Let $\mu: \mathbb{C} \to D(0,1).$ A quasiconformal map is a $W^{1}_{2,loc}-$solution to the Beltrami equation $\bar{\partial}f = \mu \partial f$. In this paper, the authors remark that one can formally solve for $f$ in this equation by the Neumann series $$\bar{\partial}f = \mu + \mu T \mu + \mu T \mu T \mu + \cdots$$ where $T$ is the Beurling transform $$Tf(z) = \partial\left(\frac{\bar{\partial}f}{\partial z}\right)^{-1}(z) = -\frac{1}{\pi} p.v \int_{\mathbb{C}}\frac{f(w) \ \mathrm dw}{(z-w)^2}$$.

I cannot see how, even formally, this is true, and no source I can find shows how to do this. What is the heuristic behind this?

Let $\mu: \mathbb{C} \to D(0,1).$ A quasiconformal map is a $W^{1}_{2,loc}-$solution to the Beltrami equation $\bar{\partial}f = \mu \partial f$. In this paper, the authors remark that one can formally solve for $f$ in this equation by the Neumann series $$\bar{\partial}f = \mu + \mu T \mu + \mu T \mu T \mu + \cdots$$ where $T$ is the Beurling transform $$Tf(z) = \partial\left(\frac{\bar{\partial}f}{\partial z}\right)^{-1}(z) = -\frac{1}{\pi} p.v \int_{\mathbb{C}}\frac{f(w) \ \mathrm dw}{(z-w)^2}. $$ I cannot see how, even formally, this is true, and no source I can find shows how to do this. What is the heuristic behind this?