Can't figure out what's wrong with this "proof" that i = -i $$i = e^{i\pi/2} = (e^{i\pi})^{1/2} = \sqrt{e^{i\pi}} = \sqrt{-1} = \pm i$$
Was it a misuse of properties of exponents? The square root, maybe? Did I use something that's not defined for all complex numbers? I would appreciate any books or resources with an in-depth explanation of what I need to know to clarify this doubt.
 A: You cannot take the square root of any complex number like that : $z\mapsto z^{1/2}$ is a multiform function since you need a complex logarithm to define it. Both $i$ and $-i$ are licit "square roots" of $-1$ because $i^2 = (-i)^2 = -1$. That does not mean they are equal.
Also, don't use the $\sqrt{}$ notation for complex numbers, it is better to only use it for positive real numbers (for the above reason).
A: Problems:  
1)  You don't actually have a problem with you conclusion.
$\sqrt{-1} = \pm i$ means $\sqrt{-1} = i$ OR $\sqrt{-1} = -i$.  It doesn't mean it equals both, or that it could mean either and it's a matter of choice (although it could-- but it doesn't have to).  It just means it is one or the other.
2) $\sqrt {moo} = \pm goo; goo^2 = moo$. is not a precise result. 
Thought experiment:  Suppose I stated to solve $2x + 3 = 9$ we can conclude $x = 3$ or $x = \sqrt{\pi}$ and you said "Square root of pi?!?!?! Where did that come from? $x =3$ and that's it!  It can't equal the square root of pi!"  And I said  "I didn't say it did equal $\sqrt{\pi}$;  I said $x = 3$ OR $x = \sqrt{\pi}$.  Do you deny it?  Are you claiming $x$ equals neither $3$ nor $\sqrt{\pi}$?"
Point is $\sqrt{x}$ is not (usually) stated to mean "the set of all numbers $y| y^2 = x$".  It usually means "$\sqrt{x}$ is a specific one of the multiple numbers such that $y^2 = x$.
In the real numbers $\sqrt{4} \ne \pm 2$.  $\sqrt{4}$ is specifically the non-negative number $y$ so that $y^2 = 4$.  So although $(-2)^2 = 4$ and $2^2 = 4$.  $\sqrt{4} \ne -2$.  $\sqrt{4} = 2$.
But it's not so universally agreed upon with complex numbers.  $\sqrt[3]{4+3i}$ has 3 complex values and, so far as I know, there is no agreement as to which of these should be THE cube root.  I'd say (and I may be oversimplifying here) that $(4+3i)^{1/3}$ is considered multi-value and the proper interpretation is the set of three possible values.  But that doesn't mean that $z^3 = 4+3i \implies z = (4+3i)^{1/3}$ means $z$ is all three values at once.  It means that $z$ is one of the potential values.
Anyway, when we say $i = \sqrt{-1}$ it is an arbitrary assignment.  It is true that $(-i)^2 = -1$ but in no way does it follow that $i = -i$, any more than $(-2)^2 = 4$ implies that $-2 = 2$.
3) $z^{1/2} = \sqrt{z}$ well.... yeah.... but....
In complex numbers $f(z) =z^{1/2}$ is a multi-valued functions (to the consternation of high-school algebra students everywhere--- for years they had it pounded into their heads that $x^2 + y^2 = 1$ is not a function because it fails the "vertical line test" and suddenly ... "Oh, it's just a multi-valued function-- that's fine"... well, I sympathize).  $\sqrt{z}$ is ... seen to be a specific value ... but no-one can agree on which one.  In general we try to avid radical symbols in complex analysis to avoid these ambiguities.
