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What's wrong with this : $$e^{i\pi} = -1$$ $$\therefore e^{2i\pi} = 1$$ $$\therefore log \left( e^{2i\pi} \right ) = log(1) = 0$$ $$\therefore 2i\pi = 0$$

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    $\begingroup$ How do you conclude that $2\pi i =0$? What rule are you using? $\endgroup$ – Thomas Jun 10 '15 at 13:40
  • $\begingroup$ @Thomas $log(e^x) = x$ $\endgroup$ – Ojas Jun 10 '15 at 13:41
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    $\begingroup$ @Thomas i suppose he is trying to argue that $e^x = 1$ implies $x=0$, which is obviously problematic in complex arithmetic. Not sure what he needs the logs for. $\endgroup$ – gt6989b Jun 10 '15 at 13:42
  • $\begingroup$ Where did you get that rule from? $\endgroup$ – Thomas Jun 10 '15 at 13:42
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    $\begingroup$ The complex logarithm is a multi-valued "function". While it is true that $e^{\log z} = z$ for any choice of $\log z$, $z$ is just one of the infinitely many possible values of $\log e^z$. $\endgroup$ – mrf Jun 10 '15 at 13:43
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The problem is that $z\mapsto e^z$ is not a bijective function, so the logarithm is not, in general, well defined. You need to specify which branch of the logarithm you are using.

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In complex numbers, $$e^a=e^b\color{red}{\implies}a=b$$ does not hold. You just found a counter-example.

But you can write $$e^a=e^b\color{green}{\implies}a=b+2k\pi i.$$

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    $\begingroup$ …where $k$ is some integer. $\endgroup$ – Akiva Weinberger Jun 10 '15 at 15:01
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The rule that you probably know is that $\log(e^x) = x$. You use this to say that if $$ e^a = e^b $$ then $$ a = b. $$ We get this exactly by taking a logarithm on both sides. And this is certainly true when we talk about real numbers. Expanding the concept of logarithms to include complex numbers isn't trivial. You can read a bit about it on the Wikipedia article. That it isn't trivial, by the way, doesn't mean that it is difficult either.

To see an example of how the above doesn't work for complex numbers you have the example that $$ e^{0} = 1 \\ e^{2\pi i} = 1. $$ (this is just what you wrote) And, as you probably know, you can put a lot of numbers in the exponent to get $1$. So taking a logarithm on both sides to bring the exponenet down isn't going to work and all of this is hidden in the definition of the complex logarithm(s).

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