= Solution
Let $A_0=f_0(0)$, $\delta=v_2-v_1$, $V=v_0+\delta$ and $K=k^2\lambda_{\mathrm{De}}^2$. Use the <Debye length> convention of the paper, $\lambda_{\mathrm{De}}=v_0/\omega_{\mathrm{pe}}$. On the candidate intervals in <2/a>, the <Landau contour> contributes no imaginary part. The piecewise constant slopes give
$$
\begin{aligned}
I(u)&=\int\frac{f_0'(v)}{v-u}\,dv\\
&=\frac{A_0}{v_0}\left[\int_{-v_0}^{0}\frac{dv}{v-u}-\int_0^{v_1}\frac{dv}{v-u}-\int_{v_2}^{V}\frac{dv}{v-u}\right]\\
&=\frac{A_0}{v_0}\left[2\ln|u|-\ln|v_0+u|-\ln|v_1-u|-\ln|V-u|+\ln|v_2-u|\right].
\end{aligned}
$$
Since $n_0=v_0A_0$, setting the <dielectric function> to zero yields
$$
\boxed{\ln\frac{u^2|v_2-u|}{|v_0+u|\,|v_1-u|\,|V-u|}=K.}
$$
The absolute values express the real response on the allowed intervals. Solving this real equation inside a sloping portion of $f_0$ would not satisfy the imaginary part of the <dielectric function>.
An exact algebraic form will also make the number of undamped branches clear. Define
$$
A=v_0^2+\delta(v_0-v_1),\qquad B=v_0v_1V,\qquad q=1-e^{-K}.
$$
On each allowed interval, the quantity inside the logarithm equals $u^2(u-v_2)/[(u+v_0)(u-v_1)(u-V)]$. Exponentiating and using
$$
(u+v_0)(u-v_1)(u-V)=u^2(u-v_2)-Au+B
$$
gives the <cubic equation>
$$
q\,u^2(u-v_2)-Au+B=0.
$$
For every $k>0$, the logarithm runs from $0$ to $+\infty$ on $(-\infty,-v_0)$, from $+\infty$ to $-\infty$ on $(v_1,v_2)$, and from $+\infty$ to $0$ on $(V,\infty)$. The <intermediate value theorem> supplies a root in each interval, and the <cubic equation> permits at most three. Thus there is exactly one undamped branch in each interval.
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