Find Möbius transformation $S$ that maps points $(-1,0,1)$ to points $(i,-1,-i)$. And what's the image of real axis and upper half of imaginary axis $\{z\in\mathbb{C}| \operatorname{Im} z \geq 0 \} $ of $S$.
Can I solve this with the implicit form formula:
\begin{align} \frac{(w-i)(-1+i)}{(w+i)(-1-i)}=\frac{(z+1)(0-1)}{(z-1)(0+1)} \end{align}
I solved this and got $w=S(z)=\dfrac{z-i}{z+i}$ correct?
And for the images should I use the inverse transformation $z=S^{-1}(w)=\dfrac{-wi-i}{w-1}$.
And if we want to know the image of real axis $\operatorname{Re}(z)=x$ and solve this? \begin{align} x&=\dfrac{-wi-i}{w-1} \\ w&=\frac{x-i}{x+1} \end{align}
Or am I in the wrong direction? I'm not sure about the upper half of imaginary axis image.
First of all, your transformation is correct. It's not too hard to check one of these, once they are given to you: simply plug in the three numbers given to make sure that you get the correct output. In your case, that is true! Success.
One thing that you can use is that Mobius transformations always transform {lines, circles} into {lines, circles}. Since in your case, you want to know the image of the real line, note that this is a line that passes through $\{0, 1, -1\}$. Thus since the image of those points is $\{i, -i, -1\}$, it follows that the image of the real axis will be the line or circle that passes through $\{-1, i, -i\}$, which should not be too hard to sort out.
Similar arguments can help you figure out the image of the positive imaginary axis.