= Solution
In vacuum the <Maxwell equations> are
$$
\nabla\mathbin{\cdot}E=0,
\qquad \nabla\mathbin{\cdot}B=0,
\qquad \nabla\times E=-\frac{\partial B}{\partial t},
\qquad \nabla\times B=\frac1{c^2}\frac{\partial E}{\partial t}.
$$
Substitution of the complex <plane electromagnetic wave> amplitudes into these equations gives
$$
k\mathbin{\cdot}E_0=0,
\qquad
B_0=\frac{k\times E_0}{\omega},
\qquad
\omega^2=c^2|k|^2.
$$
Thus the electric field must be <transverse polarization>[transverse] to $k$; taking the positive-frequency branch gives $\omega=c|k|$.
For $E_0=E(1,i,0)$, $k=k\widehat z$, and $\varphi=kz-\omega t$, the real fields are
$$
E(x,t)=E(\cos\varphi,-\sin\varphi,0),
\qquad
B(x,t)=\frac Ec(\sin\varphi,\cos\varphi,0).
$$
The electric vector rotates with constant magnitude, so this is <circular polarization>. Its <Poynting vector> is
$$
\boxed{S=\frac1{\mu_0}E\times B
=\frac{E^2}{\mu_0c}\widehat z}.
$$
The Poynting vector is the electromagnetic <energy flux>: its direction is the direction of propagation and its magnitude is the power crossing unit area.
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