As $\sqrt{x^{2}} \neq x$ for $x \in \mathbb{R}$, is $\sqrt{i^{2}} \neq i$? More generally, $\sqrt{z^{2}} \neq z$ for $ z \in \mathbb{C} $?
Because if we write $ \ z^{2} = r^{2}\cdot e^{2i\theta} \ (z\in\mathbb{C}, \theta\in [0, 2\pi[, r\in\mathbb{R}^{+}) \ $ then $ \ \sqrt{z^{2}} = r\cdot \sqrt{e^{2i\theta}} \ $ and I become unsure about simplifying the root over the exponential there.
And if $\sqrt{i^{2}}$ is indeed $i$, would the correct way to solve an equation like $z^{2}=-k \ , \ k \in\mathbb{R}^{+}$ be first writing $z^{2}= i^{2}\cdot k$ and then taking the root to end up with $\sqrt{z^{2}}$ ($z$?) $= i\cdot \sqrt{k} \ $? It'd feel great never to see $\sqrt{-k}$.
First math question on a forum ever, sorry if I made any mistake in terms of manners.
When we're talking about roots in $\mathbb{C}$ we have to be really careful since the functions are multi-valued (if you're interested more in this topic check out Roots of Unity) but the idea is a function of the form $w=z^n$ where w and z are complex and n is a non-negative integer, we're going to have n solutions. For example consider $4=z^2 ==> 4=r^2e^{ix}$ from here we observe that $r=2$ and $e^{ix}=1$, recalling both x and r are real numbers we see that $z=2e^{i*0}=2$ and $z=2e^{i\pi}=-2$ as expected. We can generalize this to any positive integer.