Sin inverse of a complex number Is it possible to calculate the value of $\delta$ from the relation
$\delta=\sin^{-1}(5.4i)$ ? where $i=\sqrt{-1}$
 A: If $\sin(\delta)=5.4i$, then 
$$\begin{align}
\sin(\delta)&=i\sinh(\sinh^{-1}(5.4))\\
&=\sin(i\sinh^{-1}(5.4))\\
\end{align}
$$
because it is a fact that $\sin(ix)=i\sinh(x)$. If that is unfamiliar to you, apply the identities $\sin(x)=\frac{\exp(ix)-\exp(-ix)}{2i}$ and $\sinh(x)=\frac{\exp(x)-\exp(-x)}2$.
So $\delta$ would be $$i\sinh^{-1}(5.4)+2\pi n$$ or $$\pi-i\sinh^{-1}(5.4)+2\pi n$$
If you want a firmly defined $\sin^{-1}$, then probably $i\sinh^{-1}(5.4)$ is what you would go with, since that would be consistent with a $\sin^{-1}$ whose range is the part of the complex plane with $-\pi/2\leq\Re(z)\leq\pi/2$. 
A: Let $\delta=x+iy$
So, $\sin(x+iy)=5.4i$
Now, $\sin(x+iy)=\sin x\cos(iy)+\cos x\sin(iy)=\sin x\cosh y+i\cos x\sinh y$
Equating the real parts, $\sin x\cosh y=0\implies \sin x=0$ as $\cosh y\ge 1$ for real $y$
So, $\cos x=\pm1$
If $\cos x=1, x=2m\pi$ where $m$  is any integer 
and $\sinh y=5.4\implies \frac{e^y-e^{-y}}2=\frac{27}5\implies 5(e^y)^2-54e^y-5=0$ 
Solve for $e^y$ which is $>0$ for real $y$
If $\cos x=-1, x=2(n+1)\pi$ where $n$  is any integer 
and $\sinh y=-5.4\implies \frac{e^y-e^{-y}}2=-\frac{27}5\implies 5(e^y)^2+54e^y-5=0$ 
Solve for $e^y$ which is $>0$ for real $y$
