to $G_\mu^j$ on $D_j$ our first instinct is to take a partition of unity and paste these together. In doing this we first add to the domains $D_j$ to generate overlap. Specifically, let us define $$\tilde D_0 = \bigg(\mathbb D\setminus\bigcup_1^4D_j\bigg)\cup\bigcup_{z_j\in S}B\left(z_j,\frac{1-|z_j|}{4A^2}\right)$$ and $$\tilde D_j = D_j\cup\bigcup_{z_k\in S\cap\gamma_j}B\left(z_j,\frac{1-|z_j|}{4A^2}\right)$$ for $1\le j\le4$. For ease of notation in what follows, we will denote the region of overlap, $\bigcup_1^4(\tilde D_j\cap\tilde D_0)$, by $U$. Let $\psi_0,\ldots,\psi_4$ be a partition of unity subordinate to $\tilde D_0,\ldots,\tilde D_4$ satisfying $|\nabla\psi_j(z)|\lesssim (1-|z|)^{-1}$. Our initial pasting is then $$\tilde g_\mu:=\psi_0g_\mu+\sum_{j=1}^4\psi_jG_\mu^j.$$

To obtain a bounded analytic solution from this we now take an approach much like that above. In particular, we wish to find functions $a_{\mu\nu}\in L^\infty(\mathbb D)$ such that $$\bar\partial a_{\mu\nu}=\tilde g_\mu\bar\partial\tilde g_\nu$$ and $$|a_{\mu\nu}(z)|\lesssim n^{-6}\left(\frac{d(z,\{\zeta_1,\zeta_2\})}{{{\rm{diam}}}(L_j)}\right)^2,$$ for $z\in L_j$, where $L_j$ is the lens domain bounded by $E_j$ and the circular arc meeting the endpoints $\zeta_1(j),\zeta_2(j)$ of $E_j$ in angle $\tilde\alpha<\frac\pi2$.

\[lem3\] If $B\in L^\infty(U)$ and $b(z)=\frac{B(z)}{1-|z|}\chi_U$ then there is $F\in L^\infty(\mathbb D)$ such that $$\bar\partial F=b$$ in the sense of distributions on $\mathbb D$, $$\Vert F\Vert_\infty\lesssim\Vert B\Vert_\infty,$$ and $$|F(z)|\lesssim n^{-6}\left(\frac{d(z,\{\zeta_1,\zeta_2\})}{{{\rm{diam}}}(L_j)}\right)^2,\qquad z\in L_j,$$ where $L_j$ is the lens domain bounded by $E_j$ and the circular arc meeting the endpoints 