Is it possible to make a proper limit of $\ln x$ in $\mathbb{C}$

Question

So it's known that $\displaystyle \lim_{x \to 0^+} \ln x=-\infty$

This one can't even be defined properly as a limit because in $\mathbb{R}$ it's not defined on the other side.

But here's the thought. What if you extended the view to complex numbers. Like if you had a circle centred at the origin with radius $r$ then every complex number on that circle would have the form $re^{i\theta}, \theta=(0,2\pi)$.

What $\ln x$ would do to this circle would be to project each point onto the vertical line at $\ln r$ where every point ends up on the imaginary unit $i\theta$ on this line. And excluding periodicity you would have all the outputs between $\ln r$ and $\ln r +2i\pi$

Then you reduce $r$ to start approaching 0. That vertical line representing all the outputs would slide to the left without bound. Then maybe you could finally say the limit of the complex function is $-\infty$

Is this a logical deduction or am I missing something?

Answer 1

It is impossible to define the $\ln$ function on the whole of the complex plane so that it is analytic.

The way that this is handled is to cut the plane along a line from $0$ to (usually, in this case) $-\infty$. Then you define $\ln$ on $(0,\infty)$ in the usual way, and extend it to the cut plane by analytic continuation $-$ there is a unique complex-valued function from the cut plane that is analytic on the cut plane and agrees with the function $\ln:(0,\infty)\to\Bbb R$ on the positive real line.

In fact we can define this function explicitly as $\ln(re^{i\theta})=\ln r+i\theta$, where we choose $\theta\in(-\pi,\pi)$ (the end-points of this interval, $\pm\pi$, represent points on the cut itself, so our function is undefined there).

You can cut the plane along any half-infinite line starting at $0$, and define a function $\ln$ on the resulting cut plane that is analytic, and satisfies $\exp(\ln z)=z$, for all $z$ in the cut plane. The choice of cut determines the range of $\theta$.

But such a function doesn't satisfy $\lim_{z\to 0}\ln z=-\infty$, because of that $i\theta$ term.

Notes:

• The cut doesn't have to be a straight line $-$ essentially, it just has to prevent the possibility of drawing a circle centred on $0$.
• Another way to resolve the problem is to define a function $\ln$ on a Riemann surface, which in the case of the $\ln$ function looks like a bit like an infinite corkscrew:

(Image credit: Gifts and Merchandise UK)

• Caveat: To be honest, I'm not even sure what it means to say that $\lim_{z\to 0}f(z)=-\infty$ for a complex-valued function $f$. Does it mean that as $z\to 0$, $f$ gets arbitrarily close to the real line, and the real part of $f$ tends to $-\infty$? That is what I assume in my final paragraph. But perhaps it has no generally accepted meaning.