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I was using Euler's formula $ e^{ix} = \cos(x) + i \sin(x)$ with the case when $ x = 2\pi$. $$ e^{i2\pi} = \cos(2\pi) + i\sin(2\pi) = 1 + 0i $$ So I end up with the identity $ e^{i2\pi} = 1$. Now, if I apply $\ln$ it goes $$ \ln e^{i2\pi}=\ln1$$ $$ i2\pi = \ln1$$ How $\ln1=0$, Should be that $i2\pi=0$ its true, or not? The first implication I found its that you can divide any imaginary number by zero. $$ {ai\over 0} = {ai\over i2\pi} = {a\over 2\pi}$$ What don you think?

poyea
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Daniel
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  • The real part is $0$, so it's purely imaginary. Also real, of course. – saulspatz May 23 '20 at 17:29
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    Once you enter the complex realm, the logarithm becomes a multi-valued function. (See Wikipedia's "Complex logarithm" entry.) So, you cannot go from $e^{i2\pi}=1$ to $i2\pi=0$. – Blue May 23 '20 at 17:30
  • @Blue But why you can go from $ e^{i\pi} = -1 $ to $ {i\pi} = \ln -1 $? – Daniel May 23 '20 at 18:15
  • Who says that you can? You may be confusing $\text{Ln}$ with $\ln$ or you may be confusing what domain or standard representations are allowed when discussing $\ln$. You've seen how $\sqrt{x^2}$ doesn't necessarily equal $x$, right? (for example $\sqrt{(-2)^2}=2\neq -2$) Or how $\arcsin(\sin(\theta))$ doesn't necessarily equal $\theta$? (for example $\arcsin(\sin(6\pi))=0\neq 6\pi$). Something similar is happening here. – JMoravitz May 23 '20 at 18:35
  • Just because $e^a = e^b$ for complex numbers that doesn't mean that $a=b$. In particular just becaues $e^{2\pi i}=e^0$ that doesn't mean that $2\pi i = 0$. They are most certainly different numbers. As for $\ln(e^z)$ compared to $z$, as above these are not necessarily the same. If talking about $\text{Ln}$, the principal logarithm, this will only be true when $z$ is in a certain range of values, $2\pi i$ falls outside of that range, but $0$ lies within it. – JMoravitz May 23 '20 at 18:38
  • oh, i didnt know that Ln and ln where different. So its just about the formalization. – Daniel May 23 '20 at 18:40
  • Yes, thanks. @Blue – Daniel May 23 '20 at 18:42

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You are assuming that there is a function $\ln\colon\Bbb C\setminus\{0\}\longrightarrow\Bbb C$ such that, if $z\in\Bbb C$, $\ln\bigl(e^z\bigr)=z$. But there is no such function.

José Carlos Santos
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