Choosing a branch of the square root so that this rewriting works

Solution 1:

In THIS ANSWER, I discuss the equality

$$\log(z_1z_2)=\log(z_1)+\log(z_2) \tag 1$$

The equality in $(1)$ is interpreted as a set equivalence. It means that any value of $\log(z_1z_2)$ can be expressed as the sum of some value of $\log(z_1)$ and some value of $\log(z_2)$. And conversely, it means that the sum of any value of $\log(z_1)$ and any value of $\log(z_2)$ can be expressed as some value of $\log(z_1z_2)$.

Now, using the definition $z^c=e^{c\log(z)}$ along with $(1)$, we can write

$$\begin{align} \sqrt{z-1/z}&=\left(\frac{(z+1)(z-1)}{z}\right)^{1/2}\\\\ &=e^{\frac12 \left(\log(z+1)+\log(z-1)-\log(z)\right)}\\\\ &e^{\frac12 \log(z+1)}e^{\frac12 \log(z-1)}e^{-\frac12 \log(z)}\\\\ &=\sqrt{z+1}\sqrt{z-1}/\sqrt{z}\tag 2 \end{align}$$

The equality in $(2)$ is also interpreted as a set equivalence. For any value of $\sqrt{z-1/z}$, there is some value of $\sqrt{z+1}$, some value of $\sqrt{z-1}$, and some value of $\sqrt{z}$ for which $(2)$ is true. And for any values of $\sqrt{z+1}$, $\sqrt{z-1}$, and $\sqrt{z}$, there is some value of $\sqrt{z-1/z}$ for which $(2)$ is true.

Let's look at an example. Suppose we take the principal branches to define $\sqrt{z+1}$, $\sqrt{z-1}$, and $\sqrt{z}$ for which $-\pi<\arg(z+1)\le \pi$, $-\pi<\arg(z-1)\le \pi$, and $-\pi<\arg(z)\le \pi$. At $z=i$, we see that

$$\begin{align} \color{blue}{\sqrt{i+1}}\color{red}{\sqrt{i-1}}/\color{green}{\sqrt{i}}&=\color{blue}{\sqrt{2}e^{i\pi/8}}\color{red}{\sqrt{2}e^{i3\pi/8}}/\color{green}{1e^{i\pi/4}}\\\\ &=\sqrt{2}e^{i\pi/4} \end{align}$$

We then must take the branch of $\sqrt{z-1/z}=\sqrt{2}\sqrt{i}$ for which $\sqrt{i}=e^{i\pi/4}$. And we are free to do so.