Aug 24, 2016 · 1 Indeed what you are proving is that in the complex numbers you don't have (in general) $$\sqrt {xy}=\sqrt {x}\sqrt {y}$$ Because you find a counterexample.
Aug 24, 2016 · 1 Indeed what you are proving is that in the complex numbers you don't have (in general) $$\sqrt {xy}=\sqrt {x}\sqrt {y}$$ Because you find a counterexample.
