Construct a Mobius transformation that maps the interior of the circle $\{z \in \mathbb{C}:|z-3|=2\}$ bijectively onto the half plane $\{z \in \mathbb{C}:Re(z)<-1 \}$.
I drew the two graphs but I don't know what to do next.
I'm having trouble in starting the problem. Any help is appreciated.
2026-03-26 16:06:06.1774541166
Mobius transformation that maps interior of a circle to a half plane bijectively
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Take $1$ to $\infty$ so that the circle is sent to the line $Re(z)=-1$. Then we need to send two symmetric points to symmetric point of the line. Hence we may choose to send $4$ to $-2$ and $7$ to $0$. Thus for $T(z)=\frac{az+b}{cz+d}$, we have $c=-d$ from the first point, $4a+b=-8c-2d$ from the second point, and $7a+b=0$. From here it is all algebra.