Simplification of Sqrt[Exp[u]] Sqrt[Exp[-u]]

(Michael already gave a good answer, but I'll leave mine here as it includes a few extra observations)

This is a tricky question. First, here's a counterexample:

expr = Sqrt[Exp[-z/(4*Sqrt[2])]] Sqrt[Exp[z/(4*Sqrt[2])]];

FullSimplify[expr /. z->4 Sqrt[2] Pi I]

-1

Next, using PowerExpand as @Ulrich did:

PowerExpand[expr, Assumptions->True]
% //TeXForm

E^(I π (Floor[1/2 - Im[z]/(8 Sqrt[2] π)] + Floor[1/2 + Im[z]/(8 Sqrt[2] π)]))

$\exp \left(i \pi \left(\left\lfloor \frac{1}{2}-\frac{\Im(z)}{8 \sqrt{2} \pi }\right\rfloor +\left\lfloor \frac{\Im(z)}{8 \sqrt{2} \pi }+\frac{1}{2}\right\rfloor \right)\right)$

The arguments of the Floor expressions are always real. Here is a plot of them:

Plot[{Floor[1/2 - x], Floor[1/2 + x]}, {x, -3, 3}]

It would seem that the sum of the two Floor expressions should be 0. However, it turns out that the endpoints are not consistent. For example:

Reduce[Floor[1/2 - x] == 1, x, Reals]
Reduce[Floor[1/2 + x] == -1, x, Reals]

-(3/2) < x <= -(1/2)

-(3/2) <= x < -(1/2)

So, at the points n+1/2 the sum is not 0. Another way to see this is to use NumberLinePlot:

NumberLinePlot[Floor[1/2-x] + Floor[1/2+x] == 0, {x, -3, 3}]

Summarizing, your expression is almost always equal to 1, except when the imaginary part of z is $8 \sqrt{2} \pi \left(n+\frac{1}{2}\right)$ for integer $n$.

For $z = 4 \sqrt{2} \pi i$, we have $$ \sqrt{e^{-z/4 \sqrt{2}}} = \sqrt{e^{-i\pi}} = \sqrt{-1} = i, $$ and $$ \sqrt{e^{z/4 \sqrt{2}}} = \sqrt{e^{i\pi}} = \sqrt{-1} = i. $$ So the product of the two square roots is $-1$ in this one case. Note that in this case, $$ \arg e^{-z/4 \sqrt{2}} \neq -\frac{\Im(z)}{4 \sqrt{2}}, $$ since $\arg (e^{-i \pi}) = \arg (-1) = \pi$, not $-\pi$.

In general, your desired simplification is true so long as $$ \frac{\Im(z)}{4 \sqrt{2} \pi} \neq 1 \mod 2. $$ Unfortunately, I haven't been able to find a way to use Assumptions to tell Mathematica to assume this. As b.gatessucks notes in the comments, we can write

Simplify[Sqrt[Exp[-z/(4*Sqrt[2])]] Sqrt[Exp[z/(4*Sqrt[2])]], 
         Assumptions -> {-4 Sqrt[2] Pi < Im[z] < 4 Sqrt[2] Pi}]

which does simplify to 1. But this should also simplify to one if {4 Sqrt[2] Pi < Im[z] < 12 Sqrt[2] Pi}, and using this for our Assumptions doesn't lead to a simplification. It may be that Mathematica isn't programmed to accept such a condition.