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$$e^{i} = e^{i2\pi/2\pi} = (e^{2\pi i})^{1/(2\pi)} = 1^{1/(2\pi )} = 1$$

Obviously, one of my algebraic manipulations is not valid.

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    $\begingroup$ @StefanHansen, you can elaborate (not because it is insufficient but to make it worthy of being called answer.) a little and answer it . $\endgroup$ – 007resu Jan 18 '13 at 18:15
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    $\begingroup$ The proof is exactly the same idea as $-1=\sqrt{(-1)^2}=\sqrt{1}=1$. It is less obvious just because the "root" part is the $\frac{1}{2\pi}$ power... $\endgroup$ – N. S. Jan 18 '13 at 19:08
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    $\begingroup$ Or $1=1^{1/2}=(e^{i 2 \pi })^{1/2}= e^{i \pi} = -1$ $\endgroup$ – leonbloy Jan 18 '13 at 19:11
  • $\begingroup$ @leonbloy When I first learnt of Euler's Identity, this was the first thing that came to mind. It is my favourite paradox (although not true) :D $\endgroup$ – Mr Pie Feb 27 '18 at 14:17
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As said in the comments, the expressions $(a^b)^c$ and $a^{bc}$ are in fact multivalued functions; they are not a uniquely determined complex number. A classic example of a multivalued function is the complex logarithm denoted by $\log(z)$, $z\in\mathbb{C}$. The complex logarithm $\log(z)$ is any complex number $w$ satisfying $e^w=z$ (which has several solutions, see e.g. this), and hence $\log(z)$ gives rise to a whole set of complex numbers instead of just a single complex number.

Complex exponentiation such as $z^w$ for $z,w\in\mathbb{C}$ is usually defined as $$ z^w=\exp(w\log(z)), $$ where $\log(z)$ is the complex logarithm, and hence this is also a multivalued function. I hope this sheds some light on the problems with doing manipulations on complex numbers as if they were real numbers. See also this for other examples of identities which fail when using complex numbers as they were real numbers.

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I'm going to add this here as the most general answer to this common problem in solving equations, a principle which the other answers do not cover:

When asked to solve for $y$, an equation of the form:

$$y=f(x)$$

(in your example $f(x)=e^i$)

The approach commonly taken is to solve by writing a sequence of equivalent equations working towards $y=\text{the desired answer}$, in your case as follows:

$y=e^{i}\\ y= e^{i2\pi/2\pi} \\y= (e^{2\pi i})^{1/(2\pi)}\\y = 1^{1/(2\pi )}\\y = 1$

Then this method is only reliable if it is truthful to write if and only if between every consecutive pair of equivalent equations in sequence as follows:

$y=e^{i}\\\iff\\ y= e^{i2\pi/2\pi} \\\iff\\y= (e^{2\pi i})^{1/(2\pi)}\\\iff\\y = 1^{1/(2\pi )}\\\iff\\y = 1$

Due to the periodicity of complex powers, the $\iff$ statement is contradictory to the laws of arithmetic in various places in the above chain.

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$$e^{i} = e^{i + i2\pi k} = e^{(i2\pi + i(2\pi)^2k)\frac{1}{2\pi}}=\bigg(e^{(i2\pi + i(2\pi)^2k)}\bigg)^{\frac{1}{2 \pi}} = \bigg(e^{i2\pi} e^{i(2\pi)^2k}\bigg)^{\frac{1}{2 \pi}} = \bigg( e^{i(2\pi)^2k}\bigg)^{\frac{1}{2 \pi}} = e^{i2\pi k}$$

$$e^{i} \ne e^{i2\pi k}$$

maybe you are right, its still a paradox.


Could it be that the algebraic property:

$$a^{pq} = (a^p)^q$$

is not a valid manipulation for complex exponentials?

$$e^{\mathbf{i}pq} \ne (e^{\mathbf{i}p})^q$$

pulling a power out of the complex exponential has an effect of changing the frequency of oscillation of the complex exponential...

I would change your complex algebra rule book to state that pulling a power out of a complex exponential can lead to inconsistent equations and should be avoided.

complex exponential anti-rule #1: Algebraically pulling a power out of the complex exponential leads to inconsistent equations. While $a^{pq} = (a^p)^q$ is true for normal algebra, for complex exponentials: $e^{\mathbf{i}pq} \ne (e^{\mathbf{i}p})^q$. For example: $e^{i} = e^{i2\pi/2\pi} = (e^{2\pi i})^{1/(2\pi)} = 1^{1/(2\pi )} = 1$, but $e^{i} \ne 1$.


But, you can find it with Euler's formula:

$$e^x = cos x + i sin x$$

$$e^i = e^{1i}$$

$$e^i = \cos 1 + i \sin 1$$

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    $\begingroup$ If there already exists a satisfactory answer, you shouldn't add one that's essentially a guess, particularly if it's 6 years later and it's reasonable for you to end the entire thing with "this all could be bogus". Your string of equations leading to the idea of "picking the right value of k" makes the same fundamental mistake as the OP multiple times. Please try not to give the impression that math is mysterious and/or hokey-- it's patently false and not constructive for people trying to learn. See my answer at math.stackexchange.com/a/3219114/355036 for the reason OP's proof fails. $\endgroup$ – jawheele May 9 '19 at 4:12

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