Proof. Suppose $(z_{1},z_{2},z_{3})$ lies in $N_{a,b,c}\cap N_{a^{\prime},b^{\prime},c^{\prime}}$. Then $a=|z_{1}|^{2}-|z_{2}|^{2}=a^{\prime}$, so $a=a^{\prime}$. Let $x=\mathop{\rm Re}(z_{3})$ and $y=\mathop{\rm Im}(z_{1}z_{2})$. Then (54) gives

$$\mathop{\rm Re}(z_{1}z_{2})=u_{a,b}(x,y)=u_{a^{\prime},b^{\prime}}(x,y),\quad\mathop{\rm Im}(z_{3})=v_{a,b}(x,y)+c=v_{a^{\prime},b^{\prime}}(x,y)+c^{\prime}.$$

As $a=a^{\prime}$ the first equation gives $u_{a,b}(x,y)=u_{a,b^{\prime}}(x,y)$, and part (ii) of Assumption 7.1 shows that $b=b^{\prime}$. The second equation then becomes $v_{a,b}(x,y)+c=v_{a,b}(x,y)+c^{\prime}$, so $c=c^{\prime}$. $\square$