Q:

Can every possible 'imaginary number' be expressed in terms of $i$? 

The domain of the function $f(x)=\sqrt{x}$ can be extended to all real numbers by introducing a new number, $i=\sqrt{-1}$.
Can this be done for any function, say $\arcsin{x}$, or $\log{x}$? 
What is $\arcsin{2}$?  Or $\log{-3}$? 

A:

We define the set of complex numbers $\mathbb{C}$ the set of $\mathbb{R} \times \mathbb{R}$, the cartesian product of the set of real numbers on itself. 
On these Complex Numbers we define operations such as summation and multiplication by scalar. 
We define them in such a way that the couple $(0; 1)$ just so happens to have the property that:
$$ (0; 1)^2 = (-1; 0) $$
And being such an important property, it has its own name as a number. 
This can be done in many ways.  Numerical sets are constructed easily in modern mathematics.  You can define your $\log{-3}$ using complex numbers.  In fact, the logarithm function is defined in the complex plane for every non-zero complex $z$.
$$ \log{-3} = \log{i^2} + \log{3} = 2\log{i} + \log{3} $$
Let $i = e^{i(\frac{\pi}{2} + 2k\pi)}$ for any $k \in \mathbb{Z}$. The logarithm of the imaginary unit is then defined as:
$$ \log{i} = i(\frac{\pi}{2} + 2k\pi) $$
Since we get an infinite amount of answers that depend on k, we usually only take the "main value" of the logarithm: that is, whenever $k=0$.
You don't need to define any new numerical set in order to solve a negative logarithm. 
EDIT: As some have stated, the sine function in the complex plane is surjective. 
$$ \sin{x} = \frac{e^{ix} - e^{-ix}}{2i} $$
And you can solve it to be:
$$ \sin{x} = 2 \implies x = \arcsin{2} = \frac\pi2-i\log(2+\sqrt3)$$

