TITLE: Question about complex emissivity and complex permeability
QUESTION [3 upvotes]: Consider a linear, isotropic material. We can then write:
$$\textbf{D}(\textbf{r},\omega) = \epsilon(\textbf{r},\omega)\textbf{E}(\textbf{r},\omega)$$
In phasor notation. The complex emissivity $\epsilon'$is defined as: $$\epsilon' = \epsilon - \frac{j\sigma(\textbf{r})}{\omega}$$
We can also write:
$$\textbf{B}(\textbf{r},\omega) = \mu(\textbf{r},\omega)\textbf{H}(\textbf{r},\omega)$$
Where I call $\mu$ the complex permeability.
I know now that a medium is passive (i.e. always absorbs the magnetic and electric energy that are stored in the fields) if the imaginary part of $\epsilon$ and $\mu$ are negative or zero (but one of them must differ from zero). A medium is active when the imaginary parts are positive or zero.
I wonder if it is also the other way around: if the medium is passive, will the imaginary parts be negative or zero (but not both at the same time zero)? Or are the cases in which the medium is passive but the imaginary parts are not both negative (or zero)? I have the same question for active media, but I guess that is analogous.

REPLY [2 votes]: You basic idea (that the sign of the imaginary part uniquely determines whether the medium is a nett sink or source of energy) is correct, although beware of different phasor sign conventions. Your statement is right for phasors wherein $\partial_t\to i\,\omega$, but sometimes it is the other way around - particularly if Maxwell's equations are thought of the first quantized propagation equation for the photon, i.e. equivalent to Schrödinger's equation, in which case we take $e^{-i\,\omega\,t}$ to be a something whose phase advances with time (see for example my answer here to "What does the complex electric field show?" ).
The way to reason through this definitively is to look at the propagation constant, i.e. the phase of plane waves. A plane wave running in the $\pm z$ direction (there is no loss of generalness in assuming this - we simply align our co-ordinate axes appropriately) has a phase $\exp(\pm i\,\omega\,t \pm \gamma z)$. Here the two signs are independent: the sign on $\omega$ sets your phasor convention, but even if we choose $+i\omega\,t$ both the $\pm \gamma z$ solutions are admissible. To show this, we shove an assumed solution $\vec{E} = E(z) \hat{x}$, $\vec{B} = B(z) \hat{y}$ into Maxwell's equations and then find we can reduce everything to ordinary differential equations with $z$ as the independent variable of the form $\mathrm{d}_z^2 = \gamma^2$ where:
$$\gamma^2 = \pm i\, \omega \, \mu (\sigma \pm i\, \omega \, \epsilon)$$
choosing the sign according to your phasor convention. Let's chose your convention, so we then have two solutions of $\gamma$:
$$\gamma = \pm i\,\,\omega\,\sqrt{\mu\,\epsilon} \sqrt{1-\frac{i\,\sigma}{\omega\,\mu}}$$
Notice that the thing in the rightmost square root can be wr\itten:
$$\sqrt{1 + \frac{\sigma^2}{\omega^2\,\mu^2}} \exp\left(-i\,\frac{1}{2} \arctan\left(\frac{\sigma}{\omega\,\mu}\right)\right)$$
Notice that the phase of this thing has the same sign as your convention. What this means is if the wave is running in the positive $z$ direction, its phase at any given time decreases with $z$, so we choose the minus sign for the expression for $\gamma$ - the phase is decreasing by approximately $k = \omega\,\sqrt{\mu\,\epsilon}$ radians per metre. So the real part of the propagation constant has the sign of $-i\,\times- i \times \arctan(\sigma/(\omega\mu))$, i.e. it is negative for positive (absorptive) conductivity. This is correct - a wave running in the $+z$ direction dwindles in amplitude with decreasing $z$ if the medium is passive. Likewise, if we choose the plus sign for $\gamma$, the phase is increasing with increasing $z$, so the wave is running in the $-z$ direction. In this case, the sign of the propagation constant's real part is positive for positive (absorptive) conductivity - the wave grows with increasing $z$ because we are moving towards the source in this direction. Note that the real part's sign would be all wrong if we chose the other sign for the complex electric constant. You can work through the same exercise if we choose the opposite (Photon Wave Function) phasor convention.


TITLE: Let $n\in\mathbb{N}$ and $z\in\mathbb{C}$ with $|z|=1$ and $z^{2n}\neq-1$. Prove that $\frac{z^n}{1+z^{2n}}\in\mathbb{R}$.
QUESTION [0 upvotes]: Let $n\in\mathbb{N}$ and $z\in\mathbb{C}$ with $|z|=1$ and $z^{2n}\neq-1$. Prove that $\frac{z^n}{1+z^{2n}}\in\mathbb{R}$.
So I know that $z^{2n}\neq{-1}$ implies that the denominator cannot be zero.
Not sure how to prove this. Any and all help is appreciated, thanks!

REPLY [1 votes]: If $|z|=1$, there exists $\theta \in \mathbb{R}$ such that $z=e^{i\theta}$. So
$$\frac{z^n}{1+z^{2n}}=\frac{e^{in\theta}}{1+e^{2in\theta}} = \frac{e^{in\theta}}{e^{in\theta}\left( e^{-in\theta} + e^{in\theta}\right)} = \frac{1}{ e^{-in\theta} + e^{in\theta}} = \frac{1}{2\cos(n\theta)} \in \mathbb{R}$$