algebra precalculus - Is $|x| cdot |x| = |x^2| = x^2$?

Is $|x| \cdot |x| = |x^2| = x^2$ ?

I'm very sorry if this question is a duplicate but I couldn't find anything about it (most likely because it's wrong..). But I'm not sure if this is correct so I need to ask you.

$$|x| \cdot |x| = |x^2| \text{ should be alright}$$

Now my confusion starts. $x^2$ should be positive / neutral for any value. That would mean we can ignore the absolute value sign? On the other hand we could have that $|-x^2|$. But that would be a different thing than $|x^2|$, they are not equal to each other...? Please help me if I do this little thing wrong the entire task will be wrong. I got some thinking error here..

When there is the same question (I couldn't find one), please link me to it and I will delete this one immediately.

Answer

You are thinking it too hard. You could just look at the definition of the absolute value
$$
|x|:=\begin{cases}
x,&x\geq 0\\
-x,&x<0
\end{cases}

$$
and check on your own that $|x|^2=|x^2|=x^2$.

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In general, we have $|a|\cdot|b|=|ab|$, which is true also for complex numbers; but the identity $|x^2|=x^2$ is not necessarily true in the complex world.