3
$\begingroup$

Consider the function $$f(z)=\log[(z^2+1)^{1/2}],\quad z>0$$

where the branch is chosen so that $(z^2+1)^{1/2}>0$ for $z>0$ and the log denotes the principal branch. Let $R$ be the union of the negative real axis and the vertical interval $\{i\gamma:-1\le \gamma \le 1\}$.

Show that $f(z)$ has an analytic continuation to the complement of R and compute

$$\lim f(\epsilon), \lim f(\epsilon e^{(i3\pi)/4}), \lim f(\epsilon e^{(-i3\pi)/4})$$

as $\epsilon\to 0$.

Any help would be greatly appreciated. This problem is a little beyond me, so I don't even know how to get started, other than writing down some obvious things from the problem such as:

$$(z^2+1)^{1/2}=(z+i)^{1/2} (z-i)^{1/2},$$ where $z^2$ is defined as $e^{2\log(z)}$

and that the principal branch of $\log(z)$ is analytic away from the negative real axis, when we restrict the argument of $\log(z)$ to ($-\pi \le \theta \ <\pi$).

Edit: A more specific question that I have is:

Why is an analytic extension even necessary for $f(z)$? Isn't the function already analytic away from the negative real axis, and thus should be analytic to the complement of R, which is just the whole plane minus the negative real axis and the interval of the imaginary axis from $-i$ to $i$.

Is it because, in addition to the principal branch of $\log(z)$, which is already chosen for us, we need to also choose two more branches of $\log(z)$ - one branch for $(z+i)^{1/2}$, and one branch for $(z+i)^{1/2}$?

$\endgroup$
1
$\begingroup$

I think the term "analytic continuation" here refers to the possibility of single-valued analytic continuation of the function in the given domain, starting from, say, a neighborhood of $z=1$.

The function $\log((z^2+1)^{1/2})$ can be analytically continued along any path $\gamma$ in $\mathbb C\setminus \{i,-i\}$ (for definiteness, let's begin every path at $1$). However, the result of continuation depends on the path taken, not just on the endpoint: if $\gamma$ winds around $i$ or $-i$, this changes the value of the function. But in a simply-connected domain this issue would not arise; cutting the plane along $[-i,i]$ and then along $(-\infty,0]$ is one way to obtain a simply-connected domain not contaning $\pm i$. Let $\Omega$ be this domain.

Another way to see what is going on is to study the map $w=(z^2+1)^{1/2}$. It maps the upper half of $\Omega$ conformally onto the upper halfplane; by reflection across $(1,\infty)$, it also maps the lower half of $\Omega$ onto the lower halfplane. Hence, the image of $\Omega$ under this map is $\mathbb C\setminus (-\infty,1]$. Then you take the principal branch of logarithm there.

Concerning the three limits that were mentioned: the images in the $w$-plane can be described as $w=1+\epsilon$, $w=-1+i\epsilon$, and $w=-1-i\epsilon$. By this I don't mean the correspondence of points ($\epsilon$ need not be the same), but rather the correspondence of directions of approach to boundary points. The technical term is prime end. These three directions of approach yield three different values of logarithm.

$\endgroup$
1
  • $\begingroup$ Very interesting; I hadn't considered single-valued analytic continuation - thanks so much for that suggestion, @Behaviour. I studied your solution for awhile but am still a little shaky on the material. I think I'll move on to seeing other problems and hopefully do a few more in conformal mappings / analytic continuation to get a better sense of what's going on. Have a great night :) $\endgroup$ – User001 Dec 16 '14 at 22:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.