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Apologies upfront if this is a very basic question.

I don't understand why the function: $$y(x)=x^{2/3}$$ yields complex numbers with an imaginary part when $x<0$.

I can rewrite the above equation as: $$y(x)=x^{2/3}=\sqrt[3]{x^2}$$

So for example, if $x=-2<0$, the above rewritten formula gives: $$y(x)=\sqrt[3]{(-2)^2}=\sqrt[3]{4}$$ which is a real number.

Where am I going wrong? Thanks.

Jose Lopez Garcia
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  • Who said it gives imaginary? – Kenny Lau Dec 21 '17 at 14:34
  • Wolfram Alpha says so: http://www.wolframalpha.com/input/?i=graph+y%3Dx%5E(2%2F3)&t=efn01 – Jose Lopez Garcia Dec 21 '17 at 14:37
  • similar question: https://math.stackexchange.com/questions/109316/find-the-domain-of-x2-3 – Vikram Dec 21 '17 at 14:40
  • In the reals the expression is positive. In the complex, you can consider three branches, of which one is pure real. –  Dec 21 '17 at 14:44
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    Wolfie A is not a definitive reference. There are three complex solutions to $y^3=x^2$; it's a matter of taste which you prefer. I would say that this shows it is dangerous to write $a^b$ is a is not positive unless you are willing to state explicitly what actually you mean in your particular context. – Angina Seng Dec 21 '17 at 14:51
  • In fact, $x$ being negative is irrelevant to the discussion ! A number has three complex cubic roots. –  Dec 21 '17 at 15:04
  • @YvesDaoust ...unless it is zero. :) Anyway, WA sometimes gets pretty high and it isn't very reliable. – DonAntonio Dec 21 '17 at 15:07
  • Statements about WA's rigor and reliability are unfair. In all subtle instances I have seen, WA was perfectly right. It is especially good dealing with corner cases in complex numbers. –  Dec 21 '17 at 15:15
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    @JoseLopezGarcia Do you see the small notice "Assuming the principal root | Use the real‐valued root instead"? This notice has to do with branches of the complex logarithm, which is used in defining $z^w$ for complex numbers $z$ and $w$. In particular, it refers to the principal branch of the logarithm, which is the one most commonly used. If you click the link, you'll get your expected real graph.$${}$$For more information on branches and the complex logarithm, check my answer below. – Fimpellizzeri Dec 21 '17 at 15:19

4 Answers4

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Note that $\sqrt[3]{4}$ is not a real number if you are working on complex numbers.

Indeed $\sqrt[3]{4+0i}$ has three solutions in $\mathbb{C}$.

enter image description here

user
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The thing is that any nonzero complex number has $n$ distinct $n$-th roots, so that $\sqrt[n]{x}$ is not uniquely defined for $n\geq 2$, in general.

However, every positive real a number has a positive, real $n$-th root. Hence, in many contexts we write $\sqrt[n]{x}$ to denote the function on the positive reals that yields this positive real root.
When $n$ is odd, something similar holds for negative real $x$: every such $x$ has a negative real $n$-th root. Similarly, for these contexts $\sqrt[n]{x}$ is used to denote the function on the reals that yields the real $n$-th root.

Now we're $\sqrt[n]{x}$ used to writing as $x^{1/n}$, and this happens in these contexts I've pointed out above. But what if instead of $x^{1/n}$ we're dealing with $x^{c/n}$? Do we consider that as $\sqrt[n]{x^c}$ or ${\left(\sqrt[n]{x}\right)}^c$? The fact of matter is that sometimes these yield different answers, and sometimes these are undefined.

Consider $(-2)^{2/4}$. Is that $(-2)^{1/2}=\sqrt{-2}$? Is that ${\left(\sqrt[4]{-2}\right)}^2$? Is that $\sqrt[4]{(-2)^2}=\sqrt[4]{4}$?
See what we're getting at?

The thing is, except for very few particular cases of exponent $p/q$ $($those when $q=1$, or when $p=1$ and $q$ is odd$)$, there is no reasonable single-valued definition for $x^{p/q}$ on all of the reals.
Thus, unless you're only concerned about the positive reals, your best bet is realizing that this is indeed a more subtle topic: that $x^{p/q}$ is not uniquely defined, and that the theory for dealing with this involves complex numbers.

For complex numbers $z$ and $w$, one usually defines $z^w$ as $\exp(z\,\text{Ln}(w))$, where $\exp$ is the exponential and $\text{Ln}$ is the complex logarithm. The complex logarithm is not like the real one.

The real logarithm can be defined as the inverse of the exponential function $\exp:\mathbb{R}\longrightarrow[0,+\infty)$. However, the complex logarithm is not injective; it is periodic with period $2\pi i$, and hence does not admit an inverse, at least not in the classical, single-valued sense.

What we do is take branches of the logarithm, that is, we choose some specific values of its multi-valued inverse as our single-valued inverse. However, this choice is arbitrary, and moreover it cannot be done continuously throughout all of the complex plane. This latter observation is why we talk about branch cuts: these are parts of the complex plane we remove from the domain of our chosen single-valued inverse. We (once again arbitrarily!) choose some points of our would-be domain and decide that our single-valued inverse will not be defined on those points. We do so in order for the resulting inverse to be continuous (and more) in what remains of its domain.

Notice how much arbitrary choice is involved in this? In general, these choices must be specificed beforehand!

Fimpellizzeri
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Interesting that you choose to say that

$y(x)=x^{2/3}=\sqrt[3]{x^2}$

You could have said that

$y(x)=x^{2/3}=\sqrt[3]{x}^2$ instead, no?

Just asking yourself if both formulas are equal should give you hints as to why we can speak about solutions with imaginary parts.

The equality symbol can both time be put in question... of course.

Martigan
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  • But is the rewritten form that I wrote correct? Can $x^{2/3}$ be written as $\sqrt[3]{x^2}$? – Jose Lopez Garcia Dec 21 '17 at 14:38
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    @JoseLopezGarcia It can be written either way... as long as $x$ isn't negative. Non-integer exponents with negative bases are tricky. For instance, you probably know that $\frac23=\frac46$ (in other words, the two fractions represent the exact same number), but is $(-2)^{2/3}$ equal to $(-2)^{4/6}$? Should the value of $(-2)^{0.6666\ldots}$ depend on exactly which fraction you use to represent the exponent? – Arthur Dec 21 '17 at 14:42
  • No, it is true only for $x>0$. Otherwise, this way of rewriting is excluding some solutions, as the $x^2$ function is clearly a surjection. – Martigan Dec 21 '17 at 14:43
  • But in a related question(https://math.stackexchange.com/questions/109316/find-the-domain-of-x2-3) Arturo claims the Domain is all real numbers. Anyone that gives me a clarifying answer that I can upvote and pick as correct? – Jose Lopez Garcia Dec 21 '17 at 14:46
  • @JoseLopezGarcia In the post you mention, the second answer gives many details about the possibility (or not) of considering fractional exponents of negative numbers. – Martigan Dec 21 '17 at 14:53
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By the definition of the cubic root, $y$ is a solution of

$$y^3=x^2>0.$$

In the reals, $$y=\sqrt[3]{x^2}$$ is the only solution.

In the complex, there are three distinct solutions,

$$\sqrt[3]{x^2},\\\omega\sqrt[3]{x^2},\\\omega^2\sqrt[3]{x^2}$$

where $\omega$ is a complex cubic root of the unit (such as $\dfrac{-1+i\sqrt3}2$).