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Today, I was playing with numbers and I found this: $$\begin{align} \ln(1) = 0 \Rightarrow \ln((-1)^2) = 0 \Rightarrow 2\ln(-1) = 0 \Rightarrow 2\pi i = 0 \Rightarrow \boxed{i = 0} \end{align}$$

Apart from this I found another proof: $$\begin{align} \ln(1) = 0 \Rightarrow \ln(e^{2\pi i}) = 0 \Rightarrow 2\pi i = 0 \Rightarrow \boxed{i = 0} \end{align}$$

This obviously is not true and I myself don't know what's the flaw in this proof. Any help/hint is appreciated

Edit : My guess is that I am treating $i$ as a variable rather $i = \sqrt{-1}$. Is this logic correct?

Thanks

Blue
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p_square
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3 Answers3

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The rule $\log(a^x)=x\log a$ only applies when $a>0$, so even if we stay in the context of the real numbers, the first argument is flawed.

We can extend the definition of exponentiation to the complex numbers by defining $e^z$ as its Taylor series. By Euler's formula, $e^{z+2k\pi i}=e^z$ for all $k\in\Bbb{Z}$ . Hence, the function $e^z$ is periodic and so is not invertible. This makes defining $\log$ tricky. There are number of different approaches; for example, we could "restrict the domain of $e^z$", i.e. define $\log$ as the inverse of the function $f:\{z:-\pi<\Im(z)\le\pi\}\mapsto\Bbb{C}$ given by $f(z)=e^z$. If $\log$ is defined this way, then $\log(e^z)=z$ is only true when $-\pi<\Im(z)\le\pi$. This is analogous to how, in the real numbers, $\arcsin(\sin (x))=x$ is only true when $-\pi/2\le x\le\pi/2$.

Joe
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  • Thanks for your answer! Makes it clear now – p_square Jul 19 '21 at 17:25
  • @Algebrology: $\Im(z)$ denotes the imaginary part of $z$. It is defined by $\Im(x+iy)=y$. – Joe Jul 19 '21 at 17:25
  • yeah got it, I got confused by I written that way but when I saw the TeX command, I got to know what it was – p_square Jul 19 '21 at 17:26
  • @Algebrology: Also, by the time you reach the faulty conclusion that $2 \pi i=0$, this is already a false statement. Are you aware of how, in mathematics, a false statement implies anything? – Joe Jul 19 '21 at 17:28
  • Yes I am aware of that! – p_square Jul 19 '21 at 17:30
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    @Algebrology: Then consider how the statement $2\pi i = 0 \implies i=0$ is true, but $2\pi i = 0 \implies i=5$ is also true. So by this point, things have already broken down. So whether or not $i$ is a variable is essentially a moot point. – Joe Jul 19 '21 at 17:38
  • Note that in particular, while we definitely don't have $0=2\pi i$, we do have they are equivalent when plugged into polar form for complex numbers, which is basically what you observed. So if we want to allow arguments outside of $[0,2 \pi )$ we need to mod out by the equivalence relationship @Algebrology – Alan Jul 19 '21 at 17:41
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The exponential function $z\mapsto e^z$ is well-defined on the whole complex plane. But it is not 1-1, because for any $z$, the values $e^z,e^{z+2\pi i},e^{z+4\pi i}$ etc are all equal.

So it can't have a well-defined inverse on its range (which in this case is the non-zero complex numbers). But by specifying which of the multiple candidate values to choose, we can define a partial inverse (and call it $\ln$), but it will necessarily have discontinuities; and it will not satisfy $\ln ab = \ln a + \ln b$ for all non-zero complex numbers $a,b$. All you can say for sure is that $\ln ab = \ln a + \ln b +{}$some integral multipe of $2\pi i$.

So your reasoning breaks down where you claim that $\ln((-1)^2)=2\ln(-1)$.

TonyK
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The problem is that as much as you cannot deal with $\sqrt{-1}$ the same way you do with a real number you cannot understand $\ln(x)$ and $\ln(z)$, $x$ real, $z$ complex to be the same function. When you want to do such thing you always have to extend the function to a higher domain and then look how it behaves.

So for $x$ and $y$ real and positive:

$$\ln(e^{x+iy})=x + i(2\pi \left \lfloor \frac1{2}-\frac{y}{2\pi} \right \rfloor+y)$$

Then for $x=0$ $y=2\pi$ you have

$$\ln(e^{2\pi iy})= i(2\pi \left \lfloor \frac1{2}-\frac{2\pi}{2\pi} \right \rfloor+2\pi)= i(-2\pi +2\pi) = i \cdot 0 = 0$$

Therefore, it is not $i=0$, but $i \cdot 0 = 0$ which is totally ok.