How do you calculate $\arcsin (2)$?

Hint: The arcsin is the inverse of the trigonometric representation of sine (sin). The range of arcsin is from negative of $\dfrac{\pi }{2}$ to positive of $\dfrac{\pi }{2}$. The domain of arcsin is from negative one to positive one. It is a bijective function, which means it will be invertible. This property will be very useful in this question.

Complete step by step solution:
According to the question, we have to find the value of $\arcsin (2)$
But, arcsin is only defined for the range of input from a negative one to a positive one and the input given in the question is two, which is outside the domain of the function arcsin. So we have to do some operations to make the value inside the arcsin function in the domain.
Now, let’s say ${\sin ^{ - 1}}2 = x$, so
$ \Rightarrow \sin x = 2$ (It is what we are trying to find)
But sine function can’t be greater than one, so there is no real solution.
However, it is possible to define $\sin (z)$ for $z \in C$ and hence find a possible definition and then we can calculate the value of $\arcsin (2)$
We have to consider these formulas,
\[\begin{array}{l}
{e^{ix}} = cos(x) + isin(x)\\
cos( - x) = cos(x)\\
sin( - x) = - sin(x)
\end{array}\] (These are basically the concepts of complex numbers and trigonometry)

Now, we have to find
\[ \Rightarrow sin(x) = \dfrac{{{e^{ix}} - {e^{ - ix}}}}{{2i}}\] For all $x \in R$
So, we need to define
\[ \Rightarrow sin(z) = \dfrac{{{e^{iz}} - {e^{ - iz}}}}{{2i}}\]For all $z \in C$
We want to solve, that is
\[ \Rightarrow \dfrac{{{e^{iz}} - {e^{ - iz}}}}{{2i}} = 2\]
Now, let $t = i{e^{iz}}$
So the equation changes to
$ \Rightarrow \dfrac{{ - it - \dfrac{i}{t}}}{{2i}} = - \dfrac{{t + \dfrac{1}{t}}}{2} = - \dfrac{{{t^2} + 1}}{{2t}} = 2$
Multiply both sides with -2t, we will get
$ \Rightarrow {t^2} + 1 = - 4t$
$ \Rightarrow {t^2} + 1 + 4t = 0$ (Add 4t both sides)
Using the quadratic formula, its roots are
\[ \Rightarrow t = \dfrac{{ - 4 \pm \sqrt {{4^2} - 4} }}{2} = - 2 \pm \sqrt 3 \]
So, as we have considered $t = i{e^{iz}}$,
\[ \Rightarrow i{e^{iz}} = - 2 \pm \sqrt 3 \]
\[ \Rightarrow {e^{iz}} = \dfrac{{ - 2 \pm \sqrt 3 }}{i}\] (Dividing both sides with i)
\[ \Rightarrow {e^{iz}} = (2 \pm \sqrt 3 )i\]
\[ \Rightarrow {e^{iz}} = (2 \pm \sqrt 3 ){e^{i\dfrac{\pi }{2}}}\] (Iota can be expressed as ${e^{i\dfrac{\pi }{2}}}$)
\[ \Rightarrow {e^{i(z - \dfrac{\pi }{2})}} = (2 \pm \sqrt 3 )\]
Now, taking log on both sides, we will get
$ \Rightarrow i(z - \dfrac{\pi }{2}) = \ln (2 \pm \sqrt 3 )$
Hence,
$ \Rightarrow z = \dfrac{\pi }{2} \pm \ln (2 + \sqrt 3 )i$
So, this is our answer

Note: Negative root three is not considered because negative terms can’t be there in a log function. We also need knowledge of complex numbers to find the solution of these types of questions. As real solutions are not possible, we go with an imaginary solution.