – abiessu May 08 '15 at 18:25
  • @abiessu I don't get it.. – Kutu kupret May 08 '15 at 18:26
  • So we have $$(-2)^\frac 13=\left{-\sqrt[3]{2},\frac{1+i\sqrt 3}{2},\frac{1-i\sqrt 3}{2}\right}$$ of which the first value is the only value among the reals. If we ignored the two complex values, we are effectively ignoring two solutions of a cubic polynomial, much like ignoring solutions in the complex numbers to the polynomial $x^2+1$. It is the case that the solutions do not appear on the real line, but they still exist. – abiessu May 08 '15 at 18:32
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    First the number $(-2)^{e}$ has no sense without a definition, and we can see that it's a complex number and there is not a unique definition , the first definition for this element would be: the number $x$ which verifies $x^e=-2$ (not uniqu) and this is equivalent to saying that $(-2)^e$ is the element which verifies $e^{e\ln(x)}=-2$ and from here we have to define the function $x\to \ln(x)$ for negative numbers because for positive numbers $x$ $e^{e\ln(x)}>0$ but this is no easy, there is a lot of definitions for logarithms for complex numbers which gives you a lot of definitions of $(-2)^e$ – Elaqqad May 08 '15 at 18:32
  • @Elaqqad Somehow I truly believe that there's a single definite value for (-2)^e, because why not? – Kutu kupret May 08 '15 at 18:36
  • My comment contains some highlights about what I'm thinking (It's not well wrote I agree) but the value is not unique and there is no only one "acceptable" definition, and I did not write a full answer but If you think a little about the problem you will understand it – Elaqqad May 08 '15 at 18:49

3 Answers3

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Consider the value of the "number" $(-2)^e$; this is equivalent to considering

$$x=(-2)^e\\ \ln x=e\ln(-2)\\ \frac {\ln x}e=\ln(-2)\\ e^{\frac {\ln x}e}=-2\\ \frac 12e^{\frac {\ln x}e}=-1\\ e^{\frac {\ln x}e+\ln\frac 12}=-1\\ $$

Given the equation $e^{a+bi}=e^a(\cos b+i\sin b)$, we only need to find $x$ such that $\frac {\ln x}e=i\pi+2ni\pi-\ln\frac 12$ or $\ln x=ei\pi+2nei\pi+e\ln 2$ which is $$x=e^{ei\pi+2nei\pi+e\ln 2}=2^e(\cos(e\pi+2ne\pi)+i\sin(e\pi+2ne\pi)$$ which is an infinite set of unique complex values in a circle of radius $2^e$.

abiessu
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  • yes i've already realized that, but I aprreciate your neat proof :) – Kutu kupret May 08 '15 at 19:01
  • If you wish to have a unique "value" associated with such numbers, I might suggest that the form $(-2)^e$ is one such way to communicate the value under discussion. I'm only guessing here, but if what you are looking for is a different "number system" under which this particular value results in only a single "value" under the system, then you will probably need to start redefining rules before the real numbers, since the complex numbers are a "simple" algebraic extension of the reals... – abiessu May 08 '15 at 19:05
  • Abiessu Yes, that exactly what I'm talking about! :D – Kutu kupret May 08 '15 at 19:08
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The general definition of $a^x$ in complex numbers (for $a \ne 0$) is $\exp(x \log(a))$, where $\log(a)$ is any branch of the natural logarithm of $a$. In that sense, $a^x$ will have infinitely many possible values, because you can add any integer multiple of $2 \pi i$ to a value of $\log(a)$ and get another value. However, if you choose a particular branch of the logarithm to use, that specifies one value that you can think of as the value. For example, a popular choice is the "principal branch": $\text{Log}(a)$ is the branch of $\log(a)$ with imaginary part in the interval $(-\pi, \pi]$.

Thus, e.g., for negative $a$, $\text{Log}(a) = \ln(|a|) + \pi i$, and so $$a^x = \exp(x \text{Log}(a)) = |a|^x \exp(x \pi i)$$

Caution: many of the standard "laws" of exponents and logarithms are no longer always true for complex numbers. For example, $\text{Log}(a^x)$ can't be $x \text{Log}(a)$ if the imaginary part of $x \text{Log}(a)$ is outside the interval $(-\pi, \pi]$. As a result, $(a^x)^y$ is not always the same as $a^{xy}$. A simple example of this: $(-1)^3 = \exp(3 \pi i) = -1$ so $$((-1)^{3})^{1/3} = (-1)^{1/3} = \exp(\pi i/3) \ne (-1)^{(3/3)} = -1$$

Robert Israel
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  • The question appears to really be about whether it is possible to redefine the underlying system of numbers such that $\log(a)$ no longer has branches... – abiessu May 08 '15 at 19:19
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Yes. The complex number system $\Bbb C$ helps us handle things like this. Coincidentally, the first example you gave has a solution in the real numbers, $\mathbb{R}$.

Sean Henderson
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