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I've been learning Complex Analysis from George Cain's website: https://people.math.gatech.edu/~cain/winter99/complex.html

In chapter 3, Elementary Functions, it claims that the complex logarithm function is not single-valued, which I can understand.

However, I can't seem to understand how the complex exponential function exp(z) and complex power function $z^c$ are also multi-valued...

For example, at the end of the chapter, the reader is meant to feel unsatisfied when confronted with an expression like $e^{1/2}$. Its meant to be confusing since it may be suggesting two different numbers $\sqrt{e}$ and $-\sqrt{e}$.

I can't see how such exponentials and powers can be multi-valued.

SpicyDonkey
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3 Answers3

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The exponential function is not multivalued.

BUT:

We have two standard definitions:

$$\exp(z)=\sum_{n=0}^\infty\frac1{n!}z^n,$$

$$z^w=\exp(w\log(z)).$$

So $\exp(z)$ is definitely single-valued, while $z^w$ is multivalued.

The question is not whether the exponential function is multivalued; it's not, and the author did not say it was. The question is whether $e^z$ is multivalued!

At this point people object vehemently that $e^z=\exp(z)$, so no it's not multivalued. This raises the question "Is it true that $e^z=\exp(z))$?". And the answer to that question is "yes or no, depending".

In fact any time people see the notation $e^z$ they interpret that notation as meaning $\exp(z)$; this is of course a good thing because that's what the writer meant by the notation. But that's not consistent with the definition of $z^w$. The definition of $z^w$ says that $$e^z=\exp(z\log(e)),$$and that is multivalued!

If we want to say that $e^z=\exp(z)$ officially, by definition, then we need to modify the definition of $z^w$ to read $$z^w=\begin{cases} \exp(w\log(z)),&(z\ne e), \\\exp(w),&(z=e). \end{cases}$$

And that would be a really bad definition. For example, we couldn't say $z^w$ was multivalued any more, we'd have to say "$z^w$ is multivalued unless $z=e$". Similarly for anything else interesting we might say about $z^w$.

In most people's minds it's actually true that $z^w$ is multivalued except when $z=e$. But this is not for any mathematical reason; $\log(e)$ is multivalued just like any other logarithm. The reason people think $z^w$ is multivalued except for $z=e$ is just an artifact of the bad standard notation.

Summary

  1. No, the exponential function is not multivalued, and nobody said it was; the quote from the author in question does not say $\exp$ is multivalued.

  2. What people "always" mean by the notation $e^z$ is single-valued.

  3. According to the actual definitions the notation $e^z$ does denote something multivalued.

Bad notation. Very bad. Too late to change it.

Not to snipe at the other users, but perhaps to clarify - I'm saying they're wrong, but we have no disagreement about the math, just about notation:

The other two answers I see argue that the exponential is single-valued, which is a straw man; the author didn't say it was multi-valued. They're missing the point, which you can see from locutions like "the exponential function $e^z$". They seem to be saying that the author is saying the exponential is multivalued, which again is missing the point, he never said that. The author's point is not mathematical, it's a valid point about bad notation; the definition of $z^w$ says that $e^z$ is not the exponential function, although standard convention says it is.

David C. Ullrich
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  • Thank you very much for your clear explanation of my confusion and also correctly pointing out my error in stating the original problem. I have now come to realise that $e^z$ is not an equivalent expression to $exp(z)$. I am very happy to be learning complex analysis and being corrected of my misunderstanding! – SpicyDonkey Aug 11 '15 at 23:48
  • @SpicyDonkey Now that you realize that $e^z$ is not equivalent to $\exp(z)$ don't forget that $e^z$ is $\exp(z)$! Heh... – David C. Ullrich Aug 12 '15 at 00:06
  • Late to the party, but a propos of ambiguities in definitions, nagging question: since the definition $f(w)= z^w=e^{wlogz}$ is only valid in some open $V$ which does not contain $0$, some accommodation must be made to account for the fact that $f(0)=0.$ For example, take $f(z)=z^2.$ The $f$ is holomorphic in $\mathbb C$ but according to the definition, it is not even defined at $z=0.$ So, one would like to say that $f$ is bounded in some nbhd of $0$, so it has a removable singularity there. But, taking the principal branch of log, $f$ is not even defined on all of $\textit {any}$ nbhd of $0$. – Matematleta Jan 27 '19 at 17:44
  • @Matematleta Of course the usual definition of $z^2$ is just $z^2=zz$. But integers $w$ are very special in the general definition of $z^w$ above; in fact if $z\ne0$ and $\log(z)$ is any branch of $\log(z)$ then $e^{2\log(z)}=e^{\log(z)+\log(z)}=e^{\log(z)}e^{\log(z)}=zz$. You're simply wrong about $e^{2\log(z)}$ not being defined in any (deleted) nbd of $0$; in fact it's single-valued in $\Bbb C\setminus{0}$. – David C. Ullrich Jan 27 '19 at 17:56
  • @DavidC.Ullrich Your comment is all jumbled. I can barely read it. From what I gather, yes I can do the calculation and the "badness" cancels, but my question is that in the very definition, you must choose a branch of log, in which case, there is an automatic restriction of the domain (a cut) and so you have to overcome this problem, e.g. by extending in some way. – Matematleta Jan 27 '19 at 18:02
  • @Matematleta It took me a while to find the TeX error - maybe you can read it now? The principal-value logarithm is defined in all of $\Bbb C\setminus{0}$. – David C. Ullrich Jan 27 '19 at 18:05
  • @DavidC.Ullrich Yes, thanks. I meant, log is not continuous at the points of its branch cut. Ugh. OK, just to see if I get it now: when you define the power function as above, your calculation shows it is holomorphic on the punctured plane and bounded in a disk about the origin, so in fact has a removable singularity at $0.$ Is this right? – Matematleta Jan 27 '19 at 18:19
  • @Matematleta Yes. Talking about the principal-value logarithm to be sppecific: No, $\log$ is not continuous in the entire punctured plane, but $e^{2\log(z)}$ nonetheless is continuous in the punctured plane. – David C. Ullrich Jan 27 '19 at 18:23
  • @DavidC.Ullrich got it! btw: your book Complex Made Simple is wonderful. The lead-up to the Riemann Mapping Theorem (quasi-metrics, etc) is fantastic. Have you written any other books? – Matematleta Jan 27 '19 at 18:25
  • @Matematleta Glad you liked CMS. Especially glad you like the part you mention: One of the "referees" complained about my including that stuff, since there are quicker ways to Montel. Was going to be a reals book for a while, never quite happened. Stay tuned - the nonexistent reals book may become a blog when I retire in a year... – David C. Ullrich Jan 27 '19 at 18:35
  • @DavidC.Ullrich The detours taken in the book are one of the things I like most about it, because they dovetail nicely with other topics. I will be on the lookout for your next book. – Matematleta Jan 28 '19 at 02:33
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The exponential is not multi-valued. However, the power function is indeed multi-valued. The reason is that we may write

$$z^c = e^{c \log{z}} $$

so that, if $\log{z}$ is know only within a term of $i 2 \pi k$, $z^c$ is known only to within a factor of $e^{i 2 \pi c}$.

EDIT

The author of the treatise in questions states the following:

"Far more serious is the fact that we are faced with conflicting definitions of $z^c$ in case $z=e$. In the above discussion, we have assumed that $e^z$ stands for $\exp{z}$. Now we have a definition for $e^z$ that implies that $e^z$ can have many values."

Horsesh-t. The author is introducing a level of confusion over the definition of a variable. For example, $z^c$ is not multi-valued in $c$, so saying that any definition of the exponential is multi-valued is disingenuous at best. Please, forget this last paragraph - you will sleep better and be all the wiser.

Ron Gordon
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  • Could you address the OP's point that $\left(-\sqrt{\mathrm{e}}\right)^2 = \mathrm{e}^1 = \left(\sqrt{\mathrm{e}}\right)^2$. – Fly by Night Aug 10 '15 at 00:34
  • @FlybyNight: you mean that the sqrt function is ambiguous because there are two branches. That is the multi-valuedness of the sqrt, not exp. – Ron Gordon Aug 10 '15 at 00:49
  • @RonGordon The square-root function is prefectly well-defined on the positive real numbers. By definition $y = \sqrt{x}$ is the unique, positive real number for which $y^2 = x$. My comment was a suggestion for you to improve your answer by addressing the OP's point that $(-\sqrt{\mathrm{e}})^2 = \mathrm{e}^1 = (\sqrt{\mathrm{e}})^2$. – Fly by Night Aug 10 '15 at 00:58
  • @Dr.MV: not really with you. $z^c$ is indeed multi-valued. The way I view the world, I ask myself if the function returns the same value under the transformation $z \to z e^{i 2 \pi}$. So I don;t see how $e^{c z}$ is multi-valued, even the way you write it. – Ron Gordon Aug 10 '15 at 00:59
  • @RonGordon Yes, but the confusion might arise because of the incorrect way forward that a student might proceed. Yes, of course $(e^z)^c=e^{cz}$ and for the exact reason you mentioned - transform $z\to ze^{i2\pi}$ and we have equality. But the naïve might instead view $e^z$ as the complex number, say $w$. Then, we have $w^c$, which is indeed a multi-valued function of $w$. Does that clarify a bit of what confusion might ensue and that I so poorly communicated previously? – Mark Viola Aug 10 '15 at 01:12
  • @Dr.MV: and in making that leap, you introduced a new logarithm. Confused students do not change reality. – Ron Gordon Aug 10 '15 at 01:13
  • @RonGordon I agree. That is the point that I am making. But this is a way that can help the novice move forward. That is the goal of some - to help others that are at an obstacle that more seasoned folks have long ago move past (or never found a challenge). Make sense? – Mark Viola Aug 10 '15 at 01:16
  • @Dr.MV: I know where you are doing from. This is a pedagogical exercise. But as you can see, the OP has been exposed to someone who has no idea what he is talking about. Perhaps I too am a lousy teacher, but really, and you said this yourself, $z \to z e^{i 2 \pi}$ is a nice simple way to determine who is multi-valued and who isn't. I can't imagine this is over the OP's head. – Ron Gordon Aug 10 '15 at 01:20
  • @RonGordon You're absolutely right. – Mark Viola Aug 10 '15 at 01:30
  • @RonGordon Thank you very much for the answer Ron. I should have been able to figure the power function is multi-valued if I actually thought about it! I got stuck initially with the exponential function business and could not see how it could be multi-valued, so didn't continue on with the power function. I agree with you that it's really strange to argue $e^{1/2}$ as being multi-valued with his logic of applying the argument for power function. – SpicyDonkey Aug 10 '15 at 06:25
  • @RonGordon The author is not introducing confusion, he's pointing out that the notation is ambiguous. The standard notation is ambiguous. People don't realize it's ambiguous because they always read $e^z$ as synonymous with $\exp(z)$. But the definition of $z^w$ is $\exp(w\log(z))$, and that definition implies that $e^z=\exp(z\log(e))$, which is multivalued. – David C. Ullrich Aug 10 '15 at 14:41
  • @DavidC.Ullrich: my issue was...multi-valued over what? I get it, but it's like moving the goalposts. You're essentially saying that $e^z$ is multivalued in $e$. This is the result of trying to be too clever. The truth is, the exponential function $\exp{z}$ is not multi-valued in its argument. Trying to alter this for pedagogy's sake is, as you can see, a disaster. – Ron Gordon Aug 10 '15 at 18:41
  • @RonGordon I don't know what "multivalued in $e$" means. I never said that $\exp$ was multivalued! It's not multivalued. That quote from the author does not say that the exponential function is multivalued either! The issue is... I'm repeating myself. Please read the answer I posted. – David C. Ullrich Aug 10 '15 at 18:46
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    @RonGordon Regarding "trying to alter this for pedagogy's sake": The author points out, correctly, that we do at this point have two mutually inconsistent definitions of $e^z$. Pointing this fact out explicitly seems to me to be a good thing - some student is going to notice the inconsistency and then conclude he must be missing something and wonder what... – David C. Ullrich Aug 10 '15 at 18:57
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The complex exponential function $e^z$ is not "multi-valued." The power function $z^c$ is multi-valued if $c \notin \mathbb{Z}$, because of how such a function is defined, $z^c = e^{log\ z^c} = e^{c\ log\ z}$ where the indeterminacy of the complex logarithm carries over to the power functions.

I think by convention $e^{\frac{1}{2}}$ should always be interpeted as $exp{(\frac{1}{2})}$, i.e. evaluate the exponential function before the fractional exponent. Then the fact that $(e^{\frac{1}{2}})^2 = (-e^{\frac{1}{2}})^2$ doesn't contradict $e^z$ being well-defined. The examples of $e^{\frac{1}{2}} = \sqrt{e}$ or $e^{\frac{1}{2}} = -\sqrt{e}$ arising from evaluating the fractional exponent first as if we were evaluating $z^{\frac{1}{2}}$ at $e$ would make a mess of the complex exponential.

sharding4
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