I am getting $i^{-1}=\pm i$. I am trying to find $i^{-1}$. I already know that the answer is $-i$, but I can't figure out a way to determine that using math. This is what I am doing: $$i^{-1}$$ $$\frac1i$$ $$\sqrt{\left(\frac1i\right)^{2}}$$ $$\sqrt{\frac{1}{-1}}$$ $$\sqrt{-1}$$ $$\pm i$$ What am I doing wrong?
 A: Do this. 
$$i(-i) = -i^2 = -(-1) = 1.$$
A: I'm not sure where your 3rd expression comes from. In general, to find $\frac{1}{a+bi}$, where $a,b\in \mathbb{R}$, multiply numerator and denominator by the conjugate $\overline{a+bi}=a-bi$ to obtain $$\frac{1}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2+b^2}$$ In your case, take $a=0,b=1$ to obtain $\frac{1}{i}=\frac{-1i}{1}=-i$. 
A: The law $\sqrt{ab} = \sqrt{a}\sqrt{b}$ doesn't hold for complex numbers, else we would have
$$
i = \sqrt{-1} = \sqrt{\frac{1}{-1}} = \frac{\sqrt{1}}{\sqrt{-1}} = \frac{1}{i} = -i.
$$
A: You asked “what am I doing wrong”, and nobody has answered that yet.
$\sqrt{x^2}$ is not in general equal to $x$.  That identity is true only for positive real numbers.   Consider $x=-3$ for example: $\sqrt{(-3)^2} = \sqrt 9 = 3 \ne -3$.
In fact, $\sqrt x$ is not even defined except when $x$ is a positive real number.
So when you passed from $\frac1i$ to $\sqrt{\left(\frac1i\right)^2}$, what you did was invalid.  $\frac1i$ means one thing, and  $\sqrt{\left(\frac1i\right)^2}$ means nothing at all.
A: The $\sqrt{\cdot}$ function is problematic in the complex numbers.
By definition, for a non-negative real number $r$, $\sqrt{r}$ is the non-negative root,
but "negative" is not well-defined for complex numbers.
Your rewriting $\sqrt{-1}$ as $\pm i$ is actually about as sensible a result
as you could hope for: it gives you a list including every number whose square is $-1$.
A safer way to do the same calculation (avoiding the problematic square root) might be
$$z = i^{-1}$$ 
$$z = \frac1i$$ 
$$z^2 = \left(\frac1i\right)^{2}$$ 
$$z^2 = \frac{1}{-1}$$ 
$$z^2 = -1$$ 
$$z = \pm i$$
Instead of slapping a $\sqrt{\cdot}$ sign on a number whose square root isn't defined,
these equations use $z^2$ to keep track of the fact that the number we are looking
for is a number whose square is the stated number.
If you didn't already know that $-i$ is $i^{-1}$, 
this would give a good hint that it might be.
To really find out, however, 
you have to try multiplying by $i$ as shown in the answer by ncmathsadist.
