Let $G_1$ and $G_2$ be the images of the disc $\{z\in \mathbb C:|z+1|<1\}$ under the transformations I was thinking about the following problem:
Let $G_1$ and $G_2$ be the images of the disc $\{z\in \mathbb C:|z+1|<1\}$ under the transformations $w=\frac{(1-i)z+2}{(1+i)z+2}$ and $w=\frac{(1+i)z+2}{(1-i)z+2}$ respectively. Then which of the following statement is correct?

(a)  $G_1=\{w\in \mathbb C:Im(w)<0\}$ and $G_2=\{w\in \mathbb C:Im(w)>0\},$
(b)  $G_1=\{w\in \mathbb C:Im(w)>0\}$ and $G_2=\{w\in \mathbb C:Im(w)<0\},$
(c)  $G_1=\{w\in \mathbb C:|w|>2\}$ and $G_2=\{w\in \mathbb C:|w|<2\},$
(d)  $G_1=\{w\in \mathbb C:|w|<2\}$ and $G_2=\{w\in \mathbb C:|w|>2\}.$

I was trying to express $z$ in terms of $w$ and then put the value in the relation $|z+1|<1$  and replace $w$ with $u+iv$ where $u$=$Re(w)$ and $v$=$Im(w)$. But the whole process in lengthy and also i could not reach desired result. Please help.Thanks in advance for your time.
 A: Let $$f_1(z) := \frac{(1-\imath) \cdot z+2}{(1+\imath) \cdot z+2} \qquad \qquad f_2(z) := \frac{(1+\imath) \cdot z+2}{(1-\imath) \cdot z+2}$$
Now consider $z=-1$ (which is obviously in the disc $\{z \in \mathbb{C}; |z+1|<1\}$). We obtain by straight-forward calculations
$$f_1(-1) = \frac{(1-\imath) \cdot (-1)+2}{(1+\imath) \cdot (-1)+2} = \frac{\imath+1}{-\imath+1} = \frac{(\imath+1)^2}{1-\imath^2} = \ldots = \imath$$
which means that only b) or d) can be true. On the other hand
$$f_2(-1)= \frac{(1+\imath) \cdot (-1)+2}{(1-\imath) \cdot (-1)+2} = \frac{-\imath+1}{\imath+1} = \frac{1}{f_1(-1)} = \frac{1}{\imath} =-\imath$$
(in particular $|f_2(-1)|<2$), thus only b) can hold.

Edit: So if one wants to prove that b) holds, one can do it like that: Observe that $\text{Im} \, w>0 \Leftrightarrow |w-\imath| < |w+\imath|$. We apply this to $w:=f_1(z)$:
$$\text{Im} \, f_1(z) > 0 \Leftrightarrow \left| \frac{(1-\imath) \cdot z+2}{(1+\imath) \cdot z+2}-\imath \right| < \left| \frac{(1-\imath) \cdot z+2}{(1+\imath) \cdot z+2}+\imath \right| \\ \Leftrightarrow |\underbrace{(1-\imath) \cdot z+2 - \imath \cdot ((1+\imath) \cdot z+2)}_{2 (1-\imath) \cdot z+2-2\imath} | < |\underbrace{(1-\imath) \cdot z+2 + \imath \cdot ((1+\imath) \cdot z+2)}_{2+2\imath}| \\ \Leftrightarrow |2 \cdot (1-\imath) \cdot (z+1)| < |2 \cdot (1+\imath)| \\ \Leftrightarrow |z+1|<1$$
thus the first claim of b) holds. Similar proof works for the second part.
A: The image of the circle under $f_1$ is a line since $f_1(-1+i)=\infty$. The line is the real axis since $f_1(-2)=-1$, $f_1(-1-i)=0$ and $f(0)=1$.  When we trace the circel in that order the region is on our left, therefore the image of the disc is the upper half plane.
The image of the circle under $f_2$ is a line since $f_2(1-i)=\infty$. The line is the real axis since $f_2(-2)=-1$, $f_2(-1+i)=0$ and $f(0)=1$.  When we trace the circel in that order the region is on our right, therefore the image of the disc is the lower half plane. 
The correct answer is (b)
