Let $T$ be a Mobius transformation such that $T(0)=\alpha$, $T(\alpha)=0$ and $T(\infty)=-\alpha$, where $\alpha = (-1+i)/\sqrt 2$. Let $L$ dnotes the straight line passing through origin with slope $-1$ and $C$ denotes the unit circle centered at origin. Then which are true ?
(A) $T$ maps $L$ to a straight line.
(B) $T$ maps $L$ to a circle.
(C) $T^{-1}$ maps $C$ to a straight line.
(D) $T^{-1}$ maps $C$ to a circle.
Using cross ratio we get the Mobius transformation is $$z=-\alpha \frac{w-\alpha}{w+\alpha}$$
where, $w:=T(z)$.
From this, equating real and imaginary parts I can solve the problem. But that is too laborious. Does there any shorter way to solve this problem ?
But
A Möbius transformation maps every [line or circle] to some [line or circle].
$T^{-1}$ maps $\alpha,0,-\alpha\in L$ to $0,\alpha,\infty\in L$ hence $T^{-1}(L)=L,$ so A is true ($T(L)=L$) hence B is false.
$T^{-1}$ maps $-\alpha\in C$ to $\infty,$ so D is false hence C is true.