In §6.8 of Ziman's "Principles of the Theory of Solids" he derives the imaginary component of the wave vector of an electron inside an energy gap (due to action of an electric field). He starts by writing the Schrödinger equation $$ -\frac{\hbar^{2}}{2m}\nabla^{2}\psi+\left\{ \mathcal{V}\left(\mathbf{r}\right)-\mathcal{E}\right\} \psi=0 $$ where the potential $\mathcal{V}$ is periodic, i.e., $\mathcal{V}\left(\mathbf{r}+\mathbf{l}\right) = \mathcal{V} \left(\mathbf{r}\right)$. He then plugs in a solution of the form (Fourier series)
$$ \psi=\sum_{\mathbf{g}}\alpha_{\mathbf{k}-\mathbf{g}}e^{i\left(\mathbf{k}-\mathbf{g}\right)\cdot\mathbf{r}} $$
where the components of $\mathbf{k}$ are complex. He then restricts the analysis to a one-dimensional model, where $\mathbf{k}$ is in the neighborhood of the zone boundary corresponding to the wave vector $\frac{1}{2} G = \frac{\pi}{a}$ for planes spaced $a$ apart transverse to the electric field. Using perturbative analysis one arrives at the secular equation
$$ \left\{ \frac{\hbar^{2}}{2m}k^{2}-\mathcal{E}\right\} \left\{ \frac{\hbar^{2}}{2m}\left(k-G\right)^{2}-\mathcal{E}\right\} =\left|\mathcal{V}_{G}\right|^{2} \tag{1} $$
If $k$ is real, this equation has no roots for $\mathcal{E}$ in the energy gap $\left(\mathcal{E}^{0}-\left|V_{G}\right|,\mathcal{E}^{0}+\left|V_{G}\right|\right)$
He then proposes to treat $\mathcal{E}$ as an arbitrary parameter and solve for $k$. Writing $$ k=\frac{1}{2} G + \kappa, \quad \mathcal{E}=\mathcal{E}^0 + \varepsilon $$ and assuming $\kappa$ and $\varepsilon$ are small, it is presumably possible to approximate the solution as
$$ \kappa^{2}\approx\frac{2m}{\hbar^{2}}\left[\frac{\varepsilon^{2}-\left|\mathcal{V}_{G}\right|^{2}}{4\mathcal{E}^{0}}\right] \tag{2} $$
However, I can't seem to derive $(2)$ from $(1)$, despite the apparent simplicity of these expressions. I'm not sure what kind of approximation Ziman is doing here. In the naive approach, I simply expanded $(1)$ and got a quadratic equation in $\kappa^2$:
$$ \frac{\hbar^{4}}{4m^{2}}\kappa^{4}-\left(\frac{\hbar^{2}\varepsilon}{m}+\frac{\hbar^{2}\mathcal{E}^{0}}{m}+\frac{G^{2}\hbar^{4}}{8m^{2}}\right)\kappa^{2}-\frac{G^{2}\hbar^{2}\varepsilon}{4m}-\frac{G^{2}\hbar^{2}\mathcal{E}^{0}}{4m}+\frac{G^{4}\hbar^{4}}{64m^{2}}+\left(\mathcal{E}^{0}+\varepsilon\right)^{2}-\left|V_{G}\right|^{2}=0 $$
As usual, we get two roots:
$$ \kappa^2 = \frac{\frac{\hbar^{2}\varepsilon}{m}+\frac{\hbar^{2}\mathcal{E}^{0}}{m}+\frac{G^{2}\hbar^{4}}{8m^{2}}\pm\sqrt{\left(\frac{\hbar^{2}\varepsilon}{m}+\frac{\hbar^{2}\mathcal{E}^{0}}{m}+\frac{G^{2}\hbar^{4}}{8m^{2}}\right)^{2}-\frac{\hbar^{4}}{m^{2}}\left(-\frac{G^{2}\hbar^{2}\varepsilon}{4m}-\frac{G^{2}\hbar^{2}\mathcal{E}^{0}}{4m}+\frac{G^{4}\hbar^{4}}{64m^{2}}+\left(\mathcal{E}^{0}+\varepsilon\right)^{2}-\left|V_{G}\right|^{2}\right)}}{\frac{\hbar^{4}}{2m^{2}}} $$ Every approximation that I tried here didn't yield $(2)$. Any suggestions?
Edit (in response to a comment by @Sebastian Riese): my initial approach consisted of neglecting the vanishingly small $\kappa^4$ term and dividing the equation by $\mathcal{E}^0$. This leads to:
$$ -\left(\frac{\hbar^{2}\varepsilon}{m\mathcal{E}^{0}}+\frac{\hbar^{2}}{m}+\frac{G^{2}\hbar^{4}}{8m^{2}\mathcal{E}^{0}}\right)\kappa^{2}-\frac{G^{2}\hbar^{2}}{4m}\left(1+\frac{\varepsilon}{\mathcal{E}^{0}}\right)+\left(1+\frac{\varepsilon}{\mathcal{E}^{0}}\right)\left(\mathcal{E}^{0}+\varepsilon\right)+\frac{G^{4}\hbar^{4}}{64m^{2}\mathcal{E}^{0}}-\frac{1}{\mathcal{E}^{0}}\left|V_{G}\right|^{2}=0 $$
Since $\varepsilon \ll \mathcal{E}^0$, we get
$$ \left(\frac{\hbar^{2}}{m}+\frac{G^{2}\hbar^{4}}{8m^{2}\mathcal{E}^{0}}\right)\kappa^{2}=-\frac{G^{2}\hbar^{2}}{4m}+\frac{G^{4}\hbar^{4}}{64m^{2}\mathcal{E}^{0}}+\left(\mathcal{E}^{0}+\varepsilon\right)-\frac{\left|V_{G}\right|^{2}}{\mathcal{E}^{0}} $$
The desired result strongly suggests that the first two terms on the RHS should be thrown away, but I don't see any physical or mathematical justification for doing so. But even if we do throw them away, and even if we make approximations to the LHS, we still won't arrive at the correct result.
