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The problem is this: the (polar) representation of a complex number depends on your choice of branch. Once you choose a branch of the square root, you cannot simultaneously represent $1$, and $-1$, because they are $\pi$ apart, so that $-1$ and $1$ will necessarily be in different branches of $z^{1/2}$. The choice of square root you make (if you want it to be well-defined) depends on the choice of branch you are working with. You are trying to combine numbers that live in different branches; it is as if you add $1+1$ and get $1-i\pi$, since 1 can also be represented as $i\pi$ ($~1$ can actually be represented as $ik\pi$, but when you operate, you are supposed to stay within a branch, and, in your case, you are not$~$).

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In complex numbers, from the perspective of polar representations, when you multiply $z_1 \dot z_2$, you multiply the respective lengths, and add the respective angles (but you have to make up for the fact that the sum of the angles may be larger than $2\pi$ (or whatever argument-system you are working with. So in this sense, when you multiply $i$ by itself, you multiply the length of $i$ by itself, and add the argument to itself; i.e., you double the argument. In complex variables, you have many possible polar representations for a given number; specifically, given $z=re^{it}$, then $z=re^{i(t+2\pi)}$ is also a valid representation.

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You must then choose a specific representation for your $z$, specifically; you will need to specify the range of the argument you will be working with. So, say you work with the "standard" range of $[0,2\pi)$. Then the expression for $i$ (me) is as $i=1e^{i\pi/2}$, so that $i^2$= $(1)(1).e^{i(\pi/2+\pi/2)}=e^{i\pi}=-1$. But the way backwards from multiplying to taking roots is more complicated if your base is non-negative, and/or your exponent has non-zero imaginary part. When this last is the case, you *define*:

$z^{{1}/{2}} \:=e^{{\Large{log(z)}}},~$
where we define:

$\log(z):=\ln |z|+i arg(z)$

(This choice of definition has to see in part with wanting to have the complex log agree with the standard real $\log$--though this agreement is possible only for one choice of "branch", as we will see.)
But because of the infinitely-many possible choicesfor the argument of a number, the $\log(z)$ itself--defined *locally* as the inverse of $e^z$ is somewhat-ambiguously-defined, since $e^z$ does not have a *global* inverse (since it is not $1-1$, for one thing, but $e^z$ does have local inverses, e.g., by using the inverse function theorem). So when we mention $\log$, we are referring just to one of (infinitely-) many possible local inverses of $e^z$ .Each possible local inverse to $e^z$ is called a "branch" of the $\log$. So once we choose a branch for the $\log$, which is a choice of an open set (technically, it is half-open) of width $2\pi$ from which we will choose the argument we will use. So, say we choose the standard branch $(0,2\pi)$, which we call *Log(z)*. We then define :

$z^{1/2}:=e^{\large{Log(z)/2}}$

But, in this branch , $(-1)^{1/2}$ is not even defined, because the argument for $(-1)$ is $0$ , which is outside of the allowable values $(0,2\pi)$. So, in this sense, the expression $(-1)^{1/2}$ is not well-defined, i.e, does not really make sense. ($~$Note that this particular branch of $\log$ reduces to the standard one when you select an argument of $t=0~$).