We call a point $(x,0)$ in $U$ with $u(x,0)=0$ a singularity of the solution $u,v$. We call a singularity $(x,0)$ isolated if there exists $\epsilon>0$ such that the open disc $B_{\epsilon}(x,0)$ of radius $\epsilon$ about $(x,0)$ lies in $U$, and the only point $(x^{\prime},y^{\prime})$ in $B_{\epsilon}(x,0)$ with $u(x^{\prime},y^{\prime})=0$ and $v(x^{\prime},y^{\prime})=v(x,0)$ is $(x,0)$.

Let $(x,0)$ be an isolated singularity of $u,v$, and let $\epsilon$ be as above. Consider the map $\gamma:{\mathcal{S}}^{1}\rightarrow\mathbin{\mathbb{C}}$ given by

$$\gamma:{\rm e}^{i\theta}\mapsto u\bigl(x+{\textstyle\frac{1}{2}}\epsilon\cos\theta,{\textstyle\frac{1}{2}}\epsilon\sin\theta\bigr)-iv\bigl(x+{\textstyle\frac{1}{2}}\epsilon\cos\theta,{\textstyle\frac{1}{2}}\epsilon\sin\theta\bigr)+iv(x,0).$$

As $(x,0)$ is isolated we see that $\gamma$ is smooth and maps ${\mathcal{S}}^{1}\rightarrow\mathbin{\mathbb{C}}\setminus\{0\}$. Define the order of the isolated singularity $(x,0)$ to be the winding number of $\gamma$ about 0 in $\mathbin{\mathbb{C}}$. It is easy to show that the order is independent of $\epsilon$, provided $\epsilon>0$ is sufficiently small.