Background
I recently set out to derive the exponential forms of the inverse trigonemtric functions using eulers identity and demoivres theorem, deciding to start with $arcsin(x)$ I first got that:
$$e^{ix}=\cos(x)+i\sin(x) \implies e^{-ix}=\cos(x)- i\sin(x) $$
$$\therefore \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$
Deriving arcsin(x)
Letting $\sin(x)=y$
$$y=\frac{e^{ix}-e^{-ix}}{2i} \implies 2iye^{ix}=e^{2ix}-1$$
$$\implies e^{2ix}-2iye^{ix}-1=0$$
Completing the square;
$$(e^{ix}-iy)^2=1-y^2 \implies e^{ix}=iy \space ± \space \sqrt{1-y^2}$$
Then by taking the natural log and noting that it exists only for positive non zero numbers;
$$x=\frac{\ln(iy + \sqrt{1-y^2})}{i} $$
$$\therefore \text{arcsin}(x)= \frac{\ln(ix + \sqrt{1-x^2})}{i}$$
The more I studied this the more it became apparent that if I were to rewrite the complex number within the natural log in its exponential form then would it not be true that:
$$\text{arcsin}(x)=\frac{(i)(\text{arg}(ix + \sqrt{1-x^2})}{i}$$
$$\therefore \text{arcsin}(x)=\text{arg}(ix + \sqrt{1-x^2})$$
Deriving arctan(x)
I am extremely fascinated by the fact that to find the $arcsin$ of some value, you merely have to find the argument of some arbitrary complex number, upon noting this I also derived the $\text{arctan}(x)$ logarithmic form in the same way:
$$\text{arctan}(x)= \frac{\ln \left(\frac{2i}{x+i} -1 \right)}{2i}$$
So that I may plug in the value of $\frac{x}{\sqrt{1-x^2}}$ into the expression for $\text{arctan}(x)$ and find the argument.
Upon making the same observation as with $arcsin(x)$ I noted that this expression was the same as
$$\text{arctan}(x)= \frac{\text{arg} \left(\frac{2i}{x+i} -1 \right)}{2}$$
Questions and query about the implications
Again, the fact that finding the $arctan$ or $arcsim$ of a value is the same as finding the argument of seemingly arbitrary complex numbers seems extremely interesting to me and I cant wait to find out more about this and what exactly these complex numbers in question represent.
However I have not been able to find any satisfactory answers during my searching and so my question is, what do the complex numbers $ix + \sqrt{1-x^2}$ and $\frac{2i}{x+i} -1$ represent, and why do their respective arguments give values for the inverse trigonometric functions in question?
Best Answer
Point 1
If we look at a right-angled triangle with horizontal length being equal to $\sqrt{1-x^2}$, and the vertical length equal to $x$ then by Pythagoras' theorem we can see that the the length of the hypotenuse of this triangle is $1$. Consider the angle between the horizontal length and the hypotenuse, $\theta$. We can see that $$\sin\theta=\frac{x}{1}=x$$ and therefore $\theta=\arcsin x$. However, if we were to instead look at the complex plane and the complex number $ix+\sqrt{1-x^2}$, the imaginary value $ix$ corresponds to the vertical side of length $x$ and the real value $\sqrt{1-x^2}$ corresponds to the horizontal side of length $\sqrt{1-x^2}$.
Hence, it is thus easy to see that $$\theta=\arcsin x=\arg(ix+\sqrt{1-x^2})$$ Similar reasoning applies to your observation about $\arctan x$.
Point 2
The regular trigonometric functions are well known, as are their inverses, such as $\sin x$ and $\arcsin x$, $\tan x$ and $\arctan x$ and so on. However, there are also some slightly lesser-known functions known as the hyperbolic trigonometric functions, or just the hyperbolic functions. The main hyperbolic functions are: $$\sinh x=\frac{e^x-e^{-x}}{2}$$ $$\cosh x=\frac{e^x+e^{-x}}{2}$$ $$\tanh x=\frac{\sinh x}{\cosh x}=\frac{e^{2x}-1}{e^{2x}+1}$$ and their inverse functions are, respectively: $$\operatorname{arsinh}x=\ln(x+\sqrt{x^2+1})$$ $$\operatorname{arcosh}x=\ln(x+\sqrt{x^2-1})$$ $$\operatorname{artanh}x=\frac{1}{2}\ln\frac{1+x}{1-x}$$
The hyperbolic functions and the trigonometric functions have an incredible amount of similar properties; many of them are listed here . In particular, $$\arcsin x= -i\operatorname{arsinh} ix$$ $$\arctan x= -i\operatorname{artanh} ix$$ $$\arccos x= -i\operatorname{arcosh} x$$ ie $$\arcsin x= -i\ln(ix+\sqrt{(ix)^2+1}=-i\ln(ix+\sqrt{1-x^2})$$ $$\arctan x= -i\frac{1}{2}\ln\frac{1+ ix}{1-ix}$$ $$\arccos x= -i\ln(x+\sqrt{x^2-1})$$ So these complex numbers are not random: they are produced by making the argument of the inverse hyperbolic functions completely imaginary in the case of $\operatorname{arsinh}x$ and $\operatorname{artanh} x$ .
I hope that helps. If you have any questions please don't hesitate to ask :)