Consider the complex function $$f(z) = \sqrt{z^2+1}.$$

Obviously, $f(z)$ has branch points at $z = \pm i$. One way of defining a branch cut would be to exclude the points on the imaginary axis with $|z| \geq 1$. Another way of defining a branch cut appears to be to exclude the (finite) region of the imaginary axis with $|z| \leq 1$.

If we define $f(z)$ as $$f(z) = e^{1/2\log(z^2+1)},$$ the first branch cut can be arrived at by taking the principal branch of $\log(z)$ with the branch cut $(-\infty,0]$. At first I thought the second branch cut could be arrived at by taking the branch cut $[0,\infty)$ for $\log(z)$. Indeed, this would exclude imaginary $z$ with $|z| \leq 1$, but it would of course also exclude all real $z$, thus constituting a different branch cut for $f(z)$.

I think it would be possible to arrive at the second branch cut for $f(z)$ differently, by defining $$f(z) = \sqrt{r_1r_2}e^{i(\theta_1+\theta_2)/2}, $$ where $r_1 = |z-i|, r_2 = |z+i|$, $\theta_1 = \arg(z-i), \theta_2 = \arg(z+i).$ However, I don't really like this approach, since I think $$f(z) = e^{1/2\log(z^2+1)}$$ is the proper way to define $f$.

Any comments on this? Is there a branch cut for $\log(z)$ which gives the correct branch cut for $f$?


3 Answers 3


Consider the function $$ h(z) = \exp\left\{ \int_{i}^{z}\frac{w}{1+w^2}dw \right\} = \exp\left\{\frac{1}{2}\int_{i}^{z}\frac{1}{w+i}+\frac{1}{w-i}\,dw\right\}, $$ where the integral from $i$ to $z$ is taken along any simple path from $i$ to $z$ in the open region $\Omega$ obtained by subtracting from $C$ the closed segment from $-i$ to $i$. It doesn't matter which path is chosen from $i$ to $z$ because the difference of exponent integrals along two different paths $\gamma_1$, $\gamma_2$ will be a closed path integral with the same winding number around $i$ as around $-i$, resulting in a difference of integrals that is an integer multiple of $2\pi i$. Hence $h(z)$ does not depend on the path chosen from $i$ to $z$, so long as it does not cross the segment from $-i$ to $i$. So $h$ is holomorphic in $\mathbb{C}\setminus [-i,i]$, and $$ h'(z) = \exp\left\{\int_{i}^{z}\frac{w}{1+w^2}dw\right\}\frac{z}{1+z^2}=h(z)\frac{z}{1+z^2} \\ \implies \left(\frac{h(z)^2}{1+z^2}\right)'=0. $$ After multiplying by an appropriate constant, the resulting function $g(z)=Ch(z)$ will satisfy $g(z)^2=1+z^2$ everywhere in $\mathbb{C}\setminus[-i,i]$.

So even thought there isn't a logarithm for $1+z^2$ in $\mathbb{C}\setminus[-i,i]$, everything works out because there is a "logarithm modulo integer multiples of $2\pi i$." I'll leave you to make sense of that phrase in whatever way you want.

There's no problem defining an actual holomorphic logarithm in $\mathbb{C}\setminus\{ (\infty,-i]\cup[i,\infty)\}$ because this is a simply connected region where $w/(1+w^2)$ is holomorphic, which gives an antiderivative of $w/(1+w^2)$ in this region, and hence a holomorphic square root function $\sqrt{1+w^2}$ in this region.

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    $\begingroup$ @ÉtienneBézout : I clarified the last paragraph. A function which is holomorphic on an open simply-connected region has a holomorphic antiderivative, and that's how you always get a nice root $\sqrt{1+w^2}$ when the region is simply connected and does not contain $\pm i$. $\endgroup$ Oct 3, 2016 at 21:48
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    $\begingroup$ Sorry for the late reply. Thanks! Regarding my other question: I've been taught that we always define $f(z)^\alpha$ as $\exp(\alpha \log f(z))$. While it works, I guess it might not be true that we require this definition, if we want another branch cut. $\endgroup$ Oct 5, 2016 at 9:43
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    $\begingroup$ @ÉtienneBézout : That is correct. For example, if you want $\sqrt{(z-1)(z-2)(z-3)}$, you can specify a region so that any closed path has to circle none or two roots. So you could remove the line from $1$ to $2$ and a line from $3$ to $\infty$ that doesn't pass through $1$ or $2$, and that would be a valid domain. Or you could create a simply-connected domain by removing segments from $1$ to $-\infty$, from $2$ to $i\infty$ and from $3$ to $+\infty$. The first would not technically give you a log in the plane (only on a Riemann surface,) while the second would be standard. $\endgroup$ Oct 5, 2016 at 19:44
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    $\begingroup$ @ÉtienneBézout : Note also that you can have branch cuts along curves, too, and not just straight lines. There's nothing wrong with that. $\endgroup$ Oct 5, 2016 at 19:50
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    $\begingroup$ @TrialAndError : what I don't get is why $ \int_p^z \frac{w}{1+w^2} dw$ is analytic (I am assuming you meant $p$ a point in the domain, and not $i$). In fact, if $\gamma_1$ and $\gamma_2$ are two paths, then it can be seen that $\int_{\gamma_1} \frac{dw}{w+i} - \int_{\gamma_2} \frac{dw}{w+i} = 2 \pi i \mathrm{Ind}_{-i}(\gamma_1 - \gamma_2)$ and $\int_{\gamma_1} \frac{dw}{w-i} - \int_{\gamma_2} \frac{dw}{w-i} = 2 \pi i \mathrm{Ind}_i(\gamma_1 - \gamma_2)$. But from this, can does it follow that the integral is well-defined? $\endgroup$
    – Ruben
    Jan 12, 2017 at 0:17

I think the OP might have been originally looking for a construction similar to this:

Consider the following Möbius transformation:


It is not hard to verify that the image of the requested region under this transformation is: $$\phi(\mathbb{C}\setminus [-i,i])=\mathbb{C}\setminus(\infty, 0]$$ Where by $[a,b]$ I mean the straight finite line in $\mathbb{C}$ connecting $a$ and $b$.

In this region, the principal branch of $Log(z)$ is well defined, and so we can use it to take a square root $$g(z)=exp(\frac{1}{2}Log(\frac{z+i}{z-i})) \implies g(z)^2=\frac{z+i}{z-i}$$

This is almost what we need. In this case, we can receive the desired result by multiplying by a simple function:

$$f(z)=(z-i)g(z)\implies f(z)^2=(z-i)(z+i)=z^2+1$$

And $f$ is indeed continuous in the specified domain.


We will let $\Omega_1=\mathbb{C}\setminus\{ir\mid r\in\mathbb R,\left\vert r\right\vert\geq1\}$, and let $\Omega_2=\mathbb{C}\setminus\{ir\mid r\in\mathbb R,\left\vert r\right\vert\leq1\}$. Geometrically, $\Omega_1$ and $\Omega_2$ "have the same shape" on the Riemann sphere: they are related by the inversion $z\mapsto 1/z$. So it is not surprising that the inversion will come in handy. Indeed, if $f$ is a branch of $\sqrt{z^2+1}$ on $\Omega_1$, then $z\cdot f(z^{-1})$ is a branch on $\Omega_2$.


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