Finding the image of a disk Find the image of the disk $D= \{ z \in \mathbb{C}: |z|<1 \}$ under the mapping $z \to w=\frac{1}{z-1}$.
My attempt:
First I graphed the boundary. 
When $z=i, f(z)=-i/2-1/2$
When $z=-i, f(z)=i/2-1/2$
When $z=-1, f(z)=-1/2$
When $z=1, f(z)$ is undefined.
Then I graphed a few other points, $z=-1,0,\frac{1}{2}$ and saw that everything is being mapped to the left hand side of the the vertical line $x<-\frac{1}{2}$, is this statement true? 
 A: The mapping you give is an example of a fractional linear transformation. It is a fact (which I will not prove here) that such transformations map (generalized) circles to (generalized) circles. Thus, it is clear from your calculations that your map takes the unit circle to the line $\text{Re}(z) = -\frac{1}{2}$. Your other points show that indeed this map takes $D$ to the half plane left of this line.
A: Say, $z=x+iy$, then $f(z)=f(x+iy)$ becomes $$\frac{x-1}{(x-1)^2+y^2}-\frac{i y}{(x-1)^2+y^2}$$
Note that in $D$, $x$ and $y$ are between $0$ and $1$.
Now considering the real part we have $(x-1)^2+y^2 = x^2 +1 -2x +y^2 < 2-2x$. It follows $$\frac{1}{x^2 +1 -2x +y^2} > \frac{1}{2-2x}$$ which implies $$\frac{x-1}{x^2 +1 -2x +y^2} < \frac{x-1}{2-2x} $$ since $x-1$ is negative. It now follows that the real part is less than $ \displaystyle -\frac {1} {2}$
A: When i solved. I got the result that the boundary of the disc maps to the line x=1/2. Did you try the points at random? Didn't you use a systematic approach?
$Hint :$
Try parameterizing $1/(z-1)$.
Replace $z$ by $ exp(\iota \theta) $
