= Normal square root from Borel functional calculus
{title2=$S^2=T,\quad S^*S=SS^*$}
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 $s(z)=|z|^{1/2}\exp(i\operatorname{Arg}(z)/2)$, 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 $S^*S=SS^*=|s|^2(T)$. Unlike a positive square root, this square root need not be unique.
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