How to show directly that $T(G \setminus \{a\})$ is open?

Question

Let $G \subseteq \mathbb C$ be open and $a \in G.$ Let $T$ be a Möbius transformation defined by $z \mapsto (z-a)^{-1}.$ Then show that $T(G \setminus \{a\})$ is open.

I can able to conclude that by open mapping theorem since $T\ \big \rvert_{G \setminus \{a\}}$ is a non-constant analytic function and $G \setminus \{a\}$ is open. But I failed to show it directly by showing that every point of $T(G \setminus \{a\})$ is an interior point. Could anyone please help me in this regard?

Thanks for investing your valuable time in reading my question.

Answer 1

Finally I am able to do that. Take $w \in T(G \setminus \{a\}).$ Then $w \neq 0$ since $w \neq \infty$ and $T(\infty) = 0.$ So there exists a unique $z \in G \setminus \{a\}$ such that $T(z) = w.$ First assume that $z \neq 0.$ Since $G \setminus \{a\}$ is open there exists $r \gt 0$ such that $B(z,|z|^{-1}|w|^{-1}r) \subseteq G \setminus \{a\}.$ Choose $0 \lt s \lt \frac {r\ |w|} {r + |z|}.$ Then for any $w' \in B(w,s)$ we have $|w'| \gt \frac {|w|\ |z|} {r + |z|}.$ From here it is easy to show that $\frac {1} {w'} \in B \left(\frac {1} {w},|z|^{-1} |w|^{-1}r\right)$ and hence $$T^{-1} (w') = \frac {1} {w'} + a \in B \left(\frac {1} {w} + a, |z|^{-1} |w|^{-1}r \right) = B(T^{-1} (w),|z|^{-1} |w|^{-1}r) = B(z,|z|^{-1} |w|^{-1}r).$$ Therefore $w' \in T(B(z,|z|^{-1} |w|^{-1}r)) \subseteq T(G \setminus \{a\}).$ This shows that $B(w,s) \subseteq T(G \setminus \{a\}).$

If $z = 0$ then we first note that $a \neq 0.$ Now choose $0 \lt r \lt |a|$ so that $B(0,r) \subseteq G \setminus \{a\}.$ Then if we simply take $0 \lt s \lt \frac {2 |a| - r} {|a| (|a| - r)}$ then $B(w,s) \subseteq T(G \setminus \{a\}).$

Answer 2

Observe, that $T\colon \mathbb C\setminus\{a\}\to \mathbb C\setminus\{0\}$ is continuous and bijective. Further we have $f(z):=T^{-1}(z)=\frac 1z+a$, which is also continuous (and bijective for $f\colon \mathbb C\setminus\{0\}\to \mathbb C\setminus\{a\}$). Then $T(G\setminus\{a\})=f^{-1}(G\setminus\{a\})$ is the preimage of the open set $G\setminus \{a\}$.