= Solution
The odd-potential branch uses $s=\tanh(kL)$, whereas the even-potential branch uses $1/s$. The <imaginary parts> are identical because
$$
\frac{\sqrt{1/s}}{1+1/s}=\frac{\sqrt s}{1+s}.
$$
Consequently the <equality of top-hat jet parity growth rates> gives
$$
\boxed{\gamma_{\mathrm{odd}}=\gamma_{\mathrm{even}}
=\frac{k|V|\sqrt{\tanh(kL)}}{1+\tanh(kL)}
=\frac{k|V|}{2}\sqrt{1-e^{-4kL}}.}
$$
Here $\gamma=kc_i$ is the positive <temporal growth rate>. The real <phase velocities> differ: $c_{r,\mathrm{odd}}=\tfrac V2(1-e^{-2kL})$ and $c_{r,\mathrm{even}}=\tfrac V2(1+e^{-2kL})$. Thus equal amplification does not imply equal propagation. As $kL\to\infty$, both approach the isolated-<vortex sheet> <phase velocity> $V/2$ and <growth rate> $k|V|/2$.
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