10
$\begingroup$

I am a bit confused with a result I get from Euler's formula:

$e^{2\pi i} = 1$

$\sqrt[3] { e^{2\pi i} }= \sqrt[3]{ 1 }$

$(e^{2\pi i})^{\frac{1}{3}}= 1$

$e^{\frac{2}{3} \pi i} = 1$

This last result seems problematic, since the left-hand side of the previous equation should result in...:

$e^{\frac{2}{3} \pi i} = −\frac{1}{2} + \frac{\sqrt{3}}{2}i$

... given that $e^{x i} = \cos(x) + i\sin(x)$

So is there a mistake in my first manipulations? If not, how can one interpret the results?

$\endgroup$
6
  • 13
    $\begingroup$ Your manipulations are fine. In fact, we can say that $\sqrt[3]{1}$ has three different values. You should confirm that your result $z$ satisfies $z^3 = 1$. $\endgroup$
    – Ben Grossmann
    May 23, 2016 at 19:40
  • $\begingroup$ Why do you think $\;\sqrt[3]1=1\;$ ? $\endgroup$
    – DonAntonio
    May 23, 2016 at 19:40
  • $\begingroup$ @Joanpemo It is merely the norm for most. $\endgroup$
    – Simply Beautiful Art
    May 23, 2016 at 23:50
  • $\begingroup$ One cube root function outputs the unique real cube root for a real input. If that's the function you meant in equation 2, then the move from line three to line four is invalid since you can't always take the product of exponents like that. $\endgroup$
    – Mark S.
    May 24, 2016 at 1:51
  • 2
    $\begingroup$ You may find Eisenstein integers interesting. $\endgroup$
    – PM 2Ring
    May 24, 2016 at 14:03

5 Answers 5

Reset to default
28
$\begingroup$

Think about it this way: if we say that $a=-1$, it follows that $a^2=1$. From there, if you take the square root, you get that $a=1$, meaning that $-1=1$. Clearly this is ridiculous. Roots are multivalued. The square root of a positive real number has two results: one positive, one negative. Similarly, there are three complex numbers whose cube is equal to $1$. Your mistake lies in assuming that $1^{1/3}$ is well-defined. It is true that $1^3=1$ and that $(-\frac{1}{2} \pm \frac{\sqrt{3}}{2}i)^3=1$.

$\endgroup$
8
  • 2
    $\begingroup$ I see. So in fact, contrary to what some people pointed out as my manipulations being correct, the error is in the second to the third line of my equations above, when $\sqrt[3] {1}$ becomes simply $1$. Clearly that's a fallacy, as it would imply that $cos(\frac{2}{3} \pi) = 1$. $\endgroup$
    – gilbertohasnofb
    May 23, 2016 at 21:22
  • 7
    $\begingroup$ Basically, yes. It is true that $1^3=1$, but $1^{\frac{1}{3}}$ is not uniquely defined. $\endgroup$
    – Neil
    May 23, 2016 at 21:23
  • 6
    $\begingroup$ Think of it this way: if $e^{2\pi i}$=1, then $e^{4\pi i}$=1 as well. As does $e^{2k\pi i} \forall k \in \mathbb{N}$. So one value for $1^{\frac{1}{3}}$ is given by $e^{\frac{2 \pi i}{3}}$, another is given by $e^{\frac{4 \pi i}{3}}$, and the third by $e^{\frac{6 \pi i}{3}}$. If you look at $e^{\frac{8 \pi i}{3}}$, this is equivalent to $e^{\frac{6 \pi i}{3}}*e^{\frac{2 \pi i}{3}}=e^{2 \pi i}*e^{\frac{2 \pi i}{3}}=1*e^{\frac{2 \pi i}{3}}=e^{\frac{2 \pi i}{3}}$, so going beyond $2 \pi i*3$ is unnecessary. Note that $e^{\frac{6 \pi i}{3}}=e^{2 \pi i}=1$, so $1$ remains a solution here. $\endgroup$
    – Neil
    May 23, 2016 at 21:34
  • 4
    $\begingroup$ @gilberto, another way to see it is to consider the factorization $z^3-1=(z-1)(z^2+z+1)$. $\endgroup$
    – J. M. ain't a mathematician
    May 23, 2016 at 22:37
  • 2
    $\begingroup$ To correct a bit, roots are multivalued in complex numbers. $\endgroup$
    – IS4
    May 24, 2016 at 9:29
5
$\begingroup$

You are making the classical mistake of generalizing the exponentiation rule $(a^b)^c=a^{bc}$ to complex numbers: it doesn't hold !

$\endgroup$
9
  • 1
    $\begingroup$ Nice remark... :) One tends to forget... $\endgroup$
    – Umberto
    May 24, 2016 at 13:03
  • $\begingroup$ It holds pretty frequently tho. $\endgroup$
    – dbanet
    May 24, 2016 at 20:08
  • $\begingroup$ @dbanet: what makes you think that ? $\endgroup$
    – user65203
    May 24, 2016 at 21:00
  • $\begingroup$ @Yves math.stackexchange.com/questions/1347504/… $\endgroup$
    – dbanet
    May 24, 2016 at 21:11
  • $\begingroup$ @dbanet: I read that the condition is $\exp(i2\pi kc)=1$, which never occurs when $c$ has an imaginary part and $k\ne0$ (which is the rule rather than the exception). $\endgroup$
    – user65203
    May 24, 2016 at 21:23
4
$\begingroup$

No, that looks good! The three cube roots of unity are indeed

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

Picture unit vectors in the complex plane at angles $0, 120, 240$ degrees.

You may have looked at your answer and scratched your head as to how weird it looked. The polar representation is compact and has polar symmetry: $z=re^{i\theta}.$ If $r=1$ then the only thing that changes when you square $z$ is the angle.

But slap the Cartesian coordinate system on top of that number, and you're forced to evaluate the $x$ and $y$ components. You're forcing square graph paper on a circle, and stuff doesn't line up. The $x$ and $y$ components are going to be, in general, more complicated (uglier?) when you do the conversion.

$\endgroup$
4
$\begingroup$

You have used the real-valued cube root function to go from $\sqrt[3]{1}$ to $1$. You have used a complex-valued cube root function to go from $(\mathrm{e}^{2\pi\mathrm{i}})^{\frac{1}{3}} = (1)^{\frac{1}{3}}$ to $\mathrm{e}^{\frac{2}{3}\pi\mathrm{i}}$. The two functions are not the same. In particular, the first function takes reals to reals and the second takes reals to non-reals (except for zero). As you have not applied the same function to both sides of your equality, you have no reason to expect to retain equality of their results.

To more clearly see what's going on, using your complex-valued cube root function and three different ways to write "$1$": \begin{align} 1 &= \mathrm{e}^0 = \mathrm{e}^{0 \pi \mathrm{i}} \text{,} \\ 1 &= \mathrm{e}^{2 \pi \mathrm{i}} \text{, and } \\ 1 &= \mathrm{e}^{2 \pi \mathrm{i}} \cdot 1 = \mathrm{e}^{2 \pi \mathrm{i}} \cdot \mathrm{e}^{2 \pi \mathrm{i}} = \mathrm{e}^{4 \pi \mathrm{i}} \text{.} \end{align} Then \begin{align} (\mathrm{e}^{0 \pi \mathrm{i}})^{\frac{1}{3}} &= \mathrm{e}^{\frac{0}{3} \pi \mathrm{i}} = \mathrm{e}^0 = 1\text{, } \\ (\mathrm{e}^{2 \pi \mathrm{i}})^{\frac{1}{3}} &= \mathrm{e}^{\frac{2}{3} \pi \mathrm{i}} \text{, and} \\ (\mathrm{e}^{4 \pi \mathrm{i}})^{\frac{1}{3}} &= \mathrm{e}^{\frac{4}{3} \pi \mathrm{i}} \text{.} \end{align}

If we were to keep going, we would find $(\mathrm{e}^{6 \pi \mathrm{i}})^{\frac{1}{3}} = \mathrm{e}^{\frac{6}{3} \pi \mathrm{i}} = \mathrm{e}^{2 \pi \mathrm{i}} = 1$ and cube roots using subsequent multiples of $2\pi \mathrm{i}$ continue cycling through the three we wrote above.

There are two ways to look at this. One is to take the point of view that there is one cube root function but every number is an infinite collection of equivalent polar forms: $(1)^\frac{1}{3} = (\mathrm{e}^{2 k \pi \mathrm{i}})^\frac{1}{3}$ for any integer $k$. When we take this point of view, we say things like "$1$ has three cube roots and you need to specify which one you mean." It would be more accurate (but too cumbersome) to say "the cube roots of the collection of things equivalent to $1$ form a set of three elements and you should specify how to choose which element of each set you mean when you say 'cube root'." (Note that once again, zero is different -- its set of cube roots has only one element. This is an argument against this point of view: you have to treat one number differently.)

The other point of view is that there are three cube root functions (using the same idea as above to pass between them): \begin{align} \sqrt[3_{(0)}]{1} &= \sqrt[3]{1} \cdot \mathrm{e}^{\frac{0}{3} \pi \mathrm{i}} \text{,} \\ \sqrt[3_{(1)}]{1} &= \sqrt[3]{1} \cdot \mathrm{e}^{\frac{2}{3} \pi \mathrm{i}} \text{, and} \\ \sqrt[3_{(2)}]{1} &= \sqrt[3]{1} \cdot \mathrm{e}^{\frac{4}{3} \pi \mathrm{i}} \text{,} \end{align} where we have used the parenthesized subscript on the index of the radical to pick out which of the three complex cube roots we mean and the unsubscripted index "3" to mean the real-valued cube root. Equivalently, we use the parenthesized number to pick out which multiple of $2\pi\mathrm{i}$ to use in the alternate version of the radicand. When we take this point of view, we say that the roots differ by the power of a cube root of unity since the three multipliers are all powers of $\zeta_3 = \mathrm{e}^{\frac{2}{3} \pi \mathrm{i}}$. It is perhaps unfortunate that $\zeta_3$ is not the right choice to recover the real-valued cube root function, but using that choice ($\zeta_3^0 = 1$), it's powers don't get us any complex numbers. (This is an argument against this point of view: the natural cube root is not the real valued cube root.)

Neither point of view is perfect, but both tell you the same thing -- the complex cube root is not as simple as the real cube root. To pick from the three options, you need to make some sort of choice. If you make no choice, you end up with three answers (except for zero).

$\endgroup$
3
$\begingroup$

Trying to put this as simply as possible:

In the complex numbers there are three cube roots of $1$.

On the right hand side of your equation you have chosen one of them (namely, $1$), and on the left hand side of your equation you have chosen a different one (namely $e^\frac{2\pi i}{3}$). The third, which you haven't used, is $e^\frac{4\pi i}{3}$ (exercise: prove that they're not equal).

You can't do that. In effect you've said, "If two values have the same cube, then the values are equal". That's a true statement in the reals (which is probably why you guessed it could be used) but not in the complex numbers.

Lessons to take away:

  • You can apply a function to both sides of an equality and preserve the equality, but cube root is not a function unless you specify which root and always use the same one.

  • The complex $n$th roots of $1$ are $$e^{\Large \frac{2 k \pi i}{n} \normalsize } \textrm{for } k = 1 \ldots n$$

Or, if you prefer, $k = 0 \ldots n-1$ since $e^0 = e^{2\pi i} = 1$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .