Is the textbook WRONG on the branch cut?

Question

As defined in the textbook, it takes the branch cut along positive real axis. So the argument angle is $[0,2\pi)$. This is no problem. But in the example, it sets $f(z)=\frac1{\sqrt{2-z}}$ and claims

$$\frac1{\sqrt{2-z}}=\frac i{\sqrt{z-2}}$$

and the phase factor of $f(z)$ is also $-1$.

If use the branch cut from $2$ to infinity, then $2-z=(z-2)e^{i\pi}$, so

$$f(z)=\frac1{\sqrt{2-z}}=\frac1{\sqrt{(z-2)e^{i\pi}}}=\frac{-i}{\sqrt{z-2}}$$

Is the textbook wrong?

Answer 1

You're both wrong because you're ignoring subtleties. It's equivalent to think about the simpler question:

If $\sqrt{\cdot}$ is the principal square root (induced from $-\pi<\arg\le\pi$) then when is it true that: $$\frac{1}{\sqrt{z}}=\frac{i}{\sqrt{-z}}$$

Notice that this is equivalent to $\sqrt{-z}=i\sqrt{z}$. $-z$ and $z$ have the same modulus, so that doesn't need to be worried about. We would require $\exp i\frac{\arg(-z)}{2}=\exp i\frac{\pi+\arg(z)}{2}$ which happens when and only when $\arg(-z)$ and $\pi+\arg(z)$ differ by an even multiple of $2\pi$. The former is in range $(-\pi,\pi]$, the latter in range $(0,2\pi]$ so this could only happen if they were literally equal.

For this principal argument, we can check case by case that: $$\Im z>0\implies\arg(-z)=\arg(z)-\pi,\Im z<0\implies\arg(-z)=\arg(z)+\pi>0$$It follows that the stated equation holds for all $z$ with $\Im z<0$ and for $z\in(0,\infty)$.

For $z\in(-\infty,0)$ and for $\Im z>0$, it is instead true that $\frac{1}{\sqrt{z}}=\frac{-i}{\sqrt{-z}}$. So, it depends.

What you did: $$\frac{1}{\sqrt{(z-2)e^{\pi i}}}=\frac{1}{\sqrt{z-2}\cdot\sqrt{e^{\pi i}}}=-\frac{i}{\sqrt{z-2}}$$Is not valid since $\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}$ is not generally true.

Answer 2

Let's see the differences for the choices of branch cuts. By shifting the $z-2$ to the origin, the key is to find the influences to $\sqrt{-z}$ due to the choices of branch cuts. Here we illustrate it as shown in the figure.

If the point $z$ on the 1st quadrant ($z$ on the 2nd quadrant gives the same result), $z=re^{i\theta}$

For the blue case, $\arg z\in(-\pi, \pi]$, Point $A=-z=re^{i(\theta-\pi)}\Longrightarrow \sqrt{-z}=\sqrt z~e^{-i\pi/2}=-i\sqrt z$

For the red case, $\arg z\in [0, 2\pi)$, Point $B=-z=re^{i(\theta+\pi)}\Longrightarrow \sqrt{-z}=\sqrt z~e^{i\pi/2}=i\sqrt z$

If the point $z$ is on the 4th quadrant, ($z$ on the 3rd quadrant gives the same result)

For the blue case, $\arg z\in(-\pi, \pi]$, Point $C=-z=re^{i(\theta+\pi)}\Longrightarrow \sqrt{-z}=\sqrt z~e^{i\pi/2}=i\sqrt z$

For the red case, $\arg z\in [0, 2\pi)$, Point $D=-z=re^{i(\theta-\pi)}\Longrightarrow \sqrt{-z}=\sqrt z~e^{-i\pi/2}=-i\sqrt z$