<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] 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 
$$
\bigl\{w\in\mathbb{C}\mid\Im\{w\} < 0\quad\text{and}\quad \lvert w - 1/2+i/2\rvert\geq 1/\sqrt{2}\bigr\}
$$

</s>