# 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.

• You have stumbled upon the interesting fact that in the complex plane, natural logarithm is a "multi-valued function" (a self-contradictory term that is widely used). The "layers" of the natural logarithm correspond to the periods of the exponential function. – hardmath Aug 14 at 18:09
• logarithm of complex numbers is multivalued because $e^{2n\pi i}=1$ for all integers $n$ – J. W. Tanner Aug 14 at 18:09
• @hardmath Whoa, and does a regular high school student supposed to know that or am I just someone who uncovered an interesting existing fact? – Anas A. Ibrahim Aug 14 at 18:11
• Although Euler's identity $e^{i\theta} = \cos \theta + i \sin \theta$ is often covered in high school mathematics, dealing with the corresponding fact about complex logarithms is more or less deferred to college courses (and above). – hardmath Aug 14 at 18:16
• "does a regular high school student supposed to know that" A regular high school student barely knows that $2^3 = 8$ and thinks the American revolution was fought against Korea. ... But a bright high school student is supposed to know what you do (and will most likely be very confused). A bright high school student taking an advanced class focusing on complex analysis should learn this. But I think must advance classes show you $e^{\pi i}=-1$ and $e^{i\theta}=\cos \theta + i\sin \theta$ but its not explained why and the consequence are eluded. I think most would find it confusing. – fleablood Aug 14 at 18:32

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.

• Amazing! I hope that somehow the definition $\ln(zw)=\ln(z)+\ln(w)$ is somehow extended over the whole set of complex numbers. – Anas A. Ibrahim Aug 14 at 18:19
• $\ln (-1)=\{(2k+1)\pi i\}$ is an infinite set of values. And yes if $\ln (z)=\{\alpha+2k\pi i\}$ and $\ln w = \{\beta+2k\pi i\}$ then $\ln (zw)$ does equal $\{(\alpha + \beta) + 2k\pi i\}$. If you keep in mind that $e^{a+bi} = e^a e^{bi}$ then $e^{Real(z)=a}$ will be a positive real number that tells you the "size" of the complex number, and $e^{iIm(z)=bi}$ will be a complex number on a unit circle that tells you the "angle" of the complex numbers. Circles "loop around" then $e^{bi}$ loops around. So there is primary value and many consequential values. Otherwise, it all works fine. – fleablood Aug 14 at 19:14

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 $$y$$.

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

$$\{2\pi i + 2k\pi i\} = \{0 + 2k\pi i\}$$.