Problem 3:Consoder the map $z \xrightarrow\ w = f(z) = \frac{1}{z}$ and compute the image $f(D)$ of $D = \{z:|z-1|>1, |z-2|<2\}$ is mapped to.
Check if $f:D\xrightarrow\ f(D)$ is a bijection.
Answer:
Let $z=x+iy$ and $w=u+iv$.
$w=f(z)+\frac{1}{z}$, then $z=\frac{1}{w}$.
For
\begin{align*}
|z-1|&>1\\
|\frac{1}{w}-1|&>1\\
|1-w|&> |w|\\
|w-1|^2 &> |w|^2\\
|w|^2 +|1|^2 - 2Re(w \cdot 1) &> |w|^2\\
|w|^2 +1 - 2Re(w) &> |w|^2\\
1 - 2Re(w) &>0\\
2Re(w) &< 1\\
Re(w) &< \frac{1}{2}
\end{align*}
Hence, we have $u<\frac{1}{2}$ since $w=u+iv$.
For
\begin{align*}
|z-2|&<2\\
|\frac{1}{w}-2|&<2\\
|1-2w|&<2|w|\\
|2w-1|^2&<4|w|^2\\
|2w|^2 + |1|^2 - 2Re(2w \cdot 1) &< 4|w|^2\\
4|w|^2 + 1 - 4Re(w) &< 4|w|^2\\
1 - 4Re(w) &<0\\
4Re(w) &> 1\\
Re(w) &>\frac{1}{4}
\end{align*}
Hence, we have $u>\frac{1}{4}$ since $w=u+iv$.
Now, we have $\frac{1}{4}<u<\frac{1}{2}$.
The figures of domain and codomian of f are attached below. For all $w$ in $f(D)$, there exists $z$ in $D$ such that $w=\frac{1}{z}$. Thus, $f$ is onto.
Then, we will show f is one to one.
Consider $z_1,z_2 \in D,$ and $f(z_1)=f(z_2)$,
Then,
\begin{align*}
\frac{1}{z_1} &= \frac{1}{z_2} \ \ \ \ \text{$z_1,z_2 \neq 0$}\\
z_1 &= z_2\\
\end{align*}
Now, we have $f$ is one to one.
Therefore, $f:D \xrightarrow\ f(D)$ is bijection.
