Gamelin writes the following in his book:
Example. Consider $\sqrt{z-1/z}$. We rewrite this as $\sqrt{z-1}\sqrt{z+ 1}/\sqrt{z}$.
How can this rewriting actually be done?
For no branch we have universally that the square root is multiplicative, so maybe he means different branches in the four square roots in this equation? But then this equation is not really an equation since the domains of the two functions are different...
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.
$(z-1/z)^{1/2}= ((z^2 - 1)/z)^{1/2}=((z-1)(z+1)/z)^{1/2}= \dfrac{(z+1)^{1/2} (z-1)^{1/2}}{z^{1/2}}$
