currently I ask myself the following question: Suppose that one has a conformal mapping given by $g_1(z)=z+\frac{2}{z-2\sqrt{2}}$ where the domain is given by $\mathbb C - D(2\sqrt{2},\sqrt{2})$, then apparently the image of this map $\mathbb C - [0,4\sqrt{2}]$?
My reasoning: I believe I understand the easier part where we know that $g(z)=z+\frac{R^2}{z}$ on the domain $\overline{\mathbb{H} \cap B(0,R)}$ then the image of this map should be $\mathbb{H} \cup [-2R,2R]$ and since $g$ maps the real part of the domain to the real part of the domain we can extend such that $g : \overline{\mathbb C \cap B(0,R)}$ is a conformal mapping that is onto $\mathbb{H} \cup - \mathbb{H} \cup [-2R,2R]$ now translating the map by $2\sqrt{2}$ gives that by a similiar reasoning that the image of $g_1$ is given by $\mathbb{C} - [-2R+2\sqrt{2},2R+2\sqrt{2}]$ by the above, hence plugging in $R = \sqrt{2}$ gives the desired answer.
Question: Is this reasoning correct,how can maybe one argue better that the translation in the dominator in the map resolves into a translation of the set of not hitted points on the real line?