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I got an idea, but that doesn't match with Euler's theory.. So What's wrong?!

$$e^{jx} = (e^{j 2\pi})^{x/2\pi} = 1^{x/2\pi} = 1$$

anomaly
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    $1^\frac{x}{2\pi}$ is not always equal to 1 on the complex plane. – zahbaz Apr 17 '15 at 20:10
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    @zahbaz Your comment is incorrect. What happens when $x$ is real? The mistake in the step before that, i.e., what is incorrect is that $(e^{it})^s \neq e^{its}$. – Adhvaitha Apr 17 '15 at 20:13
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    This is just a different variant of the good old $1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{-1}\sqrt{-1} = i^2 = -1$. – Winther Apr 17 '15 at 20:17
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    @user17762 I disagree. If we're working on the complex plane, then for $x$ real, $1^x$ maps to the complex unit circle $|z|=1$... not just to 1. It only maps to 1 if $x$ is an integer. I do agree with the last part of your comment. – zahbaz Apr 17 '15 at 21:12
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    @user17762 Consider $1^{1/4} = (e^{i2\pi n})^{1/4} = e^{i \frac{\pi}{2}n}$. Taking one branch of these values (namely $n=0,1,2,3$) gives $1^{1/4} = 1,-1,i,-i$. Alternatively, consider $i^4$. – zahbaz Apr 18 '15 at 01:49
  • @zahbaz Yes, of course. I was being an idiot. Thanks. – Adhvaitha Apr 18 '15 at 01:50

2 Answers2

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You have just proven that $(a^b)^c=a^{bc}$ is only valide when $a$, $b$ and $c$ are reals and not for any complex numbers.

Look at that other one if we assume the identity above for complex numbers:

$$(e^{2i\pi})^{2i\pi}=e^{-4\pi^2}=1^{2i\pi}=1$$

Absurd!!

marwalix
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  • More precisely for $(a^b)^c=a^{bc}$ to be valid, $a$ has to be a positive real (for exponentiation with non-integer exponent to be well defined in the firs place) and $b$ has to be real. It is allowed that $c$ be any complex number. However it is not sufficient that instead of $b$ being real just $a^b$ is real (even though in that case both sides of the equation are well defined), as your example shows. – Marc van Leeuwen Apr 18 '15 at 08:53
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Raising a complex number of the form $e^{it}$ to another number $s$, where $s$ is not an integer is a multivalued function. Hence, your first step, i.e., $(e^{i 2\pi})^{x/2\pi} = e^{ix}$ is incorrect.

Adhvaitha
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