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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?

mathjacks
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  • I did this to show that the branch cuts for the square root term are $(-\infty,-1)$ and $(1,\infty)$. – mathjacks Oct 08 '14 at 18:26
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    You don't need a branch-cut for the logarithm. Only the one for the square root. $\mathbb{C}\setminus {t\in\mathbb{R} : \lvert t\rvert \geqslant 1}$ is simply connected, and $\sqrt{1-z^2}+iz$ is never $0$ there. – Daniel Fischer Oct 08 '14 at 18:26
  • Thanks, Daniel for your help with this question and my previous one. What is the reasoning for not needing a branch cut for the logarithm? (also: can you recommend a book that covers these concepts in more detail at an introductory level?) – mathjacks Oct 08 '14 at 18:28

2 Answers2

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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 …)

Harald Hanche-Olsen
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  • Just curious … How would one evaluate the integral $\oint_C \log(iz+\sqrt{1-z^2}),dz$ where $C$ is any rectifiable curve that encloses (one time) the branch points at $z=−1$ and $z=−1$? – Mark Viola Apr 23 '20 at 16:58
  • @MarkViola You can't: Well, you could put the branch cut for $\sqrt{1-z^2}$ along the real axis from $-1$ to $1$. But when $z$ passes around $C$, $w$ (either branch) makes a whole turn around the origin, increasing its logarithm by $2\pi i$. So the integrand has no continuous branch. – Harald Hanche-Olsen Apr 23 '20 at 22:09
  • Thank you for the quick reply. If we denote $f(z)=\log\left(z+\sqrt{z^2-1}\right)-\log(z)$, then would $\int_C f(z),dz$ still pose a challenge? It seems now that we have a new branch point at $0$. – Mark Viola Apr 23 '20 at 22:40
  • @MarkViola That would take care of it: In fact now $f(z)=\log(1+\sqrt{1-z^{-2}})$. Assuming for simplicity $|z|>1$ then $1-z^{-2}$ remains in the right half plane, so picking the “obvious” branch of the square root, so does $1+\sqrt{1-z^{-2}}$, and now you can take the log without branch cuts. – Harald Hanche-Olsen Apr 24 '20 at 08:00
  • This is a very curious result. The function $\log\left(z+\sqrt{z^2-1}\right)$ has no branch point of the logarithm since $z+\sqrt{z^2-1}$ can never be $0$. The logarithm does have a branch point at $0$. So, one would conclude that $f(z)=\log\left(z+\sqrt{z^2-1}\right)-\log(z)=\log\left(1+\sqrt{1-z^{-2}}\right)$ would also have a branch point at $z=0$. But as you mentioned, $\text{Re}\left(1+\sqrt{1-z^{-2}}\right)\ge 0$. So, how does one resolve this apparent paradox? And what type singularity is $z=0$? – Mark Viola Apr 24 '20 at 15:26
  • @MarkViola There is no paradox here. The two branch points of $\sqrt{z^2-1}$ will essentially add up to a single zero. To make it clearer, write it as $\sqrt{z^2-a^2}$, with $a=1$. Note that it has branch points at $\pm a$. Allow $a$ to shrink from $1$ down to $0$. When it reaches $0$, the two branch points have joined at the origin, and the square root is reduced to $\sqrt{z^2}=z$ (or $-z$, but let's assume the plus sign). As you can see, that is just a plain zero. (This conversation is getting a bit long for this place, though. Possibly, the site software will want to move it to a chat.) – Harald Hanche-Olsen Apr 24 '20 at 18:25
  • Well, we haven't been directed yet to "chat." So, continuing … Going back, you stated in your first comment of this thread "when $z$ passes around $C$, $w$ (either branch) makes a whole turn around the origin, increasing its logarithm by $2\pi i$" Yet, inasmuch as $z+\sqrt{z^2-1}$ can never equal $0$, the function $\log\left(. z+\sqrt{z^2-1}\right)$ does not require a branch cut in the $z$-plane. So, how does encircling the branch points of $\sqrt{z^2-1}$ result in "increasing its logarithm by $2\pi i$?" – Mark Viola Apr 24 '20 at 21:25
  • Aside, if $\log\left(z+\sqrt{z^2-1}\right)$ does not have a branch point, while surely $\log(z)$ has a branch point at $0$, then why is it wrong to suspect that the difference of $\log\left(z+\sqrt{z^2-1}\right)$ and $\log(z)$ (i.e., $f(z)=\log\left(z+\sqrt{z^2-1}\right)-\log(z)$) does indeed have a branch point at $z=0$? – Mark Viola Apr 24 '20 at 21:25
  • @MarkViola I think you keep forgetting that $\sqrt{z^2-1}$ does require a branch cut. In this case, I made it a slit along the real line between $-1$ and $+1$. As far as the logarithm goes, that slit plays the same role as the branch point at $0$ of $\log z$. So don't worry so much about branch points: Any hole in the domain can have the same effect as a branch point. – Harald Hanche-Olsen Apr 25 '20 at 08:57
  • Of course I did not forget the branch cuts for $\sqrt{z^2-1}$. The "slit" $[-1,1]$ is actually a combination of, for example, the branch cuts from $-\infty$ to $1$ and from $-\infty$ to $1$ or the branch cuts from $1$ to $\infty$ and from $-1$ to $\infty$. Where the cuts overlap, $\sqrt{z^2-1}$ is continuous and hence only the "slit" from $-1$ to $1$ remains. Now, when we add $\log(z)$, we need to the plane from $0$ to the point at $\infty$. So, it appears to me that we have both the slit $[-1,1]$ AND the branch cut from $0$ to $\infty$. – Mark Viola Apr 25 '20 at 15:38
  • @MarkViola I think it's easier to work with $w$. We lost an $i$ somehwere, (I am putting it back.) Or more precisely, work with the Riemann surface whose equation is $2iz=w+1/w$. The logarithm $\ln w$ does indeed require a branch cut, going from $0$ to $\infty$ in the $w$ sphere (I include infinity, looking at the full Riemann sphere). In $z$ coordinates, this branch cut would be a loop from $\infty$ back to $\infty$, separating $+1$ and $-1$. All is very easy and transparent in the $w$ coordinate; the trick is the $z$ coordinate and the two-to-one mapping from $z$ to $w$. – Harald Hanche-Olsen Apr 25 '20 at 21:00
  • I'll try to clarify without transforming to the $w$-plane. (1) $f_1(z)=\log\left(z+\sqrt{z^2-1}\right)$ has no branch point of the logarithm at $z=0$. It does have branch points of the square root at $\pm1$. (2) $f_2(z)=\log(z)$ has branch point at $0$. (3) $f_1(z)+f_2(z)=\log\left(z+\sqrt{z^2-1}\right)-\log(z)$ has no branch point at $0$, it has branch points only at $\pm 1$. This is the apparent paradox. How can $f_1$ have no branch point at $0$, $f_2$ has a branch point at $0$, but the difference of the two has no branch point at $0$. – Mark Viola Apr 26 '20 at 02:07
  • @MarkViola The combined function does indeed have two logarithmic branch points at $z=0$: One for each branch of $\sqrt{z^-1}$. They tend to get obscured by my choice of branch cut between $+1$ and $-1$ for $\sqrt{z^-1}$. – Harald Hanche-Olsen Apr 26 '20 at 11:17
  • @MarkViola It may be useful to rewrite $\zeta=\log(z+\sqrt{z^2-1})-\log z$ as $(e^\zeta-1)^2=1-z^{-2}$. There are branch points where $e^\zeta=1$ (due to the square on the left hand side (LHS)), i.e., at $z=\pm1$. These are “square root like” branch points, in the sense that you return to the same $\zeta$ after $z$ has traveled around the branch point twice. But there are infinitely many of them, corresponding to the infinite many solutions to $e^\zeta=1$. There is also a branch point at $z=0$, near which $e^\zeta\sim\pm z^{-1}$, where each choice of sign produces a logarithmic branch point. – Harald Hanche-Olsen Apr 26 '20 at 16:26
  • And even more to the point, there is no branch point at $z=\infty$. There are of course infinitely many values for $\zeta$ there, but they remain separate. Which is why you can travel around a large loop (large enough to surround $0$ and $\pm1$ and get the same value of $\zeta$ when you return. – Harald Hanche-Olsen Apr 26 '20 at 16:42
  • Hi Harald. I hope that you are staying safe and healthy. I found an argument that seems reasonable. Note that $\arg\left(z+\sqrt{z^2-1}\right)$ and $\arg(z)$ lie in the same quadrant when we cut the plane from $-1$ to $\infty$ and from $1$ to $\infty$. Then, on $|z|=r>1$, as $\arg(z)$ goes from $0$ to $2\pi$ the argument of $z+\sqrt{z^2-1}$ does likewise. This would imply that there is a branch point of the logarithm ($z=0$). Does this make sense? – Mark Viola May 06 '20 at 15:50
  • Sounds familiar. – Alexander Cska Jun 01 '20 at 13:11
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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.

Daniel Fischer
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  • Just curious … How would one evaluate the integral $\oint_C \log(iz+\sqrt{1-z^2}),dz$ where $C$ is any rectifiable curve that encloses (one time) the branch points at $z=−1$ and $z=−1$? – Mark Viola Apr 23 '20 at 16:58
  • Hi Daniel. I hope that you are staying safe and healthy. Just curious … How would one evaluate the integral $\oint_{C}\log\left(iz+\sqrt{1-z^2}\right),dz$ where $C$ is any rectifiable curve that encloses (one time) the branch points at $z=-1$ and $z=1$? It seems that $\arg\left(z+\sqrt{z^2-1}\right)$ and $\arg(z)$ lie in the same quadrant when we cut the plane from $-1$ to $\infty$ and from $1$ to $\infty$. Then, on $|z|=r>1$, as $\arg(z)$ goes from $0$ to $2\pi$ the argument of $z+\sqrt{z^2-1}$ does likewise. This would imply that there is a branch point of the logarithm ($z=0$). – Mark Viola May 06 '20 at 15:41
  • @MarkViola I think you should post that as a separate question. – BIRA Aug 26 '20 at 23:09