> Because square root is multivalued.
Even so, I would consider this to be wrong for
a) I don't want my students to think it is true and
b) the left hand side is two-valued while the right hand
side is four-valued and hence they do not agree as
multi-valued functions. (This objection does not cause
a problem in your two examples.)
> Consider sqrt(1-z)*sqrt(1+z)=sqrt(1-z^2) and
sqrt(z-1)*sqrt(z+1)=sqrt(z^2-1)
>
>
>
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