TITLE: Why does $\sqrt{i^4} \neq i^2$.
QUESTION [5 upvotes]: I was looking at a problem $\sqrt{x}=-3$, and I had at first thought $x=9 i^4$ was a solution. ($\sqrt{9 i^4}=3i^2=-3$)
Though I then realized that this would cause some problems.
For example using this, we would have $\sqrt{i^4}=i^2=-1$. While on the other hand $\sqrt{i^4}=\sqrt{1}=1$. 
I checked Wolfram and it says that $\sqrt{i^4} \neq i^2$ (also $(i^4)^{1/2}$). Could any one explain to me why we can't treat the exponents of $i$ this way?
Is it possible to algebraically show that $\sqrt{x}=-3$ has no solutions?
I am trying to learn some complex analysis and this made me realize that I might have some really bad intuition on complex numbers.

REPLY [4 votes]: It's the same reason as why $\sqrt{(-1)^2}\neq -1$, no need to involve complex numbers in this case (to complicate our lives). When we consider square root as function $\sqrt{\,\cdot\,}\colon\mathbb R_{\geq 0}\to \mathbb R_{\geq 0}$, we define it as a function with property $(\sqrt x)^2=x$. On the other hand, $\sqrt{x^2}\neq x$ in general, it fails whenever $x<0$. We actually have $\sqrt{x^2} = |x|$.