$$Q_X(z) = \frac{e^{-|z|^2}}{\pi} d\lambda(z)$$
$$w(z) = \frac{1}{A+z}$$
$$z(w) = \frac 1w - A$$
$$\frac{dz}{dw} = \frac 1{w^2}$$
$$\frac{d\lambda(z)}{d\lambda(w)} = \frac 1{|w|^4}$$

You can compute the density function of the new distribution by Integration by Substitution/Change of Variables:
$$Q_Y(w) = \frac{e^{-|z(w)|^2}}{\pi} \cdot \frac {d\lambda(z)}{d\lambda(w)} \cdot d\lambda(w) = \frac{e^{-\left|\frac 1w - A\right|^2}}{\pi} \cdot \frac 1{|w|^4} \cdot d\lambda(w)$$
$$Q_Y(w) = \frac{e^{-\left|\frac 1w - A\right|^2}}{\pi |w|^4} d\lambda(w)$$

$Y$'s mean is always well-defined, but its 2nd moment is never finite. In fact, its $t$'th absolute moment is finite for $t < 2$ and infinite for $t \ge 2$.

More generally, if $X: \Omega \to \mathbb C$ is a 2D random variable with a density function $\mu_X = \frac {dQ_x}{d\lambda}$ such that $\liminf_{z\to A} \mu_X(z) > 0$ and $\limsup_{z\to A} \mu_X(z) < \infty$ (for example, if $\mu_X$ is continuous and nonzero in $A$), then $Y = \frac{1}{X-A}$ will have finite $t$'th absolute moments for $t < 2$ and infinite for $t \ge 2$.

Let's say that for $|z-A|<\delta$, the density function $\mu_X(z)$ satisfies
$0 < \varepsilon < \mu_X(z) < K < \infty$. First, if $t < 2$:
$$\mathbb E(|Y|^t) = \int |w|^t dQ_Y(w) =
\int {|w|^t \cdot \frac{\mu_X(z(w))}{|w|^4} d\lambda(w)} =
\int_{r=0}^\infty {\int_{\theta=0}^{2\pi} {r^t \cdot \frac{\mu_X(z(re^{i\theta}))}{r^4} r d\theta dr} } \le
1\cdot \frac 1{\delta^t} + \int_{r=\frac 1\delta}^\infty {\int_{\theta=0}^{2\pi} {r^t \cdot \frac{K}{r^4} r d\theta dr} }=
\frac 1{\delta^t} + 2\pi\cdot K\cdot \int_{r=\frac 1\delta}^\infty {r^{t-3}dr} < \infty
;$$
however, when $t \ge 2$:
$$\mathbb E(|Y|^t) = \dots \ge
\int_{r=\frac 1\delta}^\infty {\int_{\theta=0}^{2\pi} {r^t \cdot \frac{\varepsilon}{r^4} r d\theta dr} }=
2\pi\cdot\varepsilon\cdot \int_{r=\frac 1\delta}^\infty {r^{t-3} dr} = \infty
.$$