ComplexPlot3D[
Sqrt[Exp[-z/(4*Sqrt[2])]] Sqrt[Exp[z/(4*Sqrt[2])]], {z, -2 \[Pi] -
2 \[Pi] I, 2 \[Pi] + 2 \[Pi] I}, PlotLegends -> Automatic]

Mathematica chooses the branch cut for 𝑧√ to lie along the negative real axis and the first value is {0,0}
and this is 1
.
This asks whether to trust PowerExpand
or ComlexExpand
.
Refine[Re[
ComplexExpand[
Sqrt[Exp[-(x + I y)/(4*Sqrt[2])]] Sqrt[Exp[(x + I y)/(4*Sqrt[2])]],
TargetFunctions -> {Abs, Arg}]], Element[{x, y}, Reals]]
-Im[Cos[1/2 ArcTan[Cos[y/(4 Sqrt[2])], Sin[y/(4 Sqrt[2])]]] Sin[
1/2 ArcTan[Cos[y/(4 Sqrt[2])], -Sin[y/(4 Sqrt[2])]]] +
Cos[1/2 ArcTan[Cos[y/(4 Sqrt[2])], -Sin[y/(4 Sqrt[2])]]] Sin[
1/2 ArcTan[Cos[y/(4 Sqrt[2])], Sin[y/(4 Sqrt[2])]]]] +
Re[Cos[1/2 ArcTan[Cos[y/(4 Sqrt[2])], -Sin[y/(4 Sqrt[2])]]] Cos[
1/2 ArcTan[Cos[y/(4 Sqrt[2])], Sin[y/(4 Sqrt[2])]]] -
Sin[1/2 ArcTan[Cos[y/(4 Sqrt[2])], -Sin[y/(4 Sqrt[2])]]] Sin[
1/2 ArcTan[Cos[y/(4 Sqrt[2])], Sin[y/(4 Sqrt[2])]]]]
Refine[Im[
ComplexExpand[
Sqrt[Exp[-(x + I y)/(4*Sqrt[2])]] Sqrt[Exp[(x + I y)/(4*Sqrt[2])]],
TargetFunctions -> {Abs, Arg}]], Element[{x, y}, Reals]]
Im[Cos[1/2 ArcTan[Cos[y/(4 Sqrt[2])], -Sin[y/(4 Sqrt[2])]]] Cos[
1/2 ArcTan[Cos[y/(4 Sqrt[2])], Sin[y/(4 Sqrt[2])]]] -
Sin[1/2 ArcTan[Cos[y/(4 Sqrt[2])], -Sin[y/(4 Sqrt[2])]]] Sin[
1/2 ArcTan[Cos[y/(4 Sqrt[2])], Sin[y/(4 Sqrt[2])]]]] +
Re[Cos[1/2 ArcTan[Cos[y/(4 Sqrt[2])], Sin[y/(4 Sqrt[2])]]] Sin[
1/2 ArcTan[Cos[y/(4 Sqrt[2])], -Sin[y/(4 Sqrt[2])]]] +
Cos[1/2 ArcTan[Cos[y/(4 Sqrt[2])], -Sin[y/(4 Sqrt[2])]]] Sin[
1/2 ArcTan[Cos[y/(4 Sqrt[2])], Sin[y/(4 Sqrt[2])]]]]
So the product does only depend on y
or Im[z].
Plot[{Cos[1/2 ArcTan[Cos[y/(4 Sqrt[2])], Sin[y/(4 Sqrt[2])]]] Sin[
1/2 ArcTan[Cos[y/(4 Sqrt[2])], -Sin[y/(4 Sqrt[2])]]] +
Cos[1/2 ArcTan[Cos[y/(4 Sqrt[2])], -Sin[y/(4 Sqrt[2])]]] Sin[
1/2 ArcTan[Cos[y/(4 Sqrt[2])], Sin[y/(4 Sqrt[2])]]],
Cos[1/2 ArcTan[Cos[y/(4 Sqrt[2])], -Sin[y/(4 Sqrt[2])]]] Cos[
1/2 ArcTan[Cos[y/(4 Sqrt[2])], Sin[y/(4 Sqrt[2])]]] -
Sin[1/2 ArcTan[Cos[y/(4 Sqrt[2])], -Sin[y/(4 Sqrt[2])]]] Sin[
1/2 ArcTan[Cos[y/(4 Sqrt[2])],
Sin[y/(4 Sqrt[2])]]]}, {y, -6 \[Pi], 6 \[Pi]}]

Since the solution of the others this must be the solution.
Plot[{Cos[1/2 ArcTan[Cos[y/(4 Sqrt[2])], -Sin[y/(4 Sqrt[2])]]] Cos[
1/2 ArcTan[Cos[y/(4 Sqrt[2])], Sin[y/(4 Sqrt[2])]]] -
Sin[1/2 ArcTan[Cos[y/(4 Sqrt[2])], -Sin[y/(4 Sqrt[2])]]] Sin[
1/2 ArcTan[Cos[y/(4 Sqrt[2])], Sin[y/(4 Sqrt[2])]]],
Cos[1/2 ArcTan[Cos[y/(4 Sqrt[2])], Sin[y/(4 Sqrt[2])]]] Sin[
1/2 ArcTan[Cos[y/(4 Sqrt[2])], -Sin[y/(4 Sqrt[2])]]] +
Cos[1/2 ArcTan[Cos[y/(4 Sqrt[2])], -Sin[y/(4 Sqrt[2])]]] Sin[
1/2 ArcTan[Cos[y/(4 Sqrt[2])],
Sin[y/(4 Sqrt[2])]]]}, {y, -6 \[Pi], 6 \[Pi]}]

This shows the same plot with swapped coloring.
ReImPlot[Refine[
Im[ComplexExpand[
Sqrt[Exp[-(x + I y)/(4*Sqrt[2])]] Sqrt[
Exp[(x + I y)/(4*Sqrt[2])]], TargetFunctions -> {Abs, Arg}]],
Element[x, Reals] && Element[y, Reals]], {y, -6 \[Pi], 6 \[Pi]}]

ReImPlot[Refine[
Re[ComplexExpand[
Sqrt[Exp[-(x + I y)/(4*Sqrt[2])]] Sqrt[
Exp[(x + I y)/(4*Sqrt[2])]], TargetFunctions -> {Abs, Arg}]],
Element[x, Reals] && Element[y, Reals]], {y, -6 \[Pi], 6 \[Pi]}]

separates the situation for the Im and the Re part.
So Mathematica makes a mistake in doing the branch cut. The term is indeed still an identity as if the argument were pure Reals. The last ReImPlot shows the term more complicated and delicate. The plot does not change must if the interval is taken smaller. So the value can be both 0 and 1 at the same time.
Why does Mathematica draw the graph for the solution 1 as a single line and the solution 0 as dotted? It separated the Re part a line and the Im as dotted by the default style. Mathematica did make the y
out of the Complexes again as in the result of the Refine
. But y is Reals
.
PowerExpand[
Sqrt[Exp[-(x + I y)/(4*Sqrt[2])]] Sqrt[Exp[(x + I y)/(4*Sqrt[2])]],
Assumptions -> True]
(*E^((-x - I y)/(8 Sqrt[2]) + (x + I y)/(8 Sqrt[2]) +
I \[Pi] (Floor[
1/2 - Im[x]/(8 Sqrt[2] \[Pi]) - Re[y]/(8 Sqrt[2] \[Pi])] +
Floor[1/2 + Im[x]/(8 Sqrt[2] \[Pi]) + Re[y]/(8 Sqrt[2] \[Pi])]))*)
PowerExpand[
Sqrt[Exp[-(x + I y)/(4*Sqrt[2])]] Sqrt[Exp[(x + I y)/(4*Sqrt[2])]]]
(* E^((-x - I y)/(8 Sqrt2) + (x + I y)/(8 Sqrt2)) *)
Do this assumption treat the parameters correct as Reals
each? My judgement is Yes it does and the ComplexPlot3D
is correct.
Plot3D[Floor[
1/2 - Im[x]/(8 Sqrt[2] \[Pi]) - Re[y]/(8 Sqrt[2] \[Pi])] +
Floor[1/2 + Im[x]/(8 Sqrt[2] \[Pi]) + Re[y]/(
8 Sqrt[2] \[Pi])], {x, -1, 1}, {y, -1, 1}]

PowerExpand[
Sqrt[Exp[-(x + I y)/(4*Sqrt[2])]] Sqrt[Exp[(x + I y)/(4*Sqrt[2])]],
Assumptions -> True] // FullSimplify
(* (-1)^(Floor[1/2 - (Im[x] + Re[y])/(8 Sqrt2 [Pi])] +
Floor[1/16 (8 + (Sqrt2 (Im[x] + Re[y]))/[Pi])]) *)
and not to much simplified.
There are question for simplifying trigonometrics on mathematica.stackexchange.com. .
FullSimplify[Sqrt[Exp[-z/(4*Sqrt[2])]] Sqrt[Exp[z/(4*Sqrt[2])]], Assumptions -> {-Pi <= Im[z] <= Pi}]
. $\endgroup$