# What is the right treatment for $0^i$?

I need to calculate a limit of a complex expression (had it in a physics research) that contains a term $(r-b)^p$ for $r\rightarrow b+$ where $r,b$ are reals, and $p$ is complex, let's suppose for simplicity $p=i$.

At the beginning, I put it equals to $0$ as an obvious thing, but strangely enough, "Mathematica" gives me something strange: $e^{2i\operatorname{Interval}[0,\pi]}$, then I thought maybe I was wrong with "obviousity" of the result, because we can actually put (formally) $0=e^{-\infty+i\phi}$, then if we use $a^b=e^{b\log a}$ we have $0^i=e^{-i\infty-\phi}$ (where all $2\pi k$ included in the infinity or $\phi$).

This result seems to be very strange for me (it will introduce a new parameter $\phi$ into my theory), and I found no example for this, especially that I know one needs to be careful with branches when making such tricks, but I have no clue if that is correct or wrong, any help will be appreciated.

The problem is that $x^i$ is only defined as $e^{i\log x}$.

When $x$ gets close to zero, $\log x$ is of the form $a+bi$ where $a$ is a very large negative number and $b$ can be restricted to $(-\pi,\pi]$. Then $e^{i\log x}$ is $e^{-b} e^{ia}$.Note then that if we let $x$ approach zero along a line from one side at angle $\theta$, that means the modulous of $e^{i\log x}$ will be $e^{-\theta}$, which is constant, so it does not approach zero (and it does not even converge - $x^i$ basically whizzes around a circle when $x\to 0$ along a straight line.)

Exponentiation $x^u$ near $x=0$ just is not well-behaved.

And this doesn't even take into account that $\log x$ is more naturally a multi-valued function.

Even restricting $x$ to $\mathbb R^+$ with $\log x$ the normal real natural logarithm, $$x^i=e^{i\log x}$$

So $|x^i|=1$, and as $x\to 0+$, $x^i$ spins around the unit circle clockwise and certainly does not have a limit.

• Maybe it was my mistake, I didn't mentioned that $r\rightarrow b+$ and $r,b$ are reals, then your $\theta=0$, then dose my result is right? – TMS Feb 20 '13 at 17:45
• sorry for the delay, please note that in my question, I mentioned that I included multi-value of $\log$ in the angels for purpose (for that I didn't put the restriction $(-\pi,\pi]$ as you did, also this means that $\theta!=0$ even for real $x$ ), moreover, if you calculate something like $e^{i(\log x +i2\pi)}$ for $x=0.0001$ as example, you will find that for this brunch $x^i\rightarrow 0$ really! also we get undefined limit for something like $\sin \infty$ which has no meaning, but here we get a parametrized limit, and I'm note sure that we can consider it as undefined! – TMS Feb 21 '13 at 17:52
• No, it doesn't. Basically, if you choose a continuous value of $\log x$ for $x$ real - that is $\log x = a + 2\pi n i$ for some fixed $n$ - then $e^{i\log x} = e^{-2\pi n} e^{ia}$. Now, $e^{-2\pi n}$ is constant, and $|e^{ia}|=1$. So you again are circling around a circle, just a circle with a different radius, as $x\to 0$. Now, if you let $n$ vary for each $x$, making $\log x$ discontinous, you could get $x^i\to 0$, but you could also get $x^i\to\infty$. @TMS – Thomas Andrews Feb 21 '13 at 18:17
• I found my problem, it was Mathematica bug that mislead me, you have been right! thank you for your help. – TMS Feb 21 '13 at 18:32

The complex exponential $(x+iy)^p$ is not defined uniquely. It is $(x+iy)^p = \exp(p\log(x+iy))$ where the logarithm is defined up to a multiple of $2\pi i$. Hence your formula is not uniquely determined and you cannot say that it tends to zero...

• interesting point ! Since log(x) has many sheets, you can't walk around the origin without jumping from one sheet to another. – user28921 Feb 21 '13 at 18:55
• @user28921: $x$ does not walk around the origin. It approaches $0$ from the positive side. – robjohn Feb 21 '13 at 21:36

To make $\ln z$ unique, one needs to slit the complex plane and it is common to use only the main branch on $\mathbb C\setminus\mathbb R_{\le0}$. As in your specific problem $r,b$ are always reals and $r>b$, we can write $r-b=e^t$ with $t\in \mathbb R$ and $t\to-\infty$. Then $(r-b)^p=e^{pt}$. If $\Re p=0$ (but $\Im p\ne 0$) this indeed keeps rotating around on the unit circle so that $\lim_{r\to b^+}(r-b)^p$ does not exist. If $\Re p>0$, however, then $|(r-b)^p|=|e^{pt}|=e^{t\Re p}\to 0$. And if $\Re p<0$, then $|(r-b)^p|\to\infty$. In both these latter cases we still have that rotation phenomenon if $\Im p\ne 0$.