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If we define

$$e^z=1+z+\frac{z^2}{2!}+\cdots$$

then it is single-valued.

However, if we write

$$e^z=e^{z\ln e}$$

then it is multi-valued.

Besides, $a^z$ is multi-valued in general. It is kind of strange if only when the base is $e$ that it is single-valued.

My thought: Is it true that there are two exponential functions, let's call them $\exp(z)$ and $e^z$?

Where $\exp(z)$ is defined by

$$\exp(z)=1+z+\frac{z^2}{2!}+\cdots$$

and is single-valued, while $e^z$ is defined by

$$e^z = \text{exp}(z\ln e)$$

and is multi-valued?

Here $\ln z$ is defined by $\exp(\ln z)=z$ and is multi-valued.

velut luna
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5 Answers5

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The notation $e^z$ is just shorthand for $\exp(z)$. Everybody agrees that when $e^z$ is evaluated it means $$ e^z=\exp(z)=\sum_{n\ge0}\frac{z^n}{n!} $$

Notation is not always consistent, unfortunately.

Indeed, if we would interpret $e^z$ as $\exp(z\log e)$, where $\log e$ is any determination of the logarithm, we'd get infinitely many values, unless $z$ is an integer. This is because $\log e$ can be any complex number $w$ such that $\exp(w)=e$ and it's easy to see that $w=1+2ki\pi$, for $k\in\mathbb{Z}$. Thus $\exp(zw)=\exp(z(1+2ki\pi))$. For instance, if $z=1/2$, we would have to assign $e^{1/2}$ both the values $\sqrt{e}$ and $-\sqrt{e}$.

However, as I said at the beginning, notation is a bit sloppy in this respect. Writing $e^z$ instead of $\exp(z)$ is deemed more practical and so $e^z$ used with the convention that it means the same as $\exp(z)$, mainly because single valued functions can be better manipulated algebraically.

As you note, people generally avoids using $a^z$ for $a\ne e$, when $z$ possibly varies in the complex numbers. In the case of positive real $a$, however, since $\log a$ has a well defined unique real value, there's no difficulty in defining and using $a^z=\exp(z\log a)$, where $\log a$ means that unique real value. The algebraic property $a^{z_1+z_2}=a^{z_1}a^{z_2}$ holds without restriction, with this convention (but only for positive real $a$).

Ryder Rude
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egreg
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    So correct me if I am wrong. $a^z$ is defined by $e^{z\ln a}$ for all complex number $a$, except, when $a=e$, for which $e^z$ is defined by the series. So $a^z$ is multi-valued, except when $a=e$, for which $e^z$ is single-valued. – velut luna Jun 11 '15 at 13:42
  • @Kyson No, I didn't really say that. Usually people avoid writing $a^z$ (except for $a=e$), but for positive real $a$ there is a sensible definition of $a^z$ as a single valued function. – egreg Jun 11 '15 at 14:39
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    hi egreg, so what do people write for $a^z$ when $a$ is in general complex? – velut luna Jun 11 '15 at 23:10
  • @Kyson That's not a function, so… – egreg Jun 11 '15 at 23:11
  • hi egreg, you mean there is no such function $a^z$ when $a \ne e$? – velut luna Jun 11 '15 at 23:50
  • @Kyson I can only repeat myself: if $a$ is real and positive, there is no problem in defining $a^z=\exp(z\log a)$, where $\log a$ is the standard real logarithm. Otherwise it's not possible to define a single valued function $a^z$ with the usual algebraic properties valid for $\exp$. – egreg Jun 12 '15 at 08:55
  • For $exp(w)=e$, $w=1+2ki\pi$. Your third paragraph. – Ryder Rude Oct 21 '18 at 03:59
  • @RyderRude “If we interpret $e^z$ as $\exp(z\log e)$ we get infinitely many values”. But the following text makes it clear that this is not done, but rather the “principal branch” value of $\log a$ is taken when the base $a$ is a positive real. – egreg Oct 21 '18 at 10:30
  • @egreg I'm saying that, in the third paragraph of your answer, you've written $e^w=e$ is satisfied by $w=2ki\pi$ for integer $k$. This is wrong. – Ryder Rude Oct 21 '18 at 14:10
  • @RyderRude Oh! Thanks! – egreg Oct 21 '18 at 14:12
  • @egreg Also, in your last reply, you said the principal branch value is taken when the base is positive and real. Why 'positive and real'? Can't $(-1)^z$ be simplified as $e^{z\ln{-1}}$=$e^{iz\pi}$, using the principal value of $\ln{-1}$? I think exponents of all bases can be made unique if we only use the principal value of log. – Ryder Rude Oct 21 '18 at 14:21
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    @RyderRude It depends on what “principal branch” one takes. The choice is somewhat arbitrary, except for the positive $x$-semiaxis where $0$ for the argument is at least the most natural one. Why should $\pi$ be the argument of $-1$? Why not $-\pi$? – egreg Oct 21 '18 at 14:30
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  • Indeed, unfortunately, the standard definitions of $e^z\;(z\in\mathbb C)$ as single-valued—which we generally want—are consistent, when using the logarithmic $(\log)$ definition of $e^z$, only with using the principal branch $\mathrm{Log}.$

    I.e., usually, $$e^z := 1+z+\frac{z^2}{2!}+\cdots \\= e^{z\,\mathrm{Log}(e)} \\≠ e^{z\log(e)};$$ however when wanting the function to be multi-valued (e.g. when determining $n$th roots), $$e^z := e^{z\log(e)}.$$

  • The multi-valued definition of $e^z$ (i.e., using $\log$ instead $\mathrm{Log}$ to define $e^z$) results in $$e^{x+iy} = e^{x+iy}e^{2kπ(ix-y)}$$ being multivalued—which is generally not desired. For example, this apparently contradictory result ensues: $$e^{i\pi}=-e^{2n\pi^2}\\\neq-1.$$

  • Until reading this post, I had thought that $\exp()$ and $e^{()}$ are synonymous; it turns out that the convention is that the former denotes the single-valued variant of the latter. It's indeed a good disambiguation.

ryang
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You run into problems with the second definition of the exponential, as the function $e^z$ on the complex plane is not 1-1. In $\mathbb{R}$, it is true that $e^x=e^y$ implies $x=y$, but for example, $e^0=1=e^{2\pi i}$ in $\mathbb{C}$ and $0 \neq 2\pi i$. So $e^z$ does not have an inverse--hence your second equation being multi-valued.

sharris
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Define \begin{eqnarray} \ln z &=& \text{Ln}|z| + i \arg z\\ &=& \text{Ln}|z| + i(\arg z + 2 \pi n) \end{eqnarray} Where $\text{Ln}$ is the ordinary real logarithm, clearly the complex logarithm is multi-valued as it depends on $n$.

Autolatry
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It's true that the complex map $\ln$ is multi-valued, but in the case of your second definition it doesn't matter, because $\exp$'s periodicity kills the extra arguments of $\ln$, so the definition becomes single-valued and perfectly ok.

For example, $\arg(e)=\Theta=0$ and $\ln|e|=1$, so your definition becomes:

$$e^{z\cdot \ln(e)}=e^{z\cdot (\ln|e|+(\Theta+2k\pi))i}\text{ },k\in\mathbb{Z}\Rightarrow$$ $$e^{z\cdot \ln(e)}=e^{z(1+2k\pi i)}\Rightarrow$$ $$e^{z\cdot \ln(e)}=e^z\cdot e^{z2k\pi i},k\in\mathbb{Z}\text{ (1)}$$

Correction after egreg's comment:

The following two steps are in fact wrong.

$$e^{z\cdot \ln(e)}=e^z\cdot (e^{2k\pi i})^z\Rightarrow$$ $$e^{z\cdot \ln(e)}=e^z\cdot 1^z=\exp(z)$$

They would be right, had the OP specified only the principal branch of $\ln$. Since there's no such concensus, I am correcting my answer, based on egreg's comment.

I agree then that the definition is multi-valued, and it seems the actual values are given by the last line (1).