Branch cut confusion for $\ln(1+\sqrt{1-z^2})-\sqrt{1-z^2}$

Question

I'm looking at the function $f(z)=\ln(1+\sqrt{1-z^2})-\sqrt{1-z^2}$ for complex $z$ and the square root and logarithm are defined according to their principal values. So $\sqrt{1-z^2}$ is analytic in $U=\mathbb{C}\backslash (-\infty,-1] \cup [1,\infty)$. The function inside the logarithm is never zero and $U$ is simply connected, hence $f(z)$ is also analytic in $U$. Define now

$$F(z) = (i f(z))^{\frac{2}{3}} + (f(z))^{\frac{2}{3}}$$

where $w^\frac{2}{3}=e^{\frac{2}{3} \ln w}$. I want to show that $F \equiv 0$. Since $f$ is never zero in $U$ (its only zeros seem to be at $z=\pm 1$), we can define $\ln f$ as an analytic function in $U$ and therefore we can also define $f^\frac{2}{3}$. Hence $F$ is analytic in $U$. For $y>0$, $f(iy)$ is real valued and negative, so let us write $f(iy)=-c$ with $c>0$, then

$$F(iy) = (-i c)^{\frac{2}{3}} + (-c)^{\frac{2}{3}} = c^\frac{2}{3} \left( (-i)^\frac{2}{3} + (-1)^\frac{2}{3} \right) = c^\frac{2}{3} \left( e^{-i\frac{\pi}{3}} + e^{i\frac{2\pi}{3}} \right) = 0$$

So $F(iy)=0$ for any $y>0$, then by analytic continuation $F\equiv 0$ in $U$.

Now I evaluate $F$ (using Maple) at certain points and get nonzero results, for example

$$F(3+i) \approx 0.188 + i2.66$$

Or, more generally I get $F(x+i) \neq 0$ for $x\geq 3$. I can't figure out what am I doing wrong here, is it not true that $F$ is analytic in $U$? I'll appreciate the help to settle my confusion. Thank you.

EDIT

As pointed out I'm wrong in substituting $iy$ in $F$ since then I get a negative number inside the power function $(\cdot)^{\frac{2}{3}}$. To avoid this we can just look inside the right half plane $V = \{\Re z>0\} \backslash [1,\infty)$. My question still remains, as checking numerically

$$F(0.5+iy) = 0, \ \ y>0$$

It seems analyticity is lost once we cross some vertical line(?), and I don't understand why.

Answer 1

The issue is that the complex power function is not defined, as the complex logarithm principal value on the real half-line $(-\infty, 0]$. Hence your computation

$$(-1)^\frac{2}{3} = e^{i\frac{2\pi}{3}}$$ doesn't make sense and $F$ is not defined for $iy$ with $y \gt 0$.

Regarding your EDIT new question, the fact is that $F$ is defined using complex power, hence is based on logarithm principal value. Therefore, is defined for $D = \{z \in \mathbb C \mid f(z) \notin (-\infty, 0]\}$. And $D$ is probably having a complex shape.