why cannot we say squareroot of $-1$ as $-i$ When $i^2=-1$ why can't we write $i=\pm \sqrt{-1}$ that is why can't we say $$\sqrt{-1}=-i$$
 A: We can.  Simply we say $z^2+1=0$ has two solutions, and at first we cannot tell the difference between them.  We label one of them $i$, from that point on that is $i$ and the other is $-i$.  We could have also picked the other root, but we couldn't have been able to tell the difference until we label one of them.   What travels between the two possibilities is the complex conjugate operation.
Choosing $i$ fixes a certain canonical orientation on the complex plane.  The complex conjugate then goes to the other orientation.  Orientation is the choice of direction of rotation.  Multiplying by $i$ rotates to the left and $-i$ rotates to the right.  Though of course, if we happened to have picked the other root, we'd still draw it as $(0,1)$ in the $\mathbb R^2$ plane and then it seems nothing has changed and it still seemingly rotate to the left, though we've also flipped the $\mathbb R^2$ to do it.
In complex analysis, we tend to write every holomorphic (analytic) function as a  function of $z$, while a random other function is a function of both $z$ and $\bar{z}$.  For example, an arbitrary polynomial of the real and imaginary part of $z$ can be written as a polynomial in $z$ and $\bar{z}$.  It is holomorphic, if it does not depend on $\bar{z}$.  On the other hand, if it only depends on $\bar{z}$ but not on $z$, it is so-called antiholomorphic.  Now everything had better be symmetric.  Any result you can prove about holomorphic functions must be true for antiholomorphic functions (with the bars stuck into the theorem in the correct places).  That is because the antiholomorphic functions would be holomorphic functions if we swap the role of $i$ and $-i$.
A: We could have - taking $i = \sqrt{-1}$ was always just a convention. On its own, $i$ meant nothing to begin with, so when people first started considering imaginary numbers they were free to define it as either solution to the equation $i^2 = -1$. But once that decision is made, we can't switch back unless we go back and change the definition.
The thing is, "$\sqrt{-1}$" doesn't have an independent existence in the same way as, say, $\sqrt{9}$ does. $\sqrt{9}$ has a very definite meaning: that number which is greater than zero and which, when squared, yields $9$. If we try the same definition for $\sqrt{-1}$, we get gibberish: there is no number which is greater than zero and which, when squared, yields $-1$. So if we want to talk about $\sqrt{-1}$, we have to introduce a new number to make this make sense. We could introduce the "right" number, the one that's supposed to be $\sqrt{-1}$; that's what mathematicians did historically, and they called it $i$. Another option would have been to define $\sqrt{-1}$ obliquely - we could say, for example, that $\sqrt{-1}$ is that number which is a \emph{negative} multiple of $i$ and has square $-1$, in which case we have $\sqrt{-1}$ defined to be $-i$. Heck, we could say that $i$ isn't $\sqrt{-1}$ at all, but instead $2i$ is. It was all just a matter of definition.
The thing is, though, once a definition's made, you have to use it consistently. We could have defined $\sqrt{-1}$ to be $i$, or $-i$, or $2i$ - but we can't use any two of those at once. We have to pick one, and the one we picked was $i$.
A: For the purposes of traditional analysis, I don't know if it actually even matters which root you define $i$ to be.  The properties of that quantity are all the same regardless of whether $i=-\sqrt{-1}$ or $i=\sqrt{-1}$.  For example
$i^{0+4k}=1 \quad k=0,1,2,...$
$i^{1+4k}=i \quad k=0,1,2,...$
$i^{2+4k}=-1 \quad k=0,1,2,...$
$i^{3+4k}=-i \quad k=0,1,2,...$
$\exp(i\theta) = \cos(\theta) + i\sin(\theta)$
The quaternions give an example of THREE quantities $i^2=j^2=k^2=-1$.  In Clifford algebras there are actually many such general quantities when squared give -1.
