When evaluating the function $\sin \theta$ with a complex angle $\theta$, a real value $A \geq 1$ can be obtained in two ways.
- Considering $\theta = i \log \left[ -i \left( A + \sqrt{A^2 - 1} \right) \right]$, as stated in this page. After the substitution of $\theta$,
$$\sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i} = A$$
- Considering $\theta = \frac{\pi}{2} - iB$. This way,
$$\sin \theta = \cosh B$$
I want to prove that the two methods are equivalent.
My attempt:
I applied the inverse hyperbolic cosine definition:
$$\cosh B = A \Rightarrow B = \mathrm{arccosh} A = \log \left[ A + \sqrt{A^2 - 1} \right]$$ $$\theta = \frac{\pi}{2} - i\log \left[ A + \sqrt{A^2 - 1} \right] \label{a} \tag{1}$$
In method 1, $\theta$ can be rewritten as
$$\theta = i \log \left[ -i \left( A + \sqrt{A^2 - 1} \right) \right] =\\ = i \log \left( \frac{A + \sqrt{A^2 - 1}}{j} \right) = i \left[ \log \left( A + \sqrt{A^2 - 1} \right) - \log i \right] = \\ = i \left[ \log \left( A + \sqrt{A^2 - 1} \right) - i \frac{\pi}{2} \right] =$$ $$=i \log \left( A + \sqrt{A^2 - 1} \right) + \frac{\pi}{2} \label{b} \tag{2}$$
My question:
In my attempt, \ref{b} doesn't match \ref{a}. Why? What is wrong?
This question is related to a Physics question about Snell's law.
Yes, they match. Just note that$$(\forall z\in\mathbb{C}):\sin\left(\frac\pi2-z\right)=\sin\left(\frac\pi2+z\right)=\cos(z).$$