I am currently looking at the propagation constant $\gamma\in\mathbb{C}$, which is
$$ \gamma = i\omega\sqrt{\mu\epsilon-i\,\frac{\sigma\mu}{\omega}}, $$
where $i^2 = -1$ and all other quantities are real.
The attenuation loss is calculated using the real part of $\gamma$. How do I calculate $\mathfrak{R}(\gamma)$ and, out of interest, $\mathfrak{I}(\gamma)$?
There are 2 approximations for $\gamma$
If $\frac{\sigma}{\omega}<<\epsilon$ (low loss) then : $$ \gamma \simeq i\omega\sqrt{\mu\epsilon}+\frac{1}{2}\sqrt\frac{\mu}{\epsilon}\sigma$$ We know $\frac{\sigma}{\omega\epsilon}<<1$ then, $$\gamma =i\omega\sqrt{\mu\epsilon}[1+\frac{\sigma}{i\omega\epsilon}]^{1/2}\simeq i\omega\sqrt{\mu\epsilon}[1+\frac{\sigma}{2i\omega\epsilon}]$$
And if $\frac{\sigma}{\omega}>>\epsilon$ (lossy) then :
$$\gamma\simeq\frac{1}{\sqrt 2} \sqrt{\omega\mu\sigma}(i+1) $$
For proof you can simply omit $\epsilon$, assuming it is small.
$$\require{cancel}\gamma =i\omega\sqrt{\mu}[\cancelto{0}{\epsilon}+\frac{\sigma}{i\omega}]^{1/2}\simeq\frac{1}{\sqrt 2} \sqrt{\omega\mu\sigma}(i+1)$$
Otherwise you can treat it as an ordinary complex nember:
$$\gamma = i\omega\sqrt{\mu\epsilon-i\,\frac{\sigma\mu}{\omega}}= i\omega\sqrt{\sqrt{(\mu\epsilon)^2+(\frac{\sigma\mu}{\omega})^2}\,.e^{-i\tan^{-1}{\frac{\frac{\sigma\mu}{\omega}}{\mu\epsilon}}}} $$ $$ \gamma = i\omega\sqrt[4]{(\mu\epsilon)^2+(\frac{\sigma\mu}{\omega})^2}.e^{-\frac{i}{2}\tan^{-1}{\frac{\sigma}{\omega\epsilon}}} $$