  It is known that the improved Cotlar's inequality $B^{*}f(z) \le C M(Bf)(z)$,
$z\in\mathbb C$, holds for the Beurling transform $B$, the maximal Beurling
transform $B^{*}f(z)=$ $\displaystyle\sup_{\varepsilon
>0}\left|\int_{|w|>\varepsilon}f(z-w) \frac{1}{w^2} \,dw\right|$, $z\in\mathbb
C$, and the Hardy--Littlewood maximal operator $M$. In this note we consider
the maximal Beurling transform associated with squares, namely,
$B^{*}_Sf(z)=\displaystyle\sup_{\varepsilon >0}\left|\int_{w\notin
Q(0,\varepsilon)}f(z-w) \frac{1}{w^2} \,dw \right|$, $z\in\mathbb C$,
$Q(0,\varepsilon)$ being the square with sides parallel to the coordinate axis
of side length $\varepsilon$. We prove that $B_{S}^{*}f(z) \le C M^2(Bf)(z)$,
$z\in\mathbb C$, where $M^2=M \circ M$ is the iteration of the
Hardy--Littlewood maximal operator, and $M^2$ cannot be replaced by $M$.
