sqrt(exp)

[Kouznetsov]

I do not expect any other singularities. There are only 2 branchpoints and only two cutlines.

Hm perhaps then they lie on other branches.
Those singularities need not to be branch points, they can also be isolated singularities.

But why do there have to be singularities?
First one would expect that a half iterate has the same fixed points as the function. (Is this true for your branch of \( \exp^{1/2} \), i.e. is \( L_2 \) a fixed point?)
But if so, a non-integer iterate can usually be holomorphic at most at one fixed point. This is the case for regular iteration at that fixed point. As your \( \exp^{1/2} \) is not regular at any fixed point (regular iteration yields non-real values at the real axis for \( \exp \)) it must be singular at any fixed point. However if it is non-singular at \( L_2 \) then this may be due to \( L_2 \) being not a fixed point of \( \exp^{1/2} \) in the branch that you chose.
Can you please verify this?

Henryk:
There is unigue tetration \( f \), holomorphic in \( \mathbb{C}\{ x\in \mathbb{R~:~} x\le -2 \} \).
There is inverse function \( g \) such that
\( f(g(z))=z \forall z\in F \). Function \( g \) has unavoidable branchpoints at eigenvalues
\( L \approx 0.318131505204764135312654+1.33723570143068940890116 \rm i \) abe \( L^* \) of logarithm.
I choose the cutline in such a way that function \( g \) is holomorphic in the range \( F=\mathbb{C}\{ z\in \mathbb{C~:~} \Re(z)\le \Re(L),~ |\Im(z)|=\Im(L)\} \).

In the set \( F \), function \( g \) takes each value just once, except fixed points of logarithm, i.e.,
\( L\approx 0.318131505204764135312654+1.33723570143068940890116 ~\rm i \) and its conjugation. This function preserves the signum of the imaginary part. (While the imaginary part of the argument is positive, the imaginary part of the function is also positive).

Define square root of the exponential as \( \sqrt{\exp}(z)=f(0.5+g(z)) \)
for all \( z\in F \). As function \( f \) has no singularities ourside the real axis, funciton \( \sqrt{\exp} \) also is not allowed to have singularities which would not be singulatities of function \( g \). Theregore function \( \sqrt{\exp} \) is holomorphic in \( F \).

I attach the extended figure of \( f=sqrt{exp} (z) \) in the upper part of the complex \( z \). It shows the branchpoint at eigenvalue
\( L \approx 0.3181315052+1.33723570143 ~\rm i \) of logarithm.
It is holomorphic and has no cuts in vicinity of the "first" branchpoint \( L_{{\rm e, 1}}\approx 2.062277729598284 + 7.5886311784725127 ~\rm i \) of exponential. (at elast while the imaginary part is equal to 8.
   
The grid covers the range \( -4 \le Re(z) \le 4 \), \( 0 \le Im(z) \le 0 \) with unity step.
Levels \( \Re(f)f=-3,-2,-1 \) are shown with thick red curves.
Levels \( \Re(f)f=1,2,.. 14 \) are shown with thick blue curves.
Level \( \Re(f)f=0 \) is shown with thick black curve.
Level \( \Re(f)f=\Re(L) \) is shown with thick green curve.
Levels \( \Im(f)f=\pm 0.1, \pm 0.2, .. \pm 0.9 \) are shown with thin green curves.

Level \( \Im(f)f=\Im(L) \) is shown with thick green curve.
Level \( \Im(f)f=-1,-2 \) is shown with thick red curves.
Levels \( \Im(f)f=1,2,.. 14 \) are shown with thick blue curves.
Levels \( \Im(f)f= -0.1, -0.2, .. -0.9 \) are shown with thin red curves.

The cut \( \{z \in \mathbb{C~:~} \Im(z)=\Im(L)~,~ \Re(z)<\Re(L)\} \) is shown with thick pink horizontal line.