Let $u$ and $\eta$ be real numbers such that $u<0$ and $\eta>0$. What is the value of this expression? $$\left(\frac{1}{u}\right)^{i\eta/2} u^{i\eta/2}$$
This is probably trivial, but the result I got from Mathematica is Abs[...] = $e^{-\pi \eta}$. I am not sure how to obtain this result properly.
The main "trick" is to use the definition: $$ z^w = e^{w\text{Log}(z)} $$ for any complex numbers $z,w$, where $\text{Log}(z)$ refers to the principal branch of the complex logarithm: $$ \text{Log}(z) = \ln|z| + i\text{Arg}z $$
As an example, take $z = u$ and $w = i\eta/2$. Hence, $$ u^{i\eta/2} = e^{\frac{i\eta}{2}\text{Log}(u)} $$ But, $$ \text{Log}(u) = \ln|u| + i\pi $$ since $u<0$ (thus $\text{Arg}(u)=\pi$). It follows that $$ u^{i\eta/2} = e^{\frac{i\eta}{2}\ln|u| } e^{-\frac{\eta\pi}{2}} $$ In a very similar way, and using $\ln|1/u| = -\ln|u|$ we obtain: $$ \left(\frac{1}{u}\right)^{i\eta/2} = e^{-\frac{i\eta}{2}\ln|u| } e^{-\frac{\eta\pi}{2}} $$ Thus, $$ \left(\frac{1}{u}\right)^{i\eta/2}u^{i\eta/2} = \left( e^{-\frac{i\eta}{2}\ln|u| } e^{\frac{i\eta}{2}\ln|u| }\right)\left(e^{-\frac{\eta\pi}{2}} e^{-\frac{\eta\pi}{2}}\right) = e^{-\eta\pi} $$ Hopefully I didn't make any mistake and this is helpful to you.