Simple square and square root problem gives different result depending on how it is solved? 1st problem: $(\sqrt{-4})^2$
1st method: $(\sqrt{-4})^2=(2i)^2=-4$
2nd method: $(\sqrt{-4})^2=-4$ (square and square root remove each other)
Different methods were used to complete the same problem and the results match as they should.
2nd problem: $\sqrt{(-4)^2}$
1st method: $\sqrt{(-4)^2} = -4$ (again square and square root remove each other)
2nd method: $\sqrt{(-4)^2} = \sqrt16 = 4$ (now we start with the square)
Here completing the same problem with different methods gives different results. This doesn't make sense. I must be missing something or doing something wrong. What is it?

The problem is the interpretation of the symbol $\sqrt{\phantom{X}}$. Each non-zero real or complex number has two square roots which differ by the factor $-1$. Only for non-negative real numbers $x$ there is a standard interpreaton of $\sqrt x$: One takes the non-negative square root. For negative real numbers or for non-real complex numbers $z$ there is no such standard. You could define $\sqrt z$ to be the complex number with positive real part, but as I said: It is no commonly accepted standard.
Anyway, whatever your interpretation of $\sqrt z$ may be, you always get
$$(\sqrt z)^2 = z .$$
This simply reflects the definition that a square root $w$ of $z$ is characterized by $w^2 = z$.
What about $\sqrt{w^2}$? You claim that we always get $\sqrt{w^2} = w$. But this is not true. If you give a universal interpretation to $\sqrt z$ (e.g. as above) and do not make an individual choice, you must admit that sometimes we get $\sqrt{w^2} = w$ and sometimes $\sqrt{w^2} = -w$. The latter happens certainly if $w < 0$. In fact, for real $w$ we get $\sqrt{w^2} = \lvert w \rvert$ since we take the non-negative square root.
