# EM wave polarization check

I am trying to recall the quick way to check the EM polarization theoretically, I mean if I have a wave described as (ssume capital letters are vectorial fields)

$$\ E = E_o *exp(j* \varphi)\\ \ E_o = E_r +j*E_i\\$$

I recall that there were an easy math check to verify the kind of polarization that the EM wave had, however I am not completely sure about whether the vector/cross products are as showed below or not, could anyone remind me a bit?

Lineal polarization

$$\ \require {cancel} E_r = 0 \ or \ E_i= 0 \ \wedge \ \cancel {E_r \otimes E_i=0 } \$$

Circular polarization $$\ \mid E_r\mid =\mid E_i\mid \ \wedge \ \cancel { E_r\otimes E_i=1 } \$$

Elliptical polarization $$\ None \ of \ the\ conditions\ above$$

Edited with correct answer given by simplicis veritatis, and a own example since thus I can remember it even better.

Lineal polarization

$$\ E_r = 0 \ or \ E_i= 0 \ \wedge \ E_r\centerdot E_i\neq0 \$$

Circular polarization $$\ \mid E_r\mid =\mid E_i\mid \ \wedge \ E_r\centerdot E_i=0 \$$

Elliptical polarization $$\ None \ of \ the\ conditions\ above$$

Example $$\ E = (\hat{x}+\hat{y}-j\hat{y} ) *exp(j* \varphi)\\ \ E_o = \hat{x}+\hat{y} +j*-\hat{y} = E_r +j*E_i\\ \ \mid E_r\mid = \sqrt{2} \\ \ \mid E_i\mid = 1 \\ \ \mid E_i\mid \neq \mid E_r\mid \\ \ E_r\centerdot E_i =1-j \neq 0 \Rightarrow Elliptical Polarization$$

2. The wave is circularly polarized, if phases are rotated by $90^0$ degrees with respect to each other and the amplitudes are equal. (perpendicular vectors: zero dot product)
3. The wave is elliptically polarized, if phases are rotated by $90^0$ degrees with respect to each other and the amplitudes are not equal.