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.
 A: Note that $\frac{z}{z-1}$ is a Mobius transformation, and we know that Mobius transformations map lines and circles to lines or circles. So you just need to check where the boundary of your domain is mapped (which consists of two lines), and then find the image of one interior point.
A: 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 
$$\{z: |z-\frac{1}{2}+\frac{1}{2}i| \ge \frac{\sqrt{2}}{2}, \operatorname{Im}z \le 0, \operatorname{Re}z \notin [0,1] \}$$
A: For the sector $\{z=re^{it}\colon 0<r<\infty\quad\text{and}\quad 0\leq t\leq \pi/4\}$, we have in Cartesian coordinates that $y > 0$ and $0<y\leq x$. Then let $z=x+iy$ so 
$$
w = \frac{z}{z-1} = \frac{x+iy}{x-1+iy}.
$$
Since $x,y>0$ and $y\leq x$, 
$$
(x-1)^2+y^2\leq x^2+y^2
$$
Therefore, we now know $x\geq 1/2$ and $y>0$. or we have that $\Im\{z\} > 0$ and $\Re\{z\}\geq 1/2$. Now you have already determined the real and imaginary parts of $z$ to be
\begin{align}
f(u,v) &= \frac{u^2-u+v^2}{(u-1)^2+v^2}\tag{1}\\
g(u,v) &= \frac{-v}{(u-1)^2+v^2}\tag{2}
\end{align}
From $(2)$, we are able to determine $\Im\{w\} < 0$ since 
$$
\frac{-v}{(u-1)^2+v^2} > 0\Rightarrow -v > 0\Rightarrow v < 0
$$
Now, let's look at $\lvert z\rvert^2\geq 1/4$ since $\Re\{z\}\geq 1/2$
$$
0<\frac{1}{4}\leq\frac{(u^2-u+v^2)^2+v^2}{\bigl[(u-1)^2+v^2\bigr]^2}
$$
so $0 < 1/4\leq (u^2-u+v^2)^2+v^2$. Recall that $\Im\{z\} < 0$ so $-v^2<0$ and we get $-v^2 < 1/4\leq (u^2-u+v^2)^2$. Additionally, since $v<0$, $\sqrt{v^2} = -v$ so we get
$$
\frac{1}{2}\leq u^2-u+v^2+iv\Rightarrow \frac{1}{2}\leq \Bigl(u-\frac{1}{2}\Bigr)^2 + \Bigl(v+\frac{i}{2}\Bigr)^2 -\frac{1}{4}+\frac{1}{4}
$$
Therefore, $\frac{1}{\sqrt{2}}\leq\lvert w - 1/2+i/2\rvert$. The domain of $w$ must be 
$$
\bigl\{w\in\mathbb{C}\mid\Im\{w\} < 0\quad\text{and}\quad \lvert w - 1/2+i/2\rvert\geq 1/\sqrt{2}\bigr\}
$$

