Mobius transformation fixing points $z=1,z=-1$

Question

Let $\varphi$ be a mobius transformation s.t $\varphi(1)=1,\varphi(-1)=-1$. Denote $i\mathbb{R}=C$. What is $\varphi(C)$?

My attempt: Let $\varphi(z)=\frac{az+b}{cz+d}$. Since $\varphi(1)=1,\varphi(-1)=-1$ we have that $\varphi(z)=\frac{az+b}{bz+a}$. I looked at $\varphi(i),\varphi(-i)$ and $\varphi(0)$ and tried to determine under which conditions on $a,b$ they are linearly dependent (meaning they're on the same line) or otherwise when the aren't linearly dependent (and therefore are on the same circle), but I didn't get too far.

Any help would be apprecited.

Answer 1

Set $w=\frac{az+b}{bz+a}$ and rearrange to get $z=\frac{b-aw}{wb-a}$.

Now put $w=u+iv$ and do some algebra to find the real part of $z$ in terms of $u$ and $v$. $$z=\frac{b-au-aiv}{b-ua+ibv}\times\frac{bu-a-ibv}{bu-a-ibv}$$ $$=\frac{b^2u-ab-ib^2v-abu^2+a^2u+iabuv-iabuv+a^2iv-abv^2}{(bu-a)^2+b^2v}$$ So the real part is $$\frac{b^2u-ab-abu^2+a^2u-abv^2}{(bu-a)^2+b^2v}$$

Since you want the image of the imaginary axis, set this to zero and you get the equation of a circle $$u^2+v^2-u\frac{a^2+b^2}{ab}+1=0$$

Answer 2

The mapping $\varphi$ with $\varphi(1)=1,\varphi(-1)=-1$ and $\varphi(\infty)=\alpha\neq \pm 1$ is similarly solved as $$\varphi(z)=\frac{\alpha z+1}{z+\alpha}.$$ Then $\varphi(C)$ is the circle (or line) through the three points $$\varphi(0)=\frac 1{\alpha},\varphi(i)=\frac{\alpha i+1}{i+\alpha},\varphi(\infty)=\alpha.$$ It follows that $\varphi(C)$ is the circle (or line) through $\frac 1{\alpha},\frac{\alpha i +1}{i+\alpha}$ and $\alpha$.

Remark. It can be checked directly that $\varphi(C)$ is the $y$-axis if and only if $\alpha$ is purely imaginary. Note that if $\alpha$ is real, then $\varphi(i)$ lies on the unit circle.