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So I came across this video: https://www.youtube.com/watch?v=R476CTKUIr4 in which the creator shows an incorrect proof of π = 0 and the mistake made. The video proves this using the exponent rule $(a^b)^c = a^{bc}$ and shows that it cannot be applied to imaginary numbers. Here are the first few steps from the video which lead me to prove π = 0 (a different way) and I am wondering why this way is also wrong.

e^(iπ) +1 = 0

e^(iπ) = -1

[e^(iπ)]^2 = 1

e^(2iπ) = 1 -> True, checked using WolframAlpha

e^(2iπ) = e^0

2iπ = 0

iπ = 0

So either i = 0 or π = 0, and both are incorrect. I'm assuming this is because I used the exponent rule from the title here (as the step before it is true). Why is this the case?

Thomas Andrews
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DaBest Gamer
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    There is no reason that exponentiation rules from the real numbers should apply to the complex numbers. – kipf Jul 05 '23 at 20:25
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    The complex exponential function is not injective, and this follows from the fact that $e^{2\pi i}=e^0$, in spite of the fact that $2\pi i\ne0$. This has nothing to do with the (false) rule $\left(a^b\right)^c=a^{bc}$. – José Carlos Santos Jul 05 '23 at 20:27
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    For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – Another User Jul 05 '23 at 20:28
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    $a^b=a^c$ means $a^{b-c}=1.$ The only real solution to $a^x=1$ is $x=0.$ However, there are many solutions to $a^x=1$ over the complex numbers, namely the entire unit circle. This means that we cannot conclude $x=0$ anymore. – Marius S.L. Jul 05 '23 at 20:29
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    What do you mean, the entire unit circle? The solutions $a^x=1$ does not contain the unit circle. @MariusS.L. – Thomas Andrews Jul 05 '23 at 20:33
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    @SassatelliGiulio You're right, that was sloppy. All $\dfrac{2\pi i }{\log a}\mathbb{Z}$ are the solutions. – Marius S.L. Jul 05 '23 at 20:37
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    The equality $a^b=a^c \Rightarrow b=c$ is valid for $a\in\Bbb R$ because $\ln (x)$ is a single-value function. The complex logarithm is multy-value, so if you apply yo the exponentiation, the equality became false, unless you restrain the domain, in polar form, to $\theta\in [-\pi,\pi]$; then is single-value. – Daniel Muñoz Jul 05 '23 at 20:38
  • I think in your example, you have $e^{i \pi}$ where it should be $e^{-i \pi}$. – bjcolby15 Jul 05 '23 at 20:50

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Although there is probably a less hand-wavy answer for this, the reason is because exponentiation and by extension, multiplication by complex numbers results in a different geometric effect than that of real numbers. Multiplication/exponentiation by real numbers results in a scaling of the original number, whereas the same operation only with a complex number can introduce both a ROTATION, and a scaling. What does it mean to “rotate” a number? Well this is clear if you think of numbers as now members of the two-dimensional complex plane rather than just the one-dimensional real number line. Understanding why this is requires knowing a bit of calculus and why Euler’s identity works, but a simple example here doesn’t require that:

Imagine the 2D cartesian plane, but now the vertical axis is the imaginary axis and the horizontal axis is the real axis. This makes it so all complex numbers are now points in this plane. 1 resides at (1,0), -1 resides at (-1,0), i resides at (0,1), and -i resides at (0,-1). One can see that the ordered pairs are just (real part, imaginary part). Now multiply 1 by i. Where are we? i. We have rotated 90 degrees. Now multiply i by i, and we get to -1, thus rotating another 90 degrees. Repeat this process and you will end up back at 1. So, multiplication by i results in a 90 degree rotation.

It turns out that -1 has this same property, only it is a 180 ($ \pi$ radians) degree rotation. If we take $-1=e^{i \pi}$, then it may be clear enough without a ton of calculus/Euler’s identity proof to argue that in general, any complex number of magnitude 1 (that is one that provides a rotation only and not a scaling) can be written as $e^{i \theta}$, where $\theta$ is the angle of rotation that multiplication by this number yields. This also explains why $i=e^{i \frac{\pi}{2}}$.

From this, it is clear where nonsense like $2 \pi i = 0$ can arise from since both of these will result in the same rotation. (One is 0 degrees, the other is 360 degrees). At least from this point of view, it shows why misuse of these exponent rules for complex numbers can in fact be classified as “misuse.”

STL
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