What is $\ln(-1)$? How is $\ln(x)$ defined on the negative complex numbers? At first, this question - when I asked it for my self - for me was straight forward:
$$e^{i\pi}=-1 \implies \ln(-1)=i\pi$$
Yet again, at some time I also discovered that:
$$e^{i\pi}=-1 \iff -e^{i\pi}=1 \implies \ln(-e^{i\pi})=\ln(1)=0$$
$$\implies \ln(-1)=-\ln(e^{i\pi})=-i\pi$$
From there I became confused. For a generalization one might see that
$$xe^{i\pi}=-x \implies \ln(-x)=\ln(x)+i\pi$$
But also, if we replace $x$ with $-x$ $($the identical matter with $\ln(-1)$ above$)$ we would get
$$\ln(-x)=\ln(x)-i\pi$$
For example:
$$e^{i\pi}=-1 \implies 8e^{i\pi}=-8 \implies \ln(-8)=\ln(8)+i\pi$$
However
$$e^{i\pi}=-1 \implies -8e^{i\pi}=8 \implies \ln(-8) + i\pi=\ln(8)$$
$$\iff \ln(-8)=\ln(8)-i\pi$$
So what is it $-i\pi$ or $i\pi$ or both or the whole argument is invalid because we can't use such property of the $\ln$ here:
$$e^{\ln(xy)}=e^{\ln(x)+\ln(y)}=xy$$
$$\implies \ln(xy)=\ln(x)+\ln(y) \implies \ln(x^z)=z\ln(x)$$
But does this property apply here? $($i.e. can I say $\ln(e^{i\pi})=i\pi$?$)$
Please extinguish the fire of my curiosity on this issue.
 A: The logarithm is no longer unique in the complex numbers. The reason is that in the complex numbers the exponential function is no longer one-to-one, since it is periodic with period $2\pi\mathrm i$. If we want to define the logarithm as the inverse of the exponential function, this is a problem, since inverse functions can only be defined uniquely for one-to-one functions.
So for any given $w\in\mathbb C$, there are now multiple possible values for $z$ such that $\exp z=w$. Which in turn means that there are multiple ways to define $\ln w=z$. This is usually done by restricting the domain of the exponential function so that it becomes one-to-one again. That is, we no longer define $\exp$ on all of $\mathbb C$. Instead we pretend that it's only defined on a horizontal strip $2\pi$ wide, so that if $z$ is in the strip, $z+2\pi\mathrm i$ is not in the strip. This way, its periodicity won't pose a problem. And then we define the logarithm as a function to that strip.
And we're free to choose the specific strip, resulting in different definitions of the logarithm, called branches of the logarithm. The strip $\{z\in\mathbb C~\vert~-\pi<\operatorname{Im}z<\pi\}$ results in the branch called main branch, which is often written as $\operatorname{Log}$ with a capital L to distinguish it from the others. Sadly, the usual rules for the logarithm no longer apply due to the necessity to choose a branch. For instance, $\log(xy)=\log(x)+\log(y)$ may no longer be true, but there may be different branches $\log_1$ and $\log_2$ such that $\log_1(xy)=\log_1(x)+\log_2(x)$. But those will also depend on $x$ and $y$.
A: Given a real number $x$, there is at most one number $y$ such that $x=e^y$. This number $y$, if it exists, we call $\ln x$.
The situation over the complex numbers is not so simple, because given a non-zero complex number $z$, there are an infinite number of complex numbers $y$ such that $z=e^y$. All these numbers differ by a multiple of $2\pi i$.
So there is no definition of $\ln$ over the non-zero complex numbers that satisfies the properties that we might expect, such as $\ln(zw)=\ln z+\ln w$.
We can rescue the situation somewhat by defining $\ln$ over a subset of the complex numbers, such as $\Bbb C-(-\infty,0]$. Still we don't have $\ln(zw)=\ln z+\ln w$ for all $w,z$ in this subset, but at least we get continuity and differentiability.
For more on this fascinating subject, look up Riemann surfaces.
A: Thing is: $a+ bi = re^{i\theta} = re^{i(\theta + 2k\pi)}$. So if $e^w = a+ bi$ then $e^{w + i2k\pi}$ is also $= a+bi$.
So $\log$ on complex numbers is not single value.  THere is no one $w = \ln (a+bi)$ because $e^w = a+bi$.  There is a whole set of $\ln (a+bi) = \{w + i2k\pi\}$.
We say log  is a "multivalue" function.  However there is always one primary value of $w$ $e^2= a+bi$ and  where $w = re^{i\theta}$  and where $0\le \theta < 2\pi$. And we do know that the set $\ln (a+bi)$ is the set of all complex numbers of the form $ e^{i(\theta + 2\pi k) }$.
So the answer to the question what is $\ln (a+bi)$ is.
If $a + bi = r e^{i\theta}$ then $\ln (a+bi) = \ln (r e^{i\theta}) = \ln r + \ln e^{i\theta} = K + \ln e^{i\theta} = $ the SET of all values $\{K + i(\theta + 2k\pi)| K =\ln r$ (a single realnumber as $r$ is a positive real number) $; k \in \mathbb Z\}$.
....
And $\ln (-1) = \{(2k+1)\pi i\}$.  So $\pi i$ is one of the natural logs of $-1$.  .... But $3\pi i$ is another one.
.....
Now sometime very soon, you will find some wise guy giving you a "proof" that $0 = 1$.
It goes like this:
$1 = 1$
$(-1)^2 = 1$
$(e^{\pi i})^2 = 1$
$e^{2\pi i} = 1$
$e^{2\pi i} = e^0$ so
$\ln e^{2\pi i} = \ln e^0$
$2\pi i = 0$
$\frac {2\pi i}{2\pi i} = \frac 0{2\pi i}$ so
$1 = 0$.
Dont fall for it!
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Oh, I guess I should explain why if $e^w = a+ bi$ then $e^{w+i2k\pi} = e^w\cdot e^{i2k\pi} = a+ bi$.  (Its because $e^{i2k\pi} = 1$). And I should probably explain why all $a+bi \ne 0$ can be written as $a+bi = re^{i\theta}$ where $r$ is a positive real number and $\theta$ is an angle.
First of all if $a+bi \ne 0$ then $a + bi = \sqrt{a^2 + b^2}\left(\frac a{\sqrt{a^2 + b^2}} + i\frac b{\sqrt{a^2 + b^2}}\right)= r(x + iy)$ where $r = \sqrt{a^2 + b^2} > 0$.  And $x =\frac a{\sqrt{a^2 + b^2}};y = \frac b{\sqrt{a^2 + b^2}} $.  And notice that $x^2 + y^2 = 1$.
Because $x^2 + y^2 =1$ there is some angle $\theta$ where $x =\cos \theta$ and $y = \sin \theta$.  (In fact that angle is $\arctan yx =\arctan ba$).  So $a+bi = r (\cos x + i\sin x)$.
We already have a definition for what $e^x; x\in \mathbb R$ means but what could $e^z; z\in \mathbb C$. mean?
Well, if we want $e^{w+z} = e^we^z$ then $e^{a+bi} = e^a*e^{bi}$ so we have to define what $e^{ib}$ can mean for a purely imaginary number $b i$.
And we define it as $e^{ib} = cos b + i\sin b$.  Now why do we do a ridiculous thing like that?  Well to start off we need $e^{0i} = e^0 = 1$. And we need $|e^{bi}| = 1$.  But mainly if we want $e^{bi}e^{ci} = e^{i(b+c)}$ we have $(\cos b + i\sin b)(\cos c + i\sin c)= (\cos b\cos c - \sin b \sin c)+i(\cos b\sin c + \cos c\sin b)= (\cos(b+c) + i(\sin(b+c)$ and that is ... wll, it is heaven sent is what it is!
But it does lead ththe somewhat unintuitive and strange idea that  $x^z = e^w$ does not mean $z = w$.  And as $\sin $and $\cos$ ar periodic with period $2\pi$ then $e^{i\theta}$ will be periodic with period $2\pi i$.
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Oh, you asked about if $\ln (wz ) = \ln w + \ln z$.
Yes... If we consider .... define set addition as If $\alpha = \{bunch\ of\ things\}$ and $\beta = \{other\ bunch \ of\ things\}$ then $\alpha + \beta =\{a+b|a\in \alpha; b\in \beta\}$.
Then $\ln w = \alpha = \{a + i2\pi k| e^a = w\}$ and $\ln z = \beta = \{b + i2k\pi| e^b = z\}$ and then $\ln (wz) = \alpha + \beta = \{a+b|a\in \alpha; b\in \beta\}$
This is why we can "loop around" to get the weird idea $2\pi i = 0$.  Obviously $2\pi i \ne 0$.   But $2\pi i = 0 + 2k\pi i$ for some integer $k$ (namely $k =1$).  [Another way of saying that is $2\pi i \equiv 0 \pmod{2\pi i}$].
So the fake proof works in that:
$(-1)^2 = 1$
$(e^{\pi i})^2 = e^0$
$\ln e^{\pi i})^2 = \ln e^0$
$2\pi i \equiv 0$ which is to say
$\{2\pi i + 2k\pi i\} = \{0 + 2k\pi i\}$.
Which is true.  No paradox!
