Why $e^{i(π/3)} \ne -1$? [closed]

I understand why $e^{i\pi} = -1$and as a result $e^{i2\pi}=\left(e^{i\pi}\right)^2=1.$ These results can be confirmed using Euler's formula

But why does $e^{i\pi/3}\neq -1$ as we can write it $(e^{i\pi})^{1/3}$ and as such $(-1)^{1/3}$ the cubic root of $-1$ is $-1$ itself.

Yet using Euler's formula we get $e^{i\pi/3}=\frac{1}{2}+i\frac{\sqrt 3}{2}$.

closed as unclear what you're asking by Did, José Carlos Santos, Xam, Chris Godsil, user223391 Apr 3 '18 at 2:51

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• Ok nevermind, (-1)^(1/3) has three different cubic roots, of which -1 is one – Eoin Mar 30 '18 at 12:39
• Already for squares and in the reals, your reasoning fails: $(-1)^2=1$ and as such $-1=1$ since the square root of $1$ is $1$... – Did Mar 30 '18 at 12:42
• You are using MathJaX very cumbersomely. Here's a good rule. Every occurenc of maths in your text goes between one pair of $-signs. So instead of writing e$^i^\pi$=-1, you write$e^{i \pi}=-1$. Afterall, this is one single maths expression. This will look much nicer. Notice that numbers are also shown differntly within$'s. Also, you should use MathJaX in the title of your question as well as in the body. – gebruiker Mar 30 '18 at 13:14
• Interessting question. It is worth mentioning that by this reasoning $\exp(ix/k)=-1$ for all odd numbers $k$. Which should alert you :). – MrYouMath Mar 30 '18 at 13:26
• @Eoin: Your comment starting with "ok nevermind" gives the wrong reason for your error; please read my answer carefully. – user21820 Mar 30 '18 at 13:57

I'm surprised that none of the other answers fail to point out the actual error in your question (no it is not that $-1$ has $3$ cube-roots):

It is not at all true that $(x^a)^b = x^{ab}$ for complex numbers $x,a,b$ in general, under any reasonable definition of complex exponentiation.

Counter-examples (for various conventions):

$-1 = (-1)^3 = (-1)^{6/2} \color{red}{\large\boldsymbol\ne} ((-1)^6)^{1/2} = 1^{1/2} = 1$

$i = (-1)^{1/2} = (-1)^{2/4} \color{red}{\large\boldsymbol\ne} ((-1)^2)^{1/4} = 1^{1/4} = 1$

It is wrong to think you can easily solve this using multi-valued functions. Even if $1^{1/4}$ is the set $\{1,i,-1,-i\}$, it is still not the same as just $i$.

You must be absolutely clear about the conditions under which certain identities hold. Relevant are the following:

$x^{ab} = (x^a)^b$ for any real number $x$ and natural numbers $a,b$.

If you want the more general fact for integer exponents:

$x^{ab} = (x^a)^b$ for any real number $x \ne 0$ and integers $a,b$.

In fact it turns out that 'miraculously' we have an even more general fact for real exponents:

$x^{ab} = (x^a)^b$ for any real number $x > 0$ and reals $a,b$.

Notice that all these precise statements about real exponentiation show you clearly that you must know exactly what the objects are before you can apply any operations to them, not to say claim any properties about the resulting values.

In particular, the above identities are simply meaningless if you do not specify what $x,a,b$ are!

For this reason it is actually an important question to ask whether there are corresponding rules for complex numbers.

Yes, but not as nice.

$x^{ab} = (x^a)^b$ for any complex number $x \ne 0$ and integers $a,b$. (*)

Here exponentiation is simply the result of starting from $1$ and repeatedly multiplying/dividing by $x$ where the number of times is specified by the exponent (multiplying for positive; dividing for negative). This fact holds in any structure that has invertible multiplication, including the field of rationals, the field of reals, and the field of complex numbers.

$x^{ab},x^a$ are well-defined since $x \ne 0$.

However, in general "$x^{ab} = (x^a)^b$" does not hold for complex $x$ even if $a,b$ are both rational, as already shown above.

So it's excellent that you ask whether some new structure (complex numbers) have the same properties as some other structure (real numbers) instead of just blindly assuming it does.

In general (*) holds for any field.

Suppose we have a field $S$ (such as the complex numbers) and an exponentiation operation that satisfies the following: $\def\lfrac#1#2{{\large\frac{#1}{#2}}}$

$x^0 = 1$ for every $x \in S$.

$x^{k+1} = x^k x$ for every $x \in S$ and integer $k$.

Note that any reasonable foundational system is capable of defining such an operation recursively (you need one direction for positive $k$ and another for negative $k$), and can easily prove by induction the following two theorems.

$x^{a+b} = x^a x^b$ for every nonzero $x \in S$ and integers $a,b$.

Take any nonzero $x \in S$ and integer $a$.

Then $x^{a+0} = x^a = x^a x^0$.

Given any integer $b$ such that $x^{a+b} = x^a x^b$:

$x^{a+(b+1)} = x^{(a+b)+1} = x^{a+b} x = ( x^a x^b ) x = x^a ( x^b x ) = x^a x^{b+1}$.

$x^{a+(b-1)} = x^{(a+b)-1} = x^{a+b} \div x = ( x^a x^b ) \div x = x^a ( x^b \div x ) = x^a x^{b-1}$.

Therefore by induction $x^{a+b} = x^a x^b$ for every integer $b$.

$x^{ab} = (x^a)^b$ for every nonzero $x \in S$ and integers $a,b$.

Take any nonzero $x \in S$ and integer $a$.

Then $x^{a \times 0} = x^0 = 1 = (x^a)^0$.

Given any integer $b$ such that $x^{ab} = (x^a)^b$:

$x^{a(b+1)} = x^{ab+a} = x^{ab} x^a = (x^a)^b (x^a) = (x^a)^{b+1}$.

$x^{a(b-1)} = x^{ab-a} = x^{ab} \div x^a = (x^a)^b \div (x^a) = (x^a)^{b-1}$.

Therefore by induction $x^{ab} = (x^a)^b$ for every integer $b$.

Notice that we did not use commutativity here, which in fact shows that the argument holds in any division ring. If you restrict the exponents to natural numbers, then it clearly holds in any group when "nonzero" is deleted.

• Seeing the frequency of the questions based on such confusions, and the regular inadequacy of the solutions proposed, one feels your answer should be in one way or another, "factored" to answer all of these at once. (+1) – Did Mar 30 '18 at 14:06

In the complex case a number not equal to $0$ has three cubic roots.

• To a downvoter: maybe a comment, why? – Przemysław Scherwentke Mar 30 '18 at 13:14
• (I didn't downvote) I think it requires some more explanation. Why does this answer the question? It is not obvious to me (but I don't have a very strong math background (yet) - I understand the question though) – Belle-Sophie Mar 30 '18 at 13:53
• @Belle-Sophie There are 3 roots and OP takes not $e^{i\pi/3}$ but $e^{i\pi}$. – Przemysław Scherwentke Mar 30 '18 at 13:56
• @Belle-Sophie: This answer is wrong, even though I didn't downvote it. See my answer for the proper explanation. – user21820 Mar 30 '18 at 13:56

The problematic step is when you say "the cubic root of $-1$ is $-1$ itself". Yes, $(-1)^3 = -1$, but there are other cube roots of $-1$, namely $e^{i(\pi/3)}$, for example. The problem is that $(-1)^{1/3}$ is multi-valued.

• This answer is wrong or misleading, because multi-valued functions do not 'solve' the problem, namely the error in the question remains even if you attempt to use multi-valued functions. – user21820 Mar 30 '18 at 14:00
• @user21820 I disagree. My answer does not claim that one ought to use multi-valued functions to solve the problem. The questions claims that $e^{i\pi/3}$ "can be written as" ${e^{i\pi}}^{1/3}$. Well this is true if you keep track of the right cube-root of $e^{i\pi}$ (perhaps the notation is suboptimal, but to say "it can be written as" is okay in my book). I am saying that the problem in the poster's logic seems to arise because they appear to have forgotten that the cube-root function is multivalued and have accidentally picked the 'wrong' value. – abc Mar 30 '18 at 15:08
• If you keep track of something in your head but it shows up nowhere in your answer, I think I'm justified in saying your answer is misleading. No? In any case, I disagree even with your "keeping track". What are you going to do with "$(2^{1+i})^{1+i}$" and things like that? If you don't allow them, then it's really not what complex exponentiation is about... – user21820 Mar 30 '18 at 15:27
• This question isn't asking how to evaluate $(2^{1+i})^{1+i}$. The question isn't asking for a general method for complex exponentiation. Nor is it asking for some sort of formal proof verification. It was asking how to reconcile what the poster saw as a contradiction - I think it basically reconciled by noting that writing $x^{1/3}$ is ambiguous/multi-valued. – abc Mar 30 '18 at 15:47
• My point is that you have not actually addressed the misconception in the question, which is clearly the logically unsound 'use' of a 'rule' that does not hold. In the first place, it is bad enough to introduce multi-valued functions, since that is not a proper use of the term "function", and since it will only serve to further confuse any student who is not taught properly about the current conventional definition of radicals and exponentiation (which are not multi-valued)... – user21820 Mar 30 '18 at 16:05