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So $i$ is the complex unit and $n \in \mathbb{N} $. $$e^{2 \pi \ n \ i} = 1$$

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

$$(e^{2 \pi \ n \ i})^{2 \pi \ n \ i} = e^{-4\pi^2 \ n^2}$$

$$e^{-4\pi^2 \ n^2} \neq 1$$

I’m confused with this, can someone please explain to my where the error is in the equations above, or in my thinking? The more I sit on this problem the less I understand.

graydad
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solid
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    They're not equal. This is because complex exponentiation does not define a single-valued function. What you've done here is found an alternate value for $1^{2\pi n i}$. – Dustan Levenstein Jan 27 '15 at 15:15
  • $e^z$ is defined by $1+z+\frac{z^2}{2!}+\cdots$ which is single valued. So $a^z$ is single valued only when $a=e$? – velut luna Jan 27 '15 at 15:42
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    $a^z=e^{z\log a}$, so can $e^z$ be written as $e^{z\log e}$? But $\log e=1$ or $1+2ni\pi$? – velut luna Jan 27 '15 at 15:47
  • @Kyson Your last remark is correct. The power series defines a particular selection of values for $e^z$ which happen to form an entire function. But, for example, $e^{1/2}$ has two values, given by the positive and negative square roots of $e$. The negative square root is obtained by taking $\log e = 1+2\pi i$. – Dustan Levenstein Jan 27 '15 at 16:23
  • So how should $e^z$ be defined? – velut luna Jan 27 '15 at 16:27
  • $e^{z \log e}$. – Dustan Levenstein Jan 27 '15 at 16:28
  • This is one reason we often define $\exp(z)$ in terms of the power series - to avoid our intuition about exponentiation by using the notation $e^z$. @Kyson – Thomas Andrews Jan 27 '15 at 17:36
  • @Thomas Andrews, do you mean we first define a single-valued function $\exp(z)$ by power series, then define $\log$ as its inverse, which is multi-valued, then define $a^z$ by $\exp(z\log a)$, which is multi-valued. Therefore $e^z=\exp(z \log e)$ is also multivalued? – velut luna Mar 12 '15 at 14:50
  • No, $e^z$ (or $(e^n)^z$ where $n$ is an integers - is a single-valued function always. $a^z$ is multivalued. @Kyson – Thomas Andrews Mar 12 '15 at 15:11
  • @Thomas Andrews, doesn't that sound a little bit strange? I mean $a^z$ are multivalued except for the special case of $a = e$. Also, it is not consistent with $e^z=e^{z \log e}$. We'll need to say $a^z = e^{z \log a}$ EXCEPT when $a=e$? – velut luna Mar 12 '15 at 15:46
  • Whoops, I just had a brain fault in the previous comment. It is also true that $e^z$ is multivalued. Was thinking of something else. @Kyson – Thomas Andrews Mar 12 '15 at 16:16

3 Answers3

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Exponentiation $x^y$ is generally very tricky in complex numbers. You have to give up at least one of the following:

  1. That $x^y$ is a single-valued function.
  2. That $x^y$ is continuous
  3. That $x^y$ is defined for all $x,y, x\neq 0$.

If you keep (1), then you also have to give up $(x^y)^z = x^{yz}$ as a rule.

The best thing to do is define $x^y$ as a multivalued function. Specifically, define a multivalued $\log x$, and define $x^y = e^{y\log x}$.

If you have a multi-valued $x^y$ then one of the values of $1^{2\pi n i}$ is $e^{-4\pi^2n^2}$.

If $y$ is rational in reduced form $\frac{p}{q}$ with $p,q\in\mathbb Z$, then $x^y$ has $q$ possible values. In particular, if $y$ is an integer, then $x^y$ has one value.

This is related to the fact that we usually pick the positive square root, but, for example, we can see $4^{1/2}=\pm 2$. In complex numbers, there are four values of $\sqrt[4]{1}=1^{1/4}$, namely, $\pm1, \pm i$.

If $y$ is not a rational number, $x^y$ has infinitely many values.

Thomas Andrews
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With complex exponentials, you aren't allowed to say $(a^x)^y=a^{(xy)}$, even for real $a$. You are just showing us a counter-example.

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The exponential function is periodic in $\mathbb{C}$, so it's not invertible in $\mathbb{C}$.

The confusion comes from the wrong idea that exists an inverse function $\log x$ with domain $\mathbb{C}$. An inverse function of exponential is well defined only if we take as his domain $\{z : -\pi<\mathcal{Im}(z)<\pi\}$ or a suitable Riemann surface and so we have no confusion.

In my opinion the confusion expressed in the answer is an example of how ill defined and confusing is the concept of multi-valued functions: a function must be single-value defined!

Emilio Novati
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