A Mobius transformation that sends a circle passing through -1 to a line
by Hidenori
Proposition
Show that $f(z) = \frac{z - 1}{iz + i}$ maps any circle passing through -1 to a line.
Solution
A circle centered at $a \in \mathbb{C}$ passing through -1 can be expressed as the set of $z \in \mathbb{C}$ satisfying $\abs{z - a} = \abs{-1 - a} = \abs{1 + a}$.
For any $z \in \mathbb{C}$, $w = f(z)$ implies $z = (1 + iw) / (1 - iw)$.
Let $w \in \mathbb{C}$ be given. $w = f(z)$ for some $z$ in the circle if and only if
\[\begin{align*} \abs{\frac{1 + iw}{1 - iw} - a} = \abs{1 + a}. \end{align*}\]This can be simplified to
\[\begin{align*} \abs{w - i\frac{1 - a}{1 + a}} = \abs{w + i}. \end{align*}\]In other words, $w = f(z)$ for some $z$ in the circle if and only if $w$ is equidistant from $i(1 - a)/(1 + a)$ and $-i$. Therefore, $f$ indeed maps a circle passing through -1 to a line.
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