Bounded operator on $L^2(\Bbb R^2)$ Let $\lambda\in \{z\in\Bbb C\mid \text{Re}(z)>0\quad \text{and}\quad \text{Im}(z)>0 \}$. Consider the operator $T_\lambda: L^2(\Bbb C)\to L^2(\Bbb C)$:
$$
f\mapsto T_\lambda(f)(z)=\int_{\Bbb C}f(z-w) \psi\left(  \frac{\lambda-\mu}{2\lambda},1,\frac{\lambda}{2}(|w|^2)\right)e^{-\frac{\lambda}{4}\ (|w|^2\ ) - i\frac{\lambda}{2} \text{Im}(z\overline{w})}dw
$$
where  $dw=dxdy$ is the usual measure  Lebesgue on $\Bbb C$.
How to prove que $T_\lambda(f)\in L^2(\Bbb C)$ ?
Here: we have $\psi\left(  a,c,z\right)\sim z^{-a}$ as $z\to\infty$
$\psi$ is the Kummer funtion noted sometimes $U(a,b,c)$ (https://en.wikipedia.org/wiki/Confluent_hypergeometric_function)
Thanks
 A: I will only indicate the standard procedure to deal with such questions and leave the details to you.
First, showing that this function is in $L^2(\mathbb{R}^2)$ of course consists in estimating the $L^2$-norm,
$$ ||T_\lambda f||^2\leq \int \int |f(z-w) \psi(...) \exp(...)|^2 dw dz =\int \int |I(z,w)|^2 dw dz, $$
where $I(z,w)$ is the integrand. One then usually tries to "factorize" the function $I(z,w)$, i.e. $|I(z,w)|\leq I_1(z) I_2(w)$ with $I_1,I_2\in L^2(\mathbb{R})$. For example, one standard trick is to shift the argument $z\mapsto z+w$ so that $f$ only depends on $z$. This usually works quite well in combination with the exponential function whose argument is a bilinear form for obvious reasons. Regarding $\psi$ the asymptotic $\psi(a,c,z)\sim z^{-a}$ implies the existence of a constant $C>0$ such that for $z$ large enough, $$ |\psi(a,c,z)|\leq C z^{-a}, $$ which shows that $\psi$ decreases quickly (if $a>0$, $z>0$ real) and hence should be no big problem.
I also would recommend to start with only considering $\lambda>0$ real so that one can at least guarantee that $\exp(-\lambda ||w||^2)$ falls exponentially fast and $|\exp(i\lambda r)|=1$ for any real number $r$ and go from there. Also, if you are not able to get finite estimates for your integral there is a good chance that what you have written down is not an $L^2$-function.
