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 $\log{-3}$?

Euler's formula tells us that $e^{ix} = \cos(x)+i\sin(x)$. Therefore, (in a sense) $\ln (-3) = \ln3+\pi i$, since 
$$e^{\ln{3}+\pi i} = e^{\ln 3}e^{\pi i} = 3(\cos \pi + i\sin \pi) = -3$$
(note: $\ln 3+k\pi i$ would have worked for any odd integer $k$)
Similarly, if we let $z = \frac{\pi}{2}-i\ln(2+\sqrt 3)$, then
$$\sin(z) = \sin(\pi/2)\cos(i\ln(2+\sqrt 3))+\cos(\pi/2)\sin(i\ln(2+\sqrt 3)) = \cos(i\ln(2+\sqrt 3))$$
Since $e^{ix} = \cos(x)+i\sin(x)$, we have that $\cos(x) = \frac{1}{2}(e^{ix}+e^{-ix})$, and so
$$\sin(z) = \cos(i\ln(2+\sqrt 3)) = \frac{1}{2}\left(\frac{1}{2+\sqrt 3}+(2+\sqrt 3) \right) = 2$$
However, it is not always the face that functions can be "extended" in this manner. As said in a comment, for example, there is no reasonable way to define $\ln 0$ as a complex number. 
An issue that I've glossed over is that it seems that there are multiple ways to "choose" a value for $\ln (-3)$ and similar expressions. This is similar to the issue that there are multiple solutions to, say, $\sin(x) = 1/2$, and so $\sin^{-1}(1/2)$ is defined by making a reasonable "choice" from among the solutions to the equation. 
