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In the real numbers the expression $x^x$ converges to 1 coming from "any" direction (which here means from the positive side or from the negative side). Now I feel to remember that this does not hold in the complex plane, because there the expression $z^z$ should be able to take any value (when $|z|$ goes to 0) depending on the direction/angle you're coming from.

Now I tried to make sense of that, but couldn't complete the picture. I will demonstrate how far I got. Let $z = r \cdot e^{i \phi}$ then

$z^z = (r \cdot e^{i \phi})^{r \cdot e^{i \phi}} = (r)^ {r \cdot e^{i \phi}} \cdot (e^{i \phi})^{r \cdot e^{i \phi}} = (r)^ {r \cdot e^{i \phi}} \cdot e^{i \phi \cdot (r \cdot e^{i \phi})}$.

With this parametrization, when we look at $|z| \rightarrow 0$, this just means $r \rightarrow 0$. My intuition (and Wolframalpha) would now say that the term with $r$ as the base would approach 1 as $r$ approaches 0. But now Wolframalpha also says that the second term with $e$ as the base also approaches 1 as $r$ approaches 0, which would mean the combined limit is 1, not depending on $\phi$. This would then mean that $z^z$ does indeed also approach 1 in the complex plane from whereever we are coming (as the limit is independent of $\phi$).

Now, where is the flaw?

  1. Does $z^z \rightarrow 1$ as $|z| \rightarrow 0$ for all $z \in \mathbb{C}$?
  2. Is there a flaw in my calculations? (either in an intermediate step or the reasoning at the end)
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    $\begingroup$ Are you certain that $(r \cdot e^{i \phi})^{r \cdot e^{i \phi}} = (r)^ {r \cdot e^{i \phi}} \cdot (e^{i \phi})^{r \cdot e^{i \phi}}$? This seems like a dangerous step to me. $\endgroup$
    – Arthur
    Aug 17, 2017 at 13:22
  • $\begingroup$ $(a \cdot b)^c = a^c \cdot b^c$ does hold in the complex plane as well, so I'd say yes. $\endgroup$
    – tehfurbolg
    Aug 17, 2017 at 13:23
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    $\begingroup$ Even with both exponents and bases being complex? $\endgroup$
    – Arthur
    Aug 17, 2017 at 13:25
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    $\begingroup$ @tehfurbolg You may find scipp.ucsc.edu/~haber/ph116A/clog_11.pdf useful to consult, and avoid false beliefs about the complex plane :) $\endgroup$ Aug 17, 2017 at 13:28
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    $\begingroup$ When we say "$0^0$ is an indeterminate form" we do not mean that $z^z$ may converge to different values as $z \to 0$. Instead we mean $f(z)^{g(z)}$may converge to different values for functions $f,g$ with $f(z) \to 0, g(z) \to 0$ as $z\to 0$. $\endgroup$
    – GEdgar
    Aug 17, 2017 at 14:31

2 Answers 2

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First of all, $z^z$ is not analytic at $z=0$. It is defined as $z^z=e^{z\log z}$, where $\log$ is an analytic branch of the logarithm. Take the usual branch, defined on $\Bbb C\setminus(-\infty,0]$ with $\log1=0$. If $z=r\,e^{i\theta}$, $-\pi<\theta<\pi$, then $$ |z\log z|=r\,|\log r+i\,\theta|\le r(|\log r|+\pi), $$ which goes to $0$ as $r\to0$. It follows that $z^z\to1$ as $z\to0$.

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The problem with the function $z\mapsto z^z=e^{z\ln z}$ is that it is not uniquely defined because the complex logarithm $\ln z$ isn't. However, if you fix a path in the complex plane, there is indeed a way to define $z^z$ uniquely on this path. Now if you fix path going to zero, you can ask if the limit of $z^z$ is indeed 1 along this path.

As @Julian Aguirre shows, if this path is in the subspace $\mathbb{C}\setminus (-\infty;0]$, then $z^z$ has a limit and it is $1$. However, there are paths where $z^z$ has no limits.


Let $z=re^{i\theta}$ so that $\ln z=\ln r + i\theta + 2ik\pi$. Then $$ z^z=e^{z\ln z}=e^{z(\ln r+\theta+2ik\pi)}$$ Now let's have a look at the exponent.

When $|z|\rightarrow 0$, $z\ln r\rightarrow 0$ and $zi\theta\rightarrow 0$. However $z\times 2ik\pi$ might not have a limit because you don't know how $k$ behave.


Concretely, we will fix a path in $\mathbb{C}$ going to $0$ and compute $z^z$ along this path. In fact, instead of $z^z$, we will look at $z\ln z$ which is simpler. We will show that there is some path where there is no limits.

The path is the following : $z=t\times e^{i/t}$. (Draw a picture of that path, this is spiral).

Then $\ln z=\ln t+ \frac{i}{t}$ and $z\ln z=z\ln t+ z\times \frac{i}{t}=z\ln t+i e^{i/t}$.

But then, you see that $z\ln z$ has no limit along this path. And neither has $z^z$.

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