Now let $\hat{D}$ be a $\mathbin{\rm U}(1)$-invariant holomorphic disc in $\mathbin{\mathbb{C}}^{3}$ with boundary in $N_{a,b,c}$, and let $(z_{1},z_{2},z_{3})\in\hat{D}$. We claim that $z_{1}=0$ or $z_{2}=0$. Suppose $z_{1},z_{2}\neq 0$. As $D$ contains the $\mathbin{\rm U}(1)$-orbit of $(z_{1},z_{2},z_{3})$ and is holomorphic, it must locally contain the orbit of $(z_{1},z_{2},z_{3})$ under the complexification of the $\mathbin{\rm U}(1)$-action (52). Therefore, $D$ must locally be a subset of

$$\bigl\{(uz_{1},u^{-1}z_{2},z_{3}):u\in\mathbin{\mathbb{C}}\setminus\{0\}\bigr\}.$$

But there are no $\mathbin{\rm U}(1)$-invariant holomorphic discs in this set; the best one can do is an annulus. Thus one of $z_{1},z_{2}$ must be zero.

Let $(z_{1},z_{2},z_{3})$ be a point in the boundary of $\hat{D}$. Then $(z_{1},z_{2},z_{3})\in N_{a,b,c}$, so $|z_{1}|^{2}-|z_{2}|^{2}=a>0$. This implies that $z_{1}\neq 0$, so $z_{2}=0$. It is then easy to show that $\hat{D}$ must be $\bigl\{(z,0,z_{3}):|z|^{2}\leqslant a\bigr\}$. This agrees with (58) with $x=\mathop{\rm Re}(z_{3})$ and $c=\mathop{\rm Im}(z_{3})$, and $(z_{1},0,z_{3})\in N_{a,b,c}$ implies that $u_{a,b}(x,0)=0$. So all $\mathbin{\rm U}(1)$-invariant holomorphic discs $\hat{D}$ with boundary in $N_{a,b,c}$ are as in the proposition. For $N_{-a,b,c}$ the argument works in the same way. $\square$