Division by imaginary number I ran into a problem dividing by imaginary numbers recently. I was trying to simplify:
$2 \over i$
I came up with two methods, which produced different results:
Method 1: ${2 \over i} = {2i \over i^2} = {2i \over -1} = -2i$
Method 2: ${2 \over i} = {2 \over \sqrt{-1}} = {\sqrt{4} \over \sqrt{-1}} = \sqrt{4 \over -1} = \sqrt{-4} = 2i$
I know from using the formula from this Wikipedia article that method 1 produces the correct result. My question is: why does method 2 give the incorrect result? What is the invalid step?
 A: The only soundproof way to be sure to find the right result while dividing two complex numbers
$$\frac{a+bi}{c+di}$$
is reducing it to a multiplication. The answer is of the form $x+yi$; therefore 
$$(c+di)(x+yi) = a+bi$$
and you will end up with two linear equations, one for the real coefficient and another for the imaginary one. As Simon and Casebash already wrote, taking a square root leads to problems, since you cannot be sure which value must be chosen.
A: The incorrect step is saying:
$\sqrt{4}/\sqrt{-1} = \sqrt{4/-1}$
The identity:
$\sqrt{a}/\sqrt{b} = \sqrt{a/b}$
is only justified when $a$ and $b$ are positive.
A: This is exactly the same issue as in this question. Each non-zero complex number has two numbers that square to give it, with the same magnitude, but with opposite sign. When we define the square root function, we have to decide which of the roots we want. For positive numbers, it is obvious to choose the positive root. For negative number, we choose to have the positive imaginary values, although because of symmetry the choice doesn't mean much anyway.
So, to see if the standard multiplication and division laws apply, then we have to consider domain the numbers are in. We already know they apply for non-negative real numbers. It is easy enough to verify that for negative numbers $$\sqrt{a}*\sqrt{b}=-\sqrt{ab}$$ and $$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$$ We also see that, if $a$ is positive and $b$ negative, then $$\sqrt{ab}=\sqrt{a}\sqrt{b}$$ $$\sqrt{\frac{a}{b}}=-\frac{\sqrt{a}}{\sqrt{b}}$$ and $$\sqrt{\frac{b}{a}}=\frac{\sqrt{b}}{\sqrt{a}}$$
