Using the equations $|z_{1}|^{2}=z_{1}\bar{z}_{1}$ and $|z_{2}|^{2}=z_{2}\bar{z}_{2}$ to rewrite these expressions gives the first case of (25), the third case when $z_{2}\neq 0$, and the fourth case when $z_{1}\neq 0$. If $z_{1}z_{2}=0$ on the other hand, in each of parts (i)–(iii) of Definition 5 we have $|z_{3}-c|^{2}=0$, so $c=z_{3}$, giving the second case of (25), the third case when $z_{2}=0$, and the fourth case when $z_{1}=0$.

So, if $(z_{1},z_{2},z_{3})\in N_{a,c}$ then we can recover $a$ and $c$ from $(z_{1},z_{2},z_{3})$ as in the theorem. Conversely, for any $(z_{1},z_{2},z_{3})$ in $\mathbin{\mathbb{C}}^{3}$, defining $a,c$ by (24)–(25) and reversing the proof above, we find that $(z_{1},z_{2},z_{3})\in N_{a,c}$. Hence $f^{-1}(a,\mathop{\rm Re}c,\mathop{\rm Im}c)=N_{a,c}$, and $f$ is a special Lagrangian fibration of $\mathbin{\mathbb{C}}^{3}$. $\square$