To see the main ideas, let us focus on $\eta=0$ and let $\mu_{1},\mu_{2}\to 0$. By construction $d\log|z_{1}|=V^{1j}_{(1)}d\mu_{j}+\text{Re}(\beta_{1}d\eta).$ Restricted to $\eta=0$,

$$d\log|z_{1}|=(a_{11}+v^{11})d\mu_{1}+(a_{11}+v^{12})d\mu_{2}=d(a_{11}\mu_{1}+a_{12}\mu_{2})+\alpha_{1}d\mu_{1}+\alpha_{3}d(\mu_{1}-\mu_{2}),$$

hence

$$|z_{1}|=\text{const}\cdot e^{a_{11}\mu_{1}+a_{12}\mu_{2}}\exp\left(\int^{(\mu_{1},\mu_{2})}\alpha_{1}d\mu_{1}+\alpha_{3}d(\mu_{1}-\mu_{2})\right).$$

We need to show $|z_{1}|\to 0$ as $\mu_{1},\mu_{2}\to 0$, namely $\int\alpha_{1}d\mu_{1}+\alpha_{3}d(\mu_{1}-\mu_{2})\to-\infty.$ Now because $\alpha_{1},\alpha_{3}$ are positive, this integral viewed as a function of $\mu_{1}$ and $\mu_{1}-\mu_{2}$ is an increasing functions of both variables, so it suffices to show this integral decreases to $-\infty$ as $(\mu_{1},\mu_{2})\to 0$ along the ray $\mathfrak{D}_{2}$:

$$\begin{split}\int\alpha_{1}d\mu_{1}+\alpha_{3}d(\mu_{1}-\mu_{2})=&\log|\mu_{1}|\{\frac{1}{2}+\frac{1}{2\pi}\arctan(\frac{a_{12}}{\sqrt{A}})+\frac{1}{2\pi}\arctan(\frac{-a_{11}-a_{12}}{\sqrt{A}})\}\\ =&\log|\mu_{1}|\{\frac{1}{4}+\frac{1}{2\pi}\arctan(\frac{a_{12}+a_{22}}{\sqrt{A}})\}\to-\infty,\end{split}$$

where the first equality uses that the arctan functions are constant on the ray $\mathfrak{D}_{2}$, and the second equality is an elementary trignometric identity.

In the more general case of $\eta\neq 0$ the $\arctan$ factor would no longer be exactly constant, but one can still make $|z_{1}|$ arbitrarily small for sufficiently small $|\eta|,\mu_{1},\mu_{2}$. The cases of $z_{0}$ and $z_{2}$ are completely analogous. ∎