= Solution
For an odd <velocity potential>, choose
$$
\Phi_I=C\sinh(kz),\qquad
\Phi_O(z)=\begin{cases}Ae^{-k(z-L)},&z>L,\\-Ae^{k(z+L)},&z<-L.\end{cases}
$$
At $z=L$ the two kinematic conditions give $A=icB_+$ and $C\cosh(kL)=i(V-c)B_+$. The dynamic condition is $-cA=(V-c)C\sinh(kL)$. Eliminating $A,C,B_+$ gives, with $s=\tanh(kL)$,
$$
\boxed{c^2=-s(V-c)^2,\qquad
c=\frac{Vs}{1+s}\pm i\frac{|V|\sqrt s}{1+s}.}
$$
The lower interface gives the same equation by odd parity of the <velocity potential>. The interior vertical <velocity> is even, so $B_-=B_+$: the displacement is a <sinuous mode of a planar jet> even though the potential is odd.
For $L>0$, $V\ne0$ and $k>0$, $0<s<1$ and one root has $c_i>0$. Thus \b[every nonzero <wavenumber> is unstable] in this ideal <top-hat planar jet>. The phrase “all choices” is understood with these assumptions: $k=0$ has no exponential wave growth, and $V=0$ has no shear.
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