$\ e^{i 2\pi} = 1,$
$\ e^{0} = 1$
$\Rightarrow $
$\ e^{i 2\pi} = e^{0}$
$\Rightarrow $
$\ {i 2\pi} = 0 $
$\Rightarrow $
$\ i = 0 $
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1complex exponential isn't injective, thus $e^{2\pi i}=e^0$ doesn't implies $2\pi i=0$. – Fabio Lucchini Jan 11 '18 at 21:02
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1You are missing the cases $2=0$ and $\pi=0$. – Jan 11 '18 at 21:05
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And the wrong tag too, (eulers-constant) is about $\gamma$... – Hans Lundmark Jan 12 '18 at 05:47
2 Answers
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This argument is actually an excellent proof that the laws of exponents that hold for real numbers - in particular, the law that $a^x = a^y \implies x = y$ - do not hold for complex numbers. This isn't a contradiction - there are plenty of mathematical operations that work like this.
For example, imagine that you'd never worked with negative numbers before. You probably would have learned some weird "laws"; for example, you'd think that $x^2 = y^2 \implies x = y$. This is true as long as $x$ and $y$ are both non-negative. But as soon as someone told you about negative numbers, your "law" would break down, in the same way as you just saw the exponent law break down when switching from real numbers to complex numbers.
Reese Johnston
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The complex function $\exp:\mathbb{C}\to\mathbb{C}\backslash\{0\}$ is not injective. Therefore you cannot conclude that $\ e^{i 2\pi} = e^{0}$ implies $i 2\pi=0$.
However, with the domain $\mathbb{R}$ this function is injective.
The Phenotype
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