# Phasors and propagating modes in a waveguide

I'm currently studying antennas. The notes I'm following, while discussing the EM fields at the antenna feed port, make the following statement:

It can be shown that a propagating mode in a transmission line or a waveguide can be represented by real-valued phasor vectors e and h, known as modal vectors

Following this statement, the Poynting vector is computed as: $$\bar{S} =\frac{1}{2}\int\int_S(\textbf{e}\times \textbf{h})\cdot d\textbf{S}$$ Where the conjugation of the h field is skipped since, as before stated, the "phasor is real-valued".

My doubt comes exactly from this, as I've always considered phasors as complex numbers, the note's statement is very new to me. Can someone please explain this a little further, or maybe point me in the direction of resources to better understand this?

-Thank you!

In a homogeneous waveguide, both propagating TE and TM waves, modes, can be derived from the scalar Helmholtz equation. Let $$\partial \mathcal B =0$$ denote the metal boundary of the cross section. Denoting the wavenumber in the guide by $$\beta$$ and $$\kappa_c^2 +\beta ^2 = k_0^2$$, then the harmonic wave propagating as $$e^{-\mathfrak j \beta z}$$ along the $$z$$ axis satisfies $$\nabla_{x,y}^2 f(x,y) + \kappa_c^2 f(x,y) = 0$$
For a TE wave $$\nabla^2 h_z(x,y) + \kappa_c^2 h_z(x,y) = 0 \space \mathrm{and} \space \frac{\partial h_z}{\partial n}|_{\partial \mathcal B} =0 \\ \mathbf e_z=0 \\ \mathbf h_t = -\frac{\mathfrak j \beta}{\kappa_c}\nabla h_z \space e^{-\mathfrak j \beta z}\\ \mathbf e_t=-\frac{k_0}{\beta}\sqrt{\frac{\mu_o}{\epsilon_0}} \hat z \times \mathbf h_t \space e^{-\mathfrak j \beta z}\tag{1}$$
For a TM wave $$\nabla^2 e_z(x,y) + \kappa_c^2 e_z(x,y) = 0 \space \mathrm{and} \space e_z|_{\partial \mathcal B} =0\\ \mathbf h_z=0 \\ \mathbf e_t= -\frac{\mathfrak j \beta}{\kappa_c^2}\nabla e_z \space e^{-\mathfrak j \beta z}\\ \mathbf h_t = \frac{k_0}{\beta}\sqrt{\frac{\epsilon_0}{\mu_o}} \hat z \times \mathbf e_t \space e^{-\mathfrak j \beta z} \tag{2}$$
Now notice that both $$h_z$$ for a TE and $$e_z$$ for a TM mode are real functions, therefore so are their respective gradients. Since the propagation factor $$e^{-\mathfrak j \beta z}$$ is common to all components in both TE and TM the respective transversal vector fields, $$\mathbf e_t$$ and $$\mathbf h_t$$, are in quadrature. Since the $$\mathbf e_t \perp \mathbf h_t \perp \hat z$$, you also have $$dS_z =(\mathbf e \times \mathbf h)\cdot d\mathbf a =(\mathbf e_t \times \mathbf h_t) \cdot d\mathbf a$$, where $$d\mathbf a = \hat z da$$ is the cross sectional normal pointing along the axis, thus all "pieces" of each mode are in phase.