Defining the square root of $z$ squared and determining the location of branch cuts

Question

I am asked the following:

For $\epsilon > 0$, we define $$ \sqrt{z^2} = \lim_{\epsilon \to 0} \sqrt{z^2 + \epsilon^2}\,, $$ where the principle value square root is used on the right-hand side. Determine the location of the branch cuts and show that $\sqrt{z^2} = \text{sign}(\text{Re } z) z$

I have tried the following.

Let $\epsilon > 0$ be given and l et $z-i\epsilon = r_1 e^{i\theta_1}$ and $z+i\epsilon = r_2 e^{i\theta_2}$. I know we can write $$f_\epsilon(z) := \sqrt{z^2 + \epsilon^2} = \sqrt{(z-i\epsilon)(z+i\epsilon)} = \sqrt{r_1r_2} e^{i(\theta_1 + \theta_2)/2}\,.$$ I see that the branch points are $i\epsilon$ and $-i\epsilon$. Now, following the reasoning done in this StackExchange question and this handout, we can argue that $[-i\epsilon, i\epsilon]$ would be a sufficient branch cut to make $f_\epsilon(z)$ single-valued.

The book defines the principle value square root by a branch cut along the negative real axis, corresponding to $\sqrt 1 = 1$. This means $\sqrt z = \sqrt r e^{\theta i / 2}$ where $\theta \in (-\pi, \pi]$, the principle value of the argument, i.e. $\text{Arg } z$.

In the given defition, the principle value square root is used, therefore; $\theta_1, \theta_2 \in (-\pi, \pi]$. Does this mean that, for example, reasoning about encircling $z = i\epsilon$ (and not $z = -i\epsilon$), as done in mentioned StackExchange question; $$\sqrt{r_1}e^{i(\theta_1+2\pi)/2} = \sqrt{r_1}e^{i\theta_1/2}e^{\pi i} = -\sqrt{r_1}e^{i\theta_1/2}\,,$$ is not permitted by the bounds of $\theta_1$? I'm not quite sure whether this means I am force to make the branch cuts $[i\epsilon, \infty)$ and $[-i\epsilon, -\infty)$.

As for the follow-up question; $\sqrt{z^2} = \text{sign}(\text{Re } z) z$, I have tried writing it as follows: $$ \begin{align*} \sqrt{z^2} &= e^{\log(z^2)/2} \\ &= e^{\frac{1}{2}\left( \ln |z|^2 + i \text{Arg} (z^2) \right)} \\ &= |z|e^{i \text{Arg} (z^2)}\,. \end{align*} $$ This means I have to show $e^{i \text{Arg} (z^2)} = \text{sign}(\text{Re } z)e^{i \text{Arg} z}$, which I've been unable to do thus far.

Answer 1

The function $g(z)=\sqrt{(z-a)(z-b)}$, is definable as an analytic function in any domain $U\subset \mathbb C$, with the property that $a$ and $b$ lie in the same connected component of $\mathbb C\cup\{\infty\}\setminus U$. This is due to the fact that $f(z)=\log(\frac{z-a}{z-b})$ is definable as an analytic function in a domain with the same property, since $f'(z)=\frac{1}{z-a}-\frac{1}{z-b}$, and thus, for every closed curve $\gamma \subset U$ $$ \int_\gamma f'(z)\,dz=\mathrm{Ind}_{\gamma}(a)-\mathrm{Ind}_{\gamma}(b)=0, $$ and consequently, $f$ is be well-defined (modulo $2\pi i$.) Then, $g$ is naturally definable as $$ g(z)=(z-b)\exp\big(\,f(z)/2\big). $$

Now $$f_\varepsilon(z)=\sqrt{z^2+\varepsilon^2}=\sqrt{(z+i\varepsilon)(z-i\varepsilon)} $$ is definable as an analytic function in any domain $U\subset \mathbb C$, with the property that $\pm i\varepsilon$ lie in the same connected component of $\mathbb C\cup\{\infty\}\setminus U$. Most typical choices of $U$:

$U_{1,\varepsilon}=\mathbb C\setminus[-i\varepsilon,i\varepsilon]$ and

$U_{2,\varepsilon}=\mathbb C\setminus([-i\infty,-i\varepsilon]\cup[i\varepsilon,i\infty])$.

In particular, $f_{\varepsilon,1}(z):U_{1,\varepsilon}\to\mathbb C$ is odd, while $f_{\varepsilon,2}(z):U_{2,\varepsilon}\to\mathbb C$ is even, branches chosen so that both positive for $z>0$. (Not totally trivial.)

In fact, $f_2(z)=\lim_{\varepsilon\to 0}f_{\varepsilon,2}(z)$ is still even, and it satisfies $$ f_2(z)=\left\{\begin{array}{rl} z & \text{if $\,\,\mathrm{Re}\,z>0$} \\-z & \text{if $\,\,\mathrm{Re}\,z<0$}\end{array}\right. $$