 Solution

Write $\omega =u+iv$, with $v>0$, and a chosen square root $k_{\beta }=a+ib$. Squaring givesOne 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. ConsequentlyThe bracket has the sign of $b$, proving the causal vertical wavenumber branch equivalenceThis 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.