Why $\sqrt{-1 \times -1} \neq \sqrt{-1}^2$? [duplicate]

We know $$i^2=-1 $$then why does this happen? $$ i^2 = \sqrt{-1}\times\sqrt{-1} $$ $$ =\sqrt{-1\times-1} $$ $$ =\sqrt{1} $$ $$ = 1 $$

EDIT: I see this has been dealt with before but at least with this answer I'm not making the fundamental mistake of assuming an incorrect definition of $i^2$.

From $i^2=-1$ you cannot conclude that $i=\sqrt{-1}$, just like from $(-2)^2 = 4$ you cannot conclude that $-2=\sqrt 4$. The symbol $\sqrt a$ is by definition the positive square root of $a$ and is only defined for $a\ge0$.

It is said that even Euler got confused with $\sqrt{ab} = \sqrt{a}\,\sqrt{b}$. Or did he? See Euler's "mistake''? The radical product rule in historical perspective (Amer. Math. Monthly 114 (2007), no. 4, 273–285).

Any non zero number has two distinct square roots. There's an algebraic statement which is always true : "a square root of $a$ times a square root of of $b$ equals a square root of $ab$", but this does not tell you which square root of $ab$ you get.

Now if $a$ and $b$ are positive, then the positive square root of $a$ (denoted $\sqrt{a}$) times the positive square root of $b$ (denoted $\sqrt{b}$) is a positive number. Thus, it's the positive square root of $ab$ (denoted $\sqrt{ab}$). Which yields

$$\forall a,b \ge 0, \ \sqrt{a} \sqrt{b} = \sqrt{ab}$$

In your calculation, because $i$ is a square root of $-1$, then $i^2$ is indeed a square root of $1$, but not the positive one.

$\hspace{0.2in}$ 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$~$).

$\hspace{0.2in}$ 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.

$\hspace{15.0in}$ $\hspace{0.2in}$ 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 $t=0~$).