Derive branch cuts for $\log(\sqrt{1-z^2} + iz)$ as $(-\infty,-1)$ and $(1,\infty)$?

Question

Attempt:

First, we examine $\sqrt{1-z^2}$. Note that it can be written $\sqrt{1-z}\sqrt{1+z}$, so the appropriate branch cuts are $(-\infty,-1)$ and $(1,\infty)$ for the inner square root term.

Next, we look at $\log(w)$ and note that we can define the cut for $\log(w)$ as $(-\infty,0)$. But now what? I tried setting $w= \sqrt{1-z^2} + iz$, solving for the branch point where $w=0$, but this results in $1=-z^2+z^2=0$, so I think this is the wrong approach.

What is the correct way to understand this?

Answer 1

It may be useful to consider $w=\sqrt{1-z^2}+iz$. Isolating the square root and squaring, you may note that the $z^2$ terms cancel, and you end up with $$z=\frac1{2i}\Bigl(w-\frac1w\Bigr),$$ and then subsituting this into the equation $\sqrt{1-z^2}=w-iz$ you also find $$\sqrt{1-z^2}=\frac12\Bigl(w+\frac1w\Bigr).$$ Thus both $z$ and $\sqrt{1-z^2}$ are single-valued functions of $w$, so it should be easier to analyze the given function as $\log w$.

There are two values of $w$ for each value of $z$: Replacing $w$ by $-1/w$ leaves $z$ unchanged, and flips the sign of $\sqrt{1-z^2}$.

Note that imaginary $w$ gives real $z$: Your proposed branch cuts in the $z$ plane go along the imaginary axis in the $w$ plane, from $\pm i$ to infinity in opposite directions, but also (if you pick the other branch of the square root) from $\pm i$ to $0$. Thus your branch cuts divides the $w$ plane in two halves along the imaginary axis, and you end up with not having to pick further branch cuts for the logarithm.

(In my first edition of this answer I got a little confused because I was thinking of getting the full Riemann surface for the given function. I hope I managed to fix this before confusing anybody else too much. The analysis I give here is perhaps better for understanding the Riemann surface; it could well be overkill for the branch cut question. Oh well …)

Answer 2

We don't need a branch-cut for the logarithm here.

Generally, if $U$ is a simply connected domain and $f\colon U\to \mathbb{C}$ holomorphic without zeros, then $f$ has a holomorphic logarithm on $U$, that is, there exists a holomorphic $g\colon U\to \mathbb{C}$ with $e^{g(z)} = f(z)$ for all $z\in U$ (of course, $f$ has infinitely many logarithms on $U$, any two differing by an integer multiple of $2\pi i$).

Such a $g$ is conventionally denoted by $g = \log f$, without (necessarily) meaning that $g$ is globally the composition of a branch of the logarithm with $f$.

Obviously, $g$ is locally, in a (small enough) neighbourhood $V$ of each $z\in U$, the composition of a branch of the logarithm on $f(V)$ with $f$, but, if $f$ is not injective, $f(z_1)=f(z_2)$ for some $z_1\neq z_2$, then different branches of the logarithm can be used on $f(V_1)$ and $f(V_2)$, where $V_1$ is a small neighbourhood of $z_1$ and $V_2$ one of $z_2$.

Here, however, we can write $\log (\sqrt{1-z^2} + iz)$ as the composition $\log \circ f$ of a branch of the logarithm with $f(z) = \sqrt{1-z^2}+iz$, since $f$ maps $U := \mathbb{C}\setminus \{t\in\mathbb{R}:\lvert t\rvert\geqslant 1\}$ to a domain where a branch of the logarithm exists.

The - in my opinion - easiest way to see that is to follow the mapping of $\sin$. Starting from the strip $S = \left\{z\in\mathbb{C} : \lvert \operatorname{Re} z\rvert < \frac{\pi}{2}\right\}$, from the familiar behaviour of the exponential function, we see that $z\mapsto e^{iz}$ maps the strip biholomorphically to the right half-plane. Now the map $h\colon w \mapsto \frac{1}{2i}\left(w-\frac{1}{w}\right)$ is a rational function of order $2$, hence attains each value in the sphere $\widehat{\mathbb{C}}$ exactly twice (counting multiplicity) in $\widehat{\mathbb{C}}$. It is easily seen that $h\left(-\frac{1}{w}\right) = h(w)$, so it follows that $h$ is injective on the right half-plane, and

$$h\left(\{z : \operatorname{Re} z > 0\}\right) = \widehat{\mathbb{C}} \setminus h\left(i\mathbb{R}\cup \{\infty\}\right) = \mathbb{C}\setminus \{t\in\mathbb{R} : \lvert t\rvert \geqslant 1\}.$$

So altogether, $\sin$ maps $S$ biholomorphically to $U$. Then you just need to check that $h$ and $f(z) = \sqrt{1-z^2} +iz$ are inverses of each other to see that $f$ maps $U$ to the right half-plane.