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My professor showed the following false proof, which showed that complex numbers do not exist. We were told to find the point where an incorrect step was taken, but I could not find it. Here is the proof: (Complex numbers are of the form $\rho e^{i\theta}$, so the proof begins there) $$\large\rho e^{i\theta} = \rho e^{\frac{i\theta*2\pi}{2\pi}} = \rho (e^{2\pi i})^{\frac{\theta}{2\pi}} = \rho (1)^{\frac{\theta}{2\pi}} = \rho$$ $$Note: e^{i\pi} = -1, e^{2\pi i} = (-1)^2 = 1$$ Since we started with the general form of a complex number and simplified it to a real number (namely, $\rho$), the proof can claim that only real numbers exist and complex numbers do not. My suspicion is that the error occurs in step $4$ to $5$ , but I am not sure if that really is the case.

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marked as duplicate by Matthew Towers, user21820, JonMark Perry, Simply Beautiful Art, user223391 Dec 28 '17 at 15:06

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  • $\begingroup$ $e^{2\pi i+2\pi k i}=1$, where $k\in\mathbb{Z}$ for starters. Also, what is $\left(e^{z}\right)^w$ with $z,w\in\mathbb{C}$? $\endgroup$ – Antinous Dec 27 '17 at 17:19
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    $\begingroup$ The special case $\theta=\pi$ is $-1=(-1)^{2/2}=((-1)^2)^{1/2}=1^{1/2}=1$, so negative numbers do not exist either. $\endgroup$ – Carsten S Dec 27 '17 at 19:41
  • $\begingroup$ you first find the total exponent then perform potentiation. $\endgroup$ – user499752 Dec 28 '17 at 3:31
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The error lies in assuming that $(\forall a,b\in\mathbb{C}):e^{ab}=(e^a)^b$.

It's worse than wrong; it doesn't make sense. The reason why it doesn't make sense is because $e^a$ can be an arbitrary complex number (except that it can't be $0$). And what is $z^w$, where $z,w\in\mathbb C$? A reasonable definition is that it means $e^{w\log z}$, where $\log z$ is a logarithm of $z$. Problem: every non-zero complex number has infinitely many logarithms: if a number $\omega$ is a logarithm, then every number of the form $\omega+2k\pi i$ ($k\in\mathbb Z$) is also a logarithm.

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    $\begingroup$ I think it’s worth explaining to OP why it makes no sense. When your background is real, it’s a completely natural thing to expect or assume must be okay. $\endgroup$ – Randall Dec 27 '17 at 17:21
  • $\begingroup$ I agree with Randall. Would you be able to explain why this is the case? So far most of what I have dealt with is in the set of real numbers. $\endgroup$ – Madhav Nakar Dec 27 '17 at 17:23
  • $\begingroup$ @Randall You are right. I have edited my answer. $\endgroup$ – José Carlos Santos Dec 27 '17 at 17:25
  • $\begingroup$ @MadhavNakar What do you think now? $\endgroup$ – José Carlos Santos Dec 27 '17 at 18:26
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    $\begingroup$ While the opening sentence of this answer is correct, the OP only involves a case of $(e^a)^b$ where $e^a$ is (positive) real (more precisely, it is $1$), so the second paragraph going on about the (very real, no pun intended) difficulties of defining $z^w$ when $z$ is not real (or I would add, when it is real but negative) do not really pertain to the original question. $\endgroup$ – Marc van Leeuwen Dec 28 '17 at 7:24
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Because for complex numbers, $e^{zc}\ne(e^z)^c$ for $z,c \in \mathbb{C}$.

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  • $\begingroup$ Does this imply that only one-to-one functions have this property? $\endgroup$ – Madhav Nakar Dec 27 '17 at 17:21
  • $\begingroup$ I know that one-to-one functions have this property because they are not multi-valued as explained above but I am not sure the ones that have this property is only one-to-one functions. $\endgroup$ – ArsenBerk Dec 27 '17 at 17:29
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    $\begingroup$ $f(z)=1$ is also multiplicative. $\endgroup$ – Kevin Dec 27 '17 at 18:39
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    $\begingroup$ Sorry, failure of this identity and the exponential function not being injective are entirely different matters. Maybe exponential function not being injective is a reason that defining $(e^z)^c$ is problematic and cannot be done so as to have this identity, but "in other words" is just not right. $\endgroup$ – Marc van Leeuwen Dec 28 '17 at 7:15

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