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The following identity holds for all $a$ and $x$ using the principal branch: $$ \log x^a = a\log x + 2\pi i \left\lfloor \pi-\Im (a\log x) \over 2\pi \right\rfloor $$ e.g. for $a=2$, $x=-1$:

LHS = $\log (-1)^2 = \log 1 = 0$

RHS = $2\log(-1)+ 2\pi i \left\lfloor \pi-\Im (2\log (-1)) \over 2\pi \right\rfloor$ =$2\pi i+ 2\pi i \left\lfloor \pi-\Im (2\pi i) \over 2\pi \right\rfloor = 2\pi i - 2\pi i = 0$

Everything is single valued and there is no problem.

How can I perform the same calculation using multivalued logarithms? In other words, I want to use the (multivalued) identity: $$\log x^a = a\log x$$ for $a=2$, $x=-1$. I pick the same branch for both LHS and RHS, let's pick the principal branch (so that we can reuse the values calculated above), and then I add the $2\pi i n$ term for each logarithm and get:

LHS = $\log (-1)^2 + 2\pi i n = 2\pi i n$

RHS = $2\log(-1) + 2\pi i \left\lfloor \pi-\Im (2\log (-1)) \over 2\pi \right\rfloor + 2\cdot 2\pi i m = 4\pi i m$

And we can see that LHS is not equal to RHS, otherwise $m$ would have to be half-integer.

Where did I make the mistake? How can I make LHS equal to RHS using the multivalued approach?

Ondřej Čertík
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  • In the LHS you are constructing a set like this, LHS$={\ 2(\pi i) + 2\pi in,\ n\in\mathbb{Z}},\ $ and RHS=$2{\ \pi i + 2\pi im,\ m\in\mathbb{Z}}.\ $ The fact that RHS has half the members of LHS makes as much sense as $2\mathbb{Z}$ having half of $\mathbb{Z}.$ I think you have an irrelevant mistake in RHS. – will Sep 30 '15 at 00:40
  • The problem is that there are are members of LHS (e.g. $n = 1$) that are not members of RHS (no matter what you choose for $m$). If that's the case, how can LHS be equal to RHS? Or to be specific, in what exact sense is it equal? – Ondřej Čertík Sep 30 '15 at 19:04
  • Just a thought, the principle branch of the log applies to $x\not\in\Re(-\infty,0],\ $ which excludes your test of $x=-1.\ $ Are you interested in the natural log itself as a multi-value almost analytic function, or are you using it as tool to study something else? – will Oct 01 '15 at 04:15
  • I am interested in how to prove the relation $\log x^a = a \log x$ where $\log z$ is a multi-value analytic function. If you don't like $x=-1$, then use $x=1$. The result doesn't change, there are some values on LHS that are not included in the RHS. – Ondřej Čertík Oct 01 '15 at 04:52

3 Answers3

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$\DeclareMathOperator{\Log}{Log}$

Here is what Churchill, Brown, Verhey say about this particular case (Complex Variables and Applications, p. 66, Third Edition) and I quote:

"The statement $\log(z^n)=n\log(z)$ may or may not be true for specific values of $z$ and $n$ when the multiple-valued complex logarithmic function is replaced by a single branch of it. If we use the notation $\Log$ for the principal branch of the complex logarithmic function, note for example, that $\Log[(1+i)^2]=2\Log(1+i)$ while $\Log[(-1+i)^2]\neq 2\Log(-1+i)$.",

while you write (in the comments): "I am interested in how to prove the relation $\log(x^a)=a\log(x)$ where $\log(z)$ is a multi-value analytic function."

With your case, $\Log(x^a)=a\Log(x)$ holds for the specific values $x=-1$ and $a=2$.

Consequently the values of $x$ and $a$ do matter, so you are trying to prove something that may not be true in general.

Addendum #1 (after your comment):

There is such a thing, but you have to be really careful to make a clear distinction between the two symbols: $\log(z)$ and $\Log(z)$. Because $\log$ is multivalued and indexed by $k\in\mathbb{Z}$, it is more convenient to use the following notation for this map:

$$\log(k,z)=\Log(z)+2k\pi i,\,\,k\in\mathbb{Z}\Rightarrow$$

Using the fact that:

$$\Log(e^z) = z + 2\pi i \left\lfloor \pi-\Im z \over 2\pi \right\rfloor$$

we get:

$$\log(k,e^z)=z+2\pi i\left(k + \left\lfloor \pi-\Im z \over 2\pi \right\rfloor \right),\,\,k\in\mathbb{Z}$$

If you now set $z=x^a$ to the the above, using the principal branch of the logarithm to define it as: $x^a=e^{\Log(x^a)}$, you get the (multi-valued) set constructor you are looking for:

$$\log(k,x^a)=\log(k,e^{\Log(x^a)})=\Log(x^a)+2\pi i \left(k + \left\lfloor \pi-\Im \Log(x^a) \over 2\pi \right\rfloor\right) ,k\in\mathbb{Z}$$

You cannot reduce the $\Log(x^a)$ above to $a\Log(x)$ (no matter what the rest of the expression is), because that would be using the identity $\Log(x^a)=a\Log(x)$, which as Churchill shows, might not hold for the principal branch and the particular values of $x$ and $a$ you are considering.

However, using the definition of a power $x^a = e^{a\Log x}$, we obtain: $$\log(k,x^a)=\log(k,e^{a\Log(x)})=a\Log(x)+2\pi i \left(k + \left\lfloor \pi-\Im (a\Log(x)) \over 2\pi \right\rfloor\right) ,k\in\mathbb{Z}$$ and using $a\Log x = a\left(\log(l, x) - 2\pi l i\right)$ we get: $$\log(k,x^a)=a\log(l, x)+2\pi i \left(k - la + \left\lfloor \pi-\Im (a\Log(x)) \over 2\pi \right\rfloor\right) ,k,l\in\mathbb{Z}$$

Ondřej Čertík
  • 407
  • Thanks for the answer. Yes, if you have to choose a specific branch, e.g. the principal branch with $\Log(z)$ as in your quote from the book, then the only way to have a formula valid for all $x$ and $a$ is to use the very first equation in my question: $\Log x^a = a\Log x + 2\pi i \left\lfloor \pi-\Im (a\Log x) \over 2\pi \right\rfloor$, which always holds. However, I thought that somehow you can make the "more natural" formula $\log x^a = a\log x$ work with multivalued $\log(z)$, perhaps by some convention how to evaluate it. Your answer seems to suggest that there is no such thing. – Ondřej Čertík Oct 02 '15 at 00:07
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    @OndřejČertík You can go a bit further if $a$ is a real number. In order to get the $a$ out of the $\Log$ in the last identity by Yannis, you need to keep track of "on which domain" $x^a$ ends up being, that is, how many times it crosses the negative real axis. For each of those crossings you will have to add an aditional $2\pi i$ to the RHS. If you know $x$'s initial argument $\alpha \in (-\pi,\pi)$, then you will have to add an additional $2m\pi i$ if $a\cdot\alpha \in (-\pi + 2m\pi, \pi+2m\pi).$ – dafinguzman Oct 03 '15 at 14:17
  • Yes, indeedy. Matching thus the winding number of the LHS with that of the RHS. Thanks for adding this note @dafinguzman. –  Oct 03 '15 at 15:36
  • @YiannisGalidakis I could add it to the answer if you like – dafinguzman Oct 03 '15 at 15:49
  • Please go ahead. It'd be great if you could extract some sort of general rule for specific $a$ and $x$, @dafimguzman. –  Oct 03 '15 at 15:58
  • You need to be careful, it is not true for all $z$ that $\Log(e^z) = z$, but rather it is true for all $z$ that $\Log(e^z) = z + 2\pi i \left\lfloor \pi-\Im z \over 2\pi \right\rfloor$. I have proposed edits to your answer that fixes this problem (I don't know if you can see them or not). Either way, I gave you the bounty for your work, but it seems that the answer is that it is only possible if you use the floor function. @dafinguzman, yes, your winding number is precisely the floor function in all my equations. – Ondřej Čertík Oct 03 '15 at 18:35
  • Just in case you can't see my edit, the final equation that you get, using your notation, is: $\log(k,x^a)=a\log(l, x)+2\pi i \left(k - la + \left\lfloor \pi-\Im (a\Log(x)) \over 2\pi \right\rfloor\right) ,k,l\in\mathbb{Z}$ – Ondřej Čertík Oct 03 '15 at 18:38
  • Looks good to me, but I will look at it in detail in a while. I don't have enough rep to approve but it's already approved by others, so no problem @OndrejCertik. –  Oct 03 '15 at 18:59
  • It would also be useful to add a calculational example, @OndrejCertik or at least verify that the expression works for specific $a$, $x$ and $k,l$. –  Oct 03 '15 at 19:19
  • I have a Python code that tests those identities here: http://www.theoretical-physics.net/dev/math/complex.html#testing-identities-using-computer-code – Ondřej Čertík Oct 05 '15 at 16:26
1

$log(x) = Log(x) + i2\pi n = Log(|x|) + iArg(x) + i2\pi n$ where $Log$ and $Arg$ are principal values.

$log(x) + log(x) = 2Log(|x|) + i2Arg(x) + i2\pi n + i2\pi m$

$n$ and $m$ are independent and can be merged into one variable $k$.

$alog(x) = aLog(|x|) + iaArg(x) + i2\pi k$

$$alog(x) = aLog(x) + i2\pi k\ \ \ (1)$$

$e^{alog(x)} = e^{aLog(x) + i2\pi k} = x^ae^{i2\pi k} = x^a$ (sanity check)

$$log(x^a) = Log(x^a) + i2\pi k = aLog(x) + i2\pi k\ \ \ (2)$$

Equations (1) and (2) are equal. $$log(x^a) = alog(x) = aLog(x) + i2\pi k\ \ \ (3)$$

Also introduce z for clarity.

$alog(z) = aLog(x) + i2\pi k$

$log(z) = Log(x) + i\frac{2\pi k}{a}$

Take the exponential.

$z = xe^{i\frac{2\pi k}{a}}$

$e^{i\frac{2\pi k}{a}}$ are roots of unity.

How to do the calculation:

$log((-1)^2) = Log(1) + i2\pi k = i2\pi k$

$2log(-1) = 2Log(-1) + i2\pi n = 2(i\pi) + i2\pi n = i2\pi (n+1)$

$k = n+1$

Are multi-valued functions a rigorous concept or simply a conversational shorthand?

Community
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  • I think you made a mistake in equation (1), it should rather be $a\log(x) = a(\Log(x) + 2\pi i n) = a\Log(x) + 2 a\pi i n$, so then the corrected (1) and your (2) are not equal. – Ondřej Čertík Oct 03 '15 at 17:58
  • @Ondřej Čertík - No, that's the point. Adding $log(x) + log(x) = 2Log(x) + i2\pi n + i2\pi m$ the n and m are independent.$alog(x) = aLog(x) + i2\pi (n_1+n_2+\dots+n_a)$. All $n_k$ are independent so they can be replaced with k s.t. $alog(x) = aLog(x) + i2\pi k$ –  Oct 03 '15 at 19:36
  • @Ondřej Čertík - Let $f(x)$ be a multi valued function where $f(1) =$ {$0,1$}. Then $f(1) + f(1) =$ {$0+0,0+1,1+0,1+1$} $=$ {$0,1,2$}. Not {$0,2$}. –  Oct 03 '15 at 19:52
  • I think you are right! Do you have some references for how to do arithmetic with multivalued functions? E.g. what if "a" is not an integer (I know that's not the particular example I posted)? – Ondřej Čertík Oct 03 '15 at 23:51
  • For example, what if $a={1\over2}$? What are all the possible values for ${1\over2}\log z$? – Ondřej Čertík Oct 04 '15 at 00:01
  • @Ondřej Čertík - My reference is the roots of unity and phasors in electronics. $log(z^\frac{p}{q}) = \frac{p}{q}log(z)$ and ${\frac{p}{q}}log(z) = \frac{p}{q}(Log(z) + i2\pi k) + i2\pi n$. The justification I recall is that the exponential of the LHS equals that of the RHS. –  Oct 04 '15 at 06:05
  • @Ondřej Čertík - Operations on multi value functions. Let $x^p = 1$ and $y^q = 1$ how do you evaluate $f(x,y) = x + y$? $x = e^\frac{i2\pi n}{p}$ for $n = 0\dots(p-1)$. Similarly $y = e^\frac{i2\pi m}{q}$ for $m = 0\dots(q-1)$. $f(x,y) = e^\frac{i2\pi n}{p} + e^\frac{i2\pi m}{q}$ for the same range of $n$ and $m$. –  Oct 04 '15 at 06:20
  • So if $f(1) = {0, 1}$, we established that $2f(1) = {0, 1, 2}$. What is $\frac{1}{2} f(1)$ equal to? – Ondřej Čertík Oct 28 '15 at 20:48
0

$$log(z^\frac{p}{q}) = \frac{p}{q}log(z) = \frac{p}{q}(Log(z) + i2\pi k) + i2\pi n$$ The justification I recall is that the exponential of the LHS equals that of the RHS. The derivation went along the lines of:

Replace $z$ with $ze^{i2\pi k}$. Let $y=(ze^{i2\pi k})^\frac{p}{q}$. $$log(y) = Log(y) + i2\pi n = Log((ze^{i2\pi k})^\frac{p}{q}) + i2\pi n = \frac{p}{q}Log(ze^{i2\pi k}) + i2\pi n$$ $$log(y) = \frac{p}{q}(Log(z) + i2\pi k) + i2\pi n$$