Conformal mapping on unit disk

Question

Let $\mathbb{C}_{-}^r:=\mathbb{C}\setminus(]-\infty,-\frac{1}{r}])$, $~\mathbb{C}_{-}:=\mathbb{C}\setminus(]-\infty,0])$, $~\mathbb{D}:= \{z\in \mathbb{C}\mid |z|<1\}$ and $\mathbb{G}=\{z\in \mathbb{C}\mid Re(z)>0\}$
I have to find a conformal map from $\mathbb{C}_{-}^r$ to $\mathbb{D}$ for any $r>0$
My try:

1. Apply translation $f_{1}(z):= z+\frac{1}{r} \Rightarrow f_{1}(\mathbb{C}_{-}^r) = \mathbb{C}_{-}$
2. Apply $f_{2}(z)=\sqrt{z} \Rightarrow f_{1}(\mathbb{C}_{-})=\mathbb{G}$
3. Apply $f_{3}(z)=\frac{z-1}{z+1} \Rightarrow f_{1}(\mathbb{G})=\mathbb{D}$

Therefore the conformal map $$f(z)=(f_{3} \circ f_{2} \circ f_{1})(z)=\frac{\sqrt{z+\frac{1}{r}}-1}{\sqrt{z+\frac{1}{r}}+1}$$ but the official answer is $$f(z) =\frac{\sqrt{rz+1}-1}{\sqrt{rz+1}+1}$$ What am I missing ?

Answer 1

The mapping $\varphi (z) =r\left(z+\frac{1}{r} \right)$ is also a biholomorphic mapping from $\mathbb{C}^r_- $ onto $\mathbb{C}_- $. So both the answers are good.