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The problem here is to solve the following differential equation $$2\partial_{\bar{z}} v(z,\bar{z})=[v(z,\bar{z})]^*,$$ where the solution $v(z,\bar{z})$ is required to be anti-periodic around a set of (fixed) branch points $z_1,z_2..z_n$ (i.e. $v(z,\bar{z})$ picks up a minus sign when $z$ goes around any one of these points $z_j$, for example, the function $\sqrt{(z-z_1)(z-z_2)..(z-z_n)}$ has such branch points).

For example, with $n=2$ branch points $z_1$ and $z_2=\infty$, it can be easily checked that there are two independent solutions $$v_1=i\frac{e^{-\sqrt{(z-z_1)(\bar{z}-\bar{z}_1)}}}{\sqrt{z-z_1}},~v_2=\frac{e^{\sqrt{(z-z_1)(\bar{z}-\bar{z}_1)}}}{\sqrt{z-z_1}},$$ with $v_1$ localized at $z_1$ and $v_2$ localized at $\infty$.

Now, how to generalize the above solution to two arbitrary branch points $z_1,z_2$, or even $2n$ branch points $z_1,z_2...z_{2n}$? In the latter case, it is expected that there are $2n$ independent solutions $v_1..v_{2n}$, with $v_j$ localized near $z_j$.

$\bf{Edit}$: I just realized that if $v_0(z,\bar{z})$ is a solution, then $v_m=(\partial^m_z+\partial^m_{\bar{z}})v_0$ for $m\in \mathbb{N}_{+}$ are also solutions with exactly the same branch points. (Notice that since $4\partial_z\partial_{\bar{z}}v_0=v_0$, solutions like $(\partial_z+\partial_{\bar{z}})^m$v_0 are linear combinations of $v_0..v_m$.) So the question then reduces to finding a "basic" solution $v_0(z,\bar{z})$ with two branch points at $z_1,z_2$.

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