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Find a Möbius transformation $T$ from $\{z=x+iy:x+y>0\}$ onto the disk $D(1,4)$, such that $T(1)=2$ and $T(0)=-3$.

The proof given is as follows:

Since $-i$ is symmetric to $1$ with respect to the line $x+y=0$, then $T(1)$ and $T(-i)$ is symmetric with respect to $C(1,4)$, the circle centred at $1$ with radius $4$. This gives $T(-i)=17$. Since $T$ preserves cross-ratio, $(T(z),-3,2,17)=(T(z),T(0),T(1),T(-i))=(z,0,1,-i)$. Simplifying, we get \begin{equation*} T(z)=\frac{(17-8i)z-9}{(1-4i)z+3}. \end{equation*}

My question:

I know that every Möbius transformation preserves symmetry in the sense that if $z$ is symmetric to $z^{\ast}$ with respect to a circle or line $C$, then $T(z)$ and $T(z^{\ast})$ is symmetric with respect to $T(C)$.

In the above proof, how do we know that $T(C)=T(L)$ is precisely $C(1,4)$, where $L$ is the line $x+y=0$?

More generally, how can we determine an image of circle or line under a Möbius transformation which we are supposed to find, suppose that we know some points of symmetry (as in the above)?

J. W. Tanner
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Representation
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  • https://math.stackexchange.com/questions/1257294/four-complex-numbers-z-1-z-2-z-3-z-4-lie-on-a-generalized-circle-if-and-only-i?rq=1 – Kumar Jan 12 '20 at 04:27
  • @Kumar Thanks for your comment. But can I clarify, the post in the link says T(C) is a circle iff the cross ratio of the four points is real, but how can we check it is a specific circle or line? In my question, I would like to check that $T(L)$ is exactly $C(1,4)$ but without knowing what $T$ is. – Representation Jan 12 '20 at 04:32
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    Answer to your first question: $D$ is an open disk and $x+y>0$ is an open plane. The boundary of the plane is $L:x+y=0$ and the boundary of the $D$ is $C(1,4)$. Hence, the $L$ is precisely mapped to $C$ under $T$. Moreover, you can see that the points that lie in the complement of the closed plane $x+y\geq 0$ lie outside the closed disk $D$. Furthermore, as $z\rightarrow \infty$, $T(\infty)=\frac{17-8i}{1-4i}$ which lie on the $C$. Also, $T(0)=-3$ which is precisely a point on the $C$. – Kumar Jan 12 '20 at 05:28

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