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We know that $$e^{\pi i}+1=0$$and $$e^{\pi i}=-1$$

So$$(e^{\pi i})^2=(-1)^2$$$$e^{2\pi i}=1$$

Because $1$ is the multiplicative identity,$$(e^{2\pi i})^x=1^x$$$$e^{2\pi xi} =1$$should also hold true.

But we also know that $$e^{xi}=\cos(x)+i\sin(x)$$and so$$e^{2\pi xi}=\cos(2\pi x)+i\sin(2\pi x)$$which does not equal 1 for all values of $x$.

Now I realize I probably didn't break math, so I must be making an invalid assumption. What is wrong with my reasoning?

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  • $\begingroup$ @labbhattacharjee Your link seems to say that $e^{2\pi i x}$is multi-valued. Is that right? $\endgroup$
    – Anna
    Jan 27 '14 at 12:00
  • $\begingroup$ exactly, one of the values is $1$ $\endgroup$ Jan 27 '14 at 12:17
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    $\begingroup$ Likely you known that from $(-1)^2 = 1^2$ you cannot conclude $-1=1$. What you see is not really different. $\endgroup$
    – quid
    Feb 5 '16 at 19:23
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    $\begingroup$ @hardmath I'm guessing because it preceded my question. I do think that the other question is clearer, and essentially, what I wanted to ask but didn't know at the time I posted this. Maybe this got more attention because of the "fake-proof" format/tag. If I may suggest a solution, maybe the answers below could be added to Anna's question, if such a merge is possible, and remove my question. I feel like the only thing difference between the two is that mine is unnecessarily convoluted and has an eye-catching title. $\endgroup$ Apr 28 '18 at 1:48
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The notion that $(a^b)^c=a^{bc}$ has to be abandoned in complex analysis.

Or, you have to allow that $a^b$ is a multi-valued function and then you can actually say that (one of) the values of $1^x$ is $\cos(2\pi x)+i\sin(2\pi x)$. With multi-valued functions you can say "All of the values of $a^{bc}$ are values of $(a^b)^c$," but not visa versa.

Multi-valued exponentiation can be seen as an extension of the idea that there are two "square roots," and, while we usually take $\sqrt{x}$ to be the positive one, we might sometimes prefer to think of $\sqrt{x}$ as a multi-valued function. For example, if $\sqrt{x}$ is multivalued, then you can write the quadratic formula as:

$$\frac{-b+\sqrt{b^2-4ac}}{2a}$$

and no longer have that pesky $\pm$ symbol from the usual formula, being implicit in the multi-valued $\sqrt{x}$ function. But the obvious problem with multi-valued functions is that the above "looks like" it is describing a single root, when it is describing two roots.

The other problem with multivalued functions is, what would one mean by:

$$a^{b} + a^{2b}?$$Most of the time when you see something like this, you probably don't want to pick from all values of $a^{b}$ and all values of $a^{2b}$, but rather you want to pick the same "branch," which amounts to picking the same value for $\log a$ for each term, amongst the infinitely many possible values for $\log a$.

So, in short: Exponentiation in complex numbers is irritating and no fun.

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  • $\begingroup$ It has to be abandoned in real analysis already: $((-1)^2)^{1/2} = \sqrt{1} = 1 \neq (-1)^{2 \cdot 1/2} = (-1)^1 = -1$... $\endgroup$ Feb 26 '16 at 10:02
  • $\begingroup$ @NajibIdrissi ... you’re hurting my brain... I must have missed your comment when I first skimmed over the answers. I had always assumed this to be valid for real numbers, now you’re telling me its not even true in the rationals. I could have sworn I was taught that ((a)^b)^{c} = ((a)^bc) within some set of parameters that eludes me. Are these manufactured memories? Are they common misconceptions? $\endgroup$ Aug 4 '17 at 3:51
  • $\begingroup$ @user3052786 It's all kosher if $a,b,c>0$. $\endgroup$ Aug 15 '17 at 22:59
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The property $(a^b)^c = a^{bc}$, true in the case ${\text{positive}^\text{real}}$ (with $a^b$ always positive) isn't true in the complex case. Example from the link:

$$(1-i)^{2i} \ne ((1-i)^2)^i.$$

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    $\begingroup$ This is quite a vague answer. Plus it is not true in the reals either. $\endgroup$
    – quid
    Feb 5 '16 at 19:00
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    $\begingroup$ What is true in the real case? And what even is the real case? Leaving aside that OP does never talk about a multiplicative identity (in the sense of an equality), but says that $1$ is the multiplicative identity element so that it is not even directly clear what you refer to. $\endgroup$
    – quid
    Feb 5 '16 at 19:05
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    $\begingroup$ Yes, but this is not at all what is written in OP. $\endgroup$
    – quid
    Feb 5 '16 at 19:10
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    $\begingroup$ No they have not. My issue is that OP does not use the term multiplicative identity in the way you use it at all. The "your" is thus confusing. Please read again what it says in OP. "Because $1$ is the multiplicative identity," so $1$ is OP's multiplicative identity. This cannot be wrong. Why not just quote the equation that is wrong. $\endgroup$
    – quid
    Feb 5 '16 at 19:15
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    $\begingroup$ The OP acted as if the identity equation $(a^b)^c = a^{bc}$ for positive real numbers were true for complex numbers as well. Note that even in the positive reals, that equation is not a "multiplicative identity"; the multiplicative identity is just the number $1$. kamil09875 showed how the OP incorrectly applied the identity equation from the positive reals; kamil09875 never said this is a multiplicative identity. $\endgroup$
    – David K
    Feb 5 '16 at 19:58
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In complex numbers exponentiation rules are a bit different, in this case $$(e^{2 \pi i})^x\not\equiv e^{2 \pi i x}$$

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Basically, you don't have the equality $$\left(a^b\right)^c = a^{bc}$$ for complex numbers. The way I explain this to myself is that for real values, $a^b$ is defined as the limit of $a^{q_n}$ where $q_n$ is a rational number and $q_n\rightarrow b$ as $n\rightarrow\infty$. As $a^q$ is defined through roots for rational numbers, and roots become complicated for complex numbers, it makes sense that the original rule will become shaky.

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  • $\begingroup$ Well, I realise that it doesn't work. I was wondering why :) $\endgroup$
    – Anna
    Jan 27 '14 at 12:01
  • $\begingroup$ I tried to explain my reasoning behind it. the problem is that roots are not well defined for complex numbers. If you just want to know "why is something not so", then the answer is "because it isn't" $\endgroup$
    – 5xum
    Jan 27 '14 at 12:06
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When you have $e^{2\pi i x}$ you can think about the $x$ as how much you're going to rotate the number 1 (which is $e^{2\pi i}$ in complex form) in the counter-clockwise direction.

For example, if $x\in[0,1]$, then no rotation is just $e^0=1$, a full rotation is $e^{2\pi i \cdot 1}=1$. So for every integer-$x$, we always take $e^{2\pi i x}$ back to the real line again, where all our nice algebra rules hold.

If $x$ is not an integer (like $x=0.354$), then we have rotated 1 somewhere out in the complex plane, which means it is on the form $a+bi$. We know that $1^x$ is not on the form $a+bi$ when $x$ is a real number ($1^{0.333}=1$). So we see that clearly rule doesn't hold if $x$ is some decimal number.

Intuitively you can think that the fact that complex numbers rotate breaks the exponentiation rule, which is only true for real numbers which cannot rotate, but can only flip, e.g. $-1\cdot7$ flips $7$ to $-7$.

Edit: In fact, a full rotation is the same as two flips. So when $x$ is an integer we might as well be working with real numbers where the rule holds. [Disclaimer: maybe, look at it in any way that makes it fun.]

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  • $\begingroup$ I know this is an old answer, but it recently got merged (for some reason) to my question, I think it's an interesting way to visualize the idea, but a quick explanation of polar notation might be helpful to the uninitiated (as I was when I asked this question). $\endgroup$ Apr 29 '18 at 21:49
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Assuming equality is true for all $x$ then $$e^{2\pi x i}=1$$ Using Euler's Theorem,$$\cos (2\pi x)+i\sin(2\pi x)=1$$ comparing real & imaginary parts on both the sides, $$\cos (2\pi x)=1\iff 2\pi x=2\pi k\iff x=k$$ & $$\sin(2\pi x)=0\iff 2\pi x=k\pi \iff x=\frac k 2$$

where $k$ is any integer.

Thus, $x=k$ is the solution of given equality. The above equality will hold only & if only $x$ is an integer. Hence our assumption is wrong.

hence $e^{\large 2\pi x i}=1$ is not true for all $x$

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    $\begingroup$ I think that the asker knew that the claim is not true for all $x$, and was instead interested in where the "proof" breaks down, not if the conclusion is true. $\endgroup$
    – Dylan
    Feb 5 '16 at 19:57

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