<s>[INST] Describe the image of the set $\{z=re^{it}: 0 \leq t \leq \frac{\pi}{4}, 0Describe the image of the set $\{z=re^{it}: 0 \leq t \leq \frac{\pi}{4}, 0<r< \infty\}$ under the mapping $w=\frac{z}{z-1}$.
Here is what I got so far. First I got the reverse function 
$$z= \frac{w}{w-1}=\frac{2u+2iv}{u-1+iv}$$
I did some algebra and I got $x=r \cos t=\frac{u^2 -u+v^2}{(u-1)^2 +v^2}$ and $y=r \sin t = \frac{v}{(u-1)^2 +v^2}$.
For here, I'm not sure how to use the given boundary to find the boundary of the image. I wonder if anyone would give me a hint please.
 [/INST] It's easy to see that the $x$-axis is mapped to itself, but note that the positive part is mapped to $(-\infty,0) \cup (1,\infty)$  
To see what $re^{i\frac{\pi}{4}}$ is mapped to try to work with $f(re^{i\frac{\pi}{4}})$ and get a general formula of a curve in $r$. You should get a circle around $\frac{1}{2}-\frac{1}{2}i$ with radius $r=\frac{\sqrt{2}}{2}$  
Then you should be able to figure out the image of the set.
Update:
You can show that $f(re^{i\frac{\pi}{4}}) = \frac{r^2-r\frac{\sqrt{2}}{2}}{r^2-r\sqrt{2}+1} - i\frac{r\frac{\sqrt{2}}{2}}{r^2-r\sqrt{2}+1}$
So the image curve is represented as: $$h(r)=(\frac{r^2-r\frac{\sqrt{2}}{2}}{r^2-r\sqrt{2}+1},\frac{-r\frac{\sqrt{2}}{2}}{r^2-r\sqrt{2}+1})$$
Taking the derivative gives you the tangent to the curve:
$$
h'(r) = (\frac{-\frac{\sqrt{2}}{2}r^2 + 4r-\frac{\sqrt{2}}{2}}{(r^2-r\sqrt{2}+1)^2},\frac{\frac{\sqrt{2}}{2}r^2-\frac{\sqrt{2}}{2}}{(r^2-r\sqrt{2}+1)^2})
$$
Since you know it's a circle (Mobius), looking for the tangents parallel to the $x$-axis can give you $2$ points with distance $2r$ with the center in the middle. The tangent is parallel where the second term is zero which is at $r=1,-1$
So $h(1) = \frac{1}{2}-\frac{i}{2\sqrt{2}-2}, h(-1)=\frac{1}{2}+\frac{i}{2\sqrt{2}+2}$
The distance between these points is $\sqrt{2}$ and the point $z=\frac{1}{2}-\frac{1}{2}i$ is right in the middle.
So in total $f(re^{i\frac{\pi}{4}}) = \{z: |z-\frac{1}{2}+\frac{1}{2}i|=\frac{\sqrt{2}}{2}\}$ 
Putting it all together:
The $(0,\infty)$ is mapped to $(-\infty,0) \cup (1,\infty)$ but in the other direction, meaning points in the upper half plane are mapped to the lower half plane.
$re^{i\frac{\pi}{4}}, r \gt 0$ is mapped to the part in the circle $\{z: |z-\frac{1}{2}+\frac{1}{2}i|=\frac{\sqrt{2}}{2}\}$ from $0$ to $1$ going counter clockwise, i.e. in the lower half plane, as $r$ increases, therefore points to the right of $re^{i\frac{\pi}{4}}, r \gt 0$ are mapped to points outside the circle.
And so in total we have that the image is points in the lower half plane outside the circle around $\frac{1}{2}-\frac{1}{2}i$ with radius $\frac{\sqrt{2}}{2}$ but where the real part is not between $[0,1]$ or 
$$\{z: |z-\frac{1}{2}+\frac{1}{2}i| \ge \frac{\sqrt{2}}{2}, \operatorname{Im}z \le 0, \operatorname{Re}z \notin [0,1] \}$$
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