$1/i=i$. I must be wrong but why? $$\frac{1}{i} = \frac{1}{\sqrt{-1}} = \frac{\sqrt{1}}{\sqrt{-1}} = \sqrt{\frac{1}{-1}} = \sqrt{-1} = i$$
I know this is wrong, but why? I often see people making simplifications such as $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$, and I would calculate such a simplification in the manner shown above, namely
$$\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{\sqrt{4}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}$$
 A: Even for real numbers, there are two square roots.  $(-2)^2 = 2^2  =4$, so both $2$ and $-2$ can be thought of as square roots of $4$.  In the real numbers, we have an easy way to pick one of the two:  just always pick the positive one.  So we define $\sqrt{x}$ to be the positive square root of $x$.  In the complex case, we do not have an order, and so no consistent way to pick one of the two square roots.  So $\sqrt{z}$ is not really a function:  it is a multivalued function.  You confusion arises from thinking that $\sqrt{i}$ indicates only one number.
A: What you are doing is a version of 
$$
-1=i^2=\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}=\sqrt1=1.
$$
It simply shows that for non-positive numbers, it is not always true that $\sqrt{ab}=\sqrt{a}\sqrt{b}$.
A: It is wrong because the calculation rule of square roots only works for real non negative roots.
A: Well, in general, you should, when dealing with non-positive numbers, treat $\sqrt{x}$ as a set of all numbers such that $y^2=x$. Then you get that $\sqrt{1}=\{+1,-1\}$ and $\sqrt{-1}=\{+\mathrm i,-\mathrm i\}$, and then you get
$$ \frac{1}{\sqrt{-1}} = \frac{\sqrt{1}}{\sqrt{-1}} = \sqrt{\frac{1}{-1}} = \{+\mathrm i,-\mathrm i\}=\{\mathrm i,1/\mathrm i\},$$
and everything seems to be ok now. Of course, this brings a lot vagueness into the equals sign, similar vagueness as one has with Langrange symbols like $f(x)=\mathcal O(x)$.
As pointed in the comments, you can't write $\mathrm i=\sqrt{-1}$ just because $\mathrm i^2=-1$. Actually, $i$ is defined as such a complex number that $\mathrm i^2+1=0$, but nothing more. There are two such numbers, since the polynomial is quadratic irreducible over $\mathbb Q$.
A: The inverse of i is -i
$$1/i = i/(i*i) = i/-1 = -i$$
so you take the negative result from \sqrt
A: Square roots are not that simple with complex numbers.
You'd better use exponential form: 
$$i = e^{\frac{\pi}{2}i}$$
$$\frac{1}{i} = e^{-\frac{\pi}{2}i} = e^{(2\pi-\frac{\pi}{2})i} = e^{\frac{3\pi}{2}i} = -i$$
A: $$\frac1{\sqrt{-1}}=\sqrt{-1}$$ is only true in the sense that $1$ over a square root of $-1$ is a square root of $-1$. However, there are two square roots of every non-zero complex number, so you have to make sure to pick the right one. For square roots of non-negative real numbers it works to consistently pick the non-negative square root, but no such rule exists for all complex numbers.
A: You assume $\sqrt{-1}=i$, but it's wrong to write such a thing, because the square root function is really defined as a function only for positive argument, and it has positive values. If you write everywhere $\pm \sqrt{a}$ instead of $\sqrt{a}$ (because it really means "a root of $x^2-a$" and it has two roots), then everything you wrote is "almost right".
A: the inverse of $i$ is $-i$ due to the formula of 
$$\frac{a+bi}{c+di}=\frac{(a+bi)(c-di)}{c^2+d^2}$$
fill in $a=d=1$ and $b=c=0$ and you'll see
$$\frac{1+0i}{0+1i}=\frac{(1+0i)(0-1i)}{0^2+1^2}=\frac{-1i}{1}=-i$$
this provides the hint that the issue is in the last square root you take, namely that you need to take the negative square root
