Normal square root from Borel functional calculus ($S^2=T,S^{\ast }S=S{S}^{\ast }$)

A bounded normal operator admits a normal square root even when no continuous scalar square root exists on its Banach algebra spectrum. Choose a bounded Borel branch $s(z)=|z{|}^{1/2}\exp (i\mathrm{Arg}(z)/2)$, with $s(0)=0$. The Borel functional calculus for a normal operator gives $S=s(T)$; multiplicativity gives $S^2=T$ and preservation of the adjoint gives $S^{\ast }S=S{S}^{\ast }=|s{|}^2(T)$. Unlike a positive square root, this square root need not be unique.