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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.

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    $\begingroup$ 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. $\endgroup$ – hardmath Aug 14 at 18:09
  • $\begingroup$ logarithm of complex numbers is multivalued because $e^{2n\pi i}=1$ for all integers $n$ $\endgroup$ – J. W. Tanner Aug 14 at 18:09
  • $\begingroup$ @hardmath Whoa, and does a regular high school student supposed to know that or am I just someone who uncovered an interesting existing fact? $\endgroup$ – Anas A. Ibrahim Aug 14 at 18:11
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    $\begingroup$ 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). $\endgroup$ – hardmath Aug 14 at 18:16
  • $\begingroup$ "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. $\endgroup$ – fleablood Aug 14 at 18:32
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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.

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  • $\begingroup$ Amazing! I hope that somehow the definition $\ln(zw)=\ln(z)+\ln(w)$ is somehow extended over the whole set of complex numbers. $\endgroup$ – Anas A. Ibrahim Aug 14 at 18:19
  • $\begingroup$ $\ln (-1)=\{(2k+1)\pi i\}$ 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. $\endgroup$ – fleablood Aug 14 at 19:14
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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 $y$.

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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!

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