$\sqrt{1}= |1|$ by definition. This accounts for $\sqrt{(-1)(-1)}$ giving the principal square root which is 1, as the answer
However,this can also be written as $$\sqrt{(-1)(-1)}= \sqrt{-1}.\sqrt{-1}$$ $$\sqrt{(-1)(-1)}= i. i $$ (if we consider imaginary roots, in our analysis.) But by definition $$\sqrt{1}= |1| = 1$$ which gives us, $$\sqrt{1}= i^2= 1 $$ which cannot be since $$ i^2 = (-1)$$ by definition.
Where am I wrong here? My understanding is that normal arithmetic is applicable for imaginary numbers as they are for real numbers. Can someone clear this up? Maybe I go the definitions wrong. How exactly is an imaginary number defined? Thanks in advance.
Edit- $$\sqrt{1}= 1$$ as per the answer.
The expression$$\sqrt{ab}=\sqrt a\sqrt b$$is valid when $a,b\geqslant0$. You are trying to apply it in a situation in which $a,b<0$. Besides, you state that $\sqrt1=|1|$ “by definition”. No. By definition, $\sqrt1=1$ (that is $\sqrt1$ is the only non-negative real number whose square is $1$).