Expand superspace function into component form In 2D (1,1) superconformal field theory, the invariant "distance" between two points  $Z_1=(z_1,\theta_1)$ and $Z_1=(z_1,\theta_1)$ in superspace is $$Z_{12}=z_1-z_2-\theta_1\theta_2.$$ 
My question is how to compute the quantities when the $\theta$ appears in denominator or in square root, for example, how to expand $\frac{1}{Z_{12}}$ into component form? 
And how to expand 
\begin{equation}
\sqrt{1-\frac{Z_{12}Z_{34}}{Z_{13}Z_{24}}}
\end{equation}
into component form?
 A: TL;DR: It is defined by expanding in the finite Taylor series of the $\theta$s.
More details:

*

*Recall that a supernumber $z=z_B+z_S$ has a body $z_B\in\mathbb{C}$, which is a complex number; and a soul $z_S$, which belongs to the ideal generated by Grassmann-odd generators.


*For any analytic function $f$, the supernumber $f(z)$ is defined via its formal Taylor series around the body $z_B$.
In particular, for the reciprocal function $f(z)=z^{-1}$, this prescription only make sense if the body $z_B\neq 0$ is non-zero.


*Point 2 can be generalized, so that if supernumbers $z,w$ share the same body $z_B=w_B$, then $f(w)$ may be defined as a formal Taylor series around the other supernumber $z$.
In particular for a superfield $\Phi(\theta)=\phi_0+{\cal O}(\theta)$ and its lowest component field $\phi_0$, they share the the same body $\Phi(\theta)_B=\phi_{0B}$, so that $f(\Phi(\theta))$ may be defined as a Taylor series (which happens to be finite!) around $\phi_0$.


*This generalizes in a straightforward way to several superfields.
