= Solution
Write $\omega=u+iv$, with $v>0$, and a chosen square root $k_\beta=a+ib$. Squaring gives
$$
a^2-b^2=\frac{u^2-v^2}{\beta^2}-k^2,\qquad ab=\frac{uv}{\beta^2}.
$$
One cannot have $b=0$: the second relation would force $u=0$, and then the first would require the square of a real number to be negative. Consequently
$$
\operatorname{Re}p_\beta
=\operatorname{Re}\frac{a+ib}{u+iv}
=\frac{au+bv}{|\omega|^2}
=\frac{v}{|\omega|^2}\left(b+\frac{u^2}{\beta^2b}\right).
$$
The bracket has the sign of $b$, proving the <causal vertical wavenumber branch> equivalence
$$
\boxed{\operatorname{Im}k_\beta>0\quad\Longleftrightarrow\quad\operatorname{Re}p_\beta>0.}
$$
This selects the square root analytically in the upper <frequency> half-plane. At $\omega=iv$ it is $k_\beta=i\sqrt{v^2/\beta^2+k^2}$; continuation supplies the outgoing branch for other <frequencies>. The expression $\sqrt{\beta^{-2}-p^2}$ must use this inherited branch rather than an independently chosen square root. For positive real <frequency> its limit is positive real for a propagating wave and positive imaginary for an evanescent wave.
Back to article page
