# $n^{th}$ roots and $e$.

In Churchill's book on complex variables, the $n^{th}$ root of $e$ is defined to be $e^{1/n}$. A comment is made that in this respect $e$ is treated differently than the $n^{th}$ roots of other complex numbers (in the sense that there are typically n roots of the nth root of a number in complex analysis rather than just one as in the case of $e$).

I am curious why $e$ is treated so differently. Is there an obvious reason/motivation why?

Edit: The section from Churchill is,

As anticipated earlier, we define here the exponential function $e^z$ by writing $$e^z = e^xe^{iy}\ \ \ \ \ \ (z = x + iy)\ \ \ \ \ \ \ \ \ (1)$$ where Euler's formula $$e^{iy} = \cos y + i\sin y$$ is used and $y$ is to be taken in radians. We see from this definition that $e^z$ reduces to the usual exponential function in calculus when $y=0$; and, following the convention used in calculus, we often write $\exp z$ for $e^z$.

Note that since the positive $n$th root $\sqrt[n]{e}$ of $e$ is assigned to $e^x$ when $x = 1/n$ ($n = 2,3,\ldots$), expression (1) tells us that the complex exponential function $e^z$ is also $\sqrt[n]{e}$ when $z = 1/n$ ($n = 2,3,\ldots$). This is an exception to the convention that would ordinarily require us to interpret $e^{1/n}$ as the set of $n$th roots of $e$.

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Hmmm, that doesn't sound right. What page of Churchill's book are you looking at? – Álvaro Lozano-Robledo Feb 21 '12 at 2:32
Apart from the Routine $\TeX$ edit, I have made vars "variables". Please do not make this place into a SMS chat. This is a dedicated forum for Math Q&A. Further, any signature or tagline for the posts is advised against, by the faq. So, I have removed it. – user21436 Feb 21 '12 at 2:33
The constant $e$ actually does have $n$ distinct $n$-th roots, for all $n=1,2,3,\cdots$. The only number in all of $\mathbb{C}$ without such a property is $0$. I'm not familiar with Churchill - is your parenthetical your own understanding or something specifically stated by the author? Generally in complex analysis the function $z\to z^{1/n}$ is chosen with a particular branch in mind rather than assumed multi-valued. // Could you transcribe the comment in full, if possible? – anon Feb 21 '12 at 2:34
The only explanation I can think of is that $f(z)=z^{1/n}$ is taken multi-valued while $e^w$ is defined to be the exponential, given by the usual power series, and hence takes on only one value. – anon Feb 21 '12 at 2:37
@MattBrenneman and everyone else! I've added the passage in the book. Hopefully it's everything you mentioned, and if not please feel free to add (or emphasize) any other parts! – Alex Feb 21 '12 at 23:49

The natural exponential function is defined by $$\exp(z) = \sum_{n=1}^\infty {z^n\over n!}.$$ This is an entire function. It is not hard to show it has all of the expected properties.

To define $z^w$ you must define something like $$z^w = \exp(z\log(w)).$$

Unfortunately, the exponential function is $2\pi i$-periodic. Therefore it is not 1-1, so the business of defining a logarithm function becomes tricky. You must choose a domain to restrict the exponential function to so it is 1-1. And there the trouble begins. But where the trouble begins, complex analysis begins in all of its beauty and elegance.

I quote one of my grad school professors, Sidney Graham, who said, "There are those who say that the study of complex variables is the study of the logarithm function."

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The point is that we want "the" exponential function to be single-valued. If you want to write $\exp(1/n)$ as ${\rm e}^{1/n}$, that singles out one "$n$'th root of e". There are still $n$ $n$'th roots of e, it's just that only one of them is written as ${\rm e}^{1/n}$.

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