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

Question

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 $\log{-3}$?

Answer 1

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: $$ \sin{x} = 2 \implies x = \arcsin{2} = \frac\pi2-i\log(2+\sqrt3)$$

Answer 2

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.

Answer 3

Multiplying with $i$ rotates 90 degrees in the complex plane. One way to be able to reduce that to smaller steps is:

You can use a number really close to positive 1 (just a tiny little bit imaginary). For example:

$$z = \exp\left[\frac{2 \pi \cdot i}{k}\right], \, \text{ $k$ very large integer}$$

Now $z^k$ will be $1$. So you can reach all points on the unit circle with an angular resolution of $k$ steps per rotation. Now what is needed is to get off the unit circle. We can do this with a real radius $$r\in \mathbb R : r = 1+\frac{f}{100}$$ where f is the step in growth percentage. Now we can reach any complex number on a polar grid with the power functions and integer exponents:

$$r^{e_1}\cdot z^{e_2}, \, e_1,e_2 \in \mathbb{Z}$$