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Sketch the following sets in the complex plane $\mathbb{C}$ and determine whether they are open, closed, or neither; bounded; connected. Briefly state your reason.
(a) $|z+3|<1$;
(b) $|\operatorname{Im}(z)| \geq 1$;
(c) $1 \leq|z+3|<2$.
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(a) Since $\{|z+3|<1\}=\left\{(x+3)^{2}+y^{2}-1<0\right\}$ and $f(x, y)=(x+3)^{2}+$
$y^{2}-1$ is a continuous function on $\mathbb{R}^{2}$, the set is open. It is not closed since the only sets that are both open and closed in $\mathbb{C}$ are $\emptyset$ and $\mathbb{C}$. Since
$$|z|=|z+3-3| \leq|z+3|+|-3|=|z+3|+3<4$$
for all $|z+3|<1,\{|z+3|<1\} \subset\{|z|<4\}$ and hence it is bounded. It is connected since it is a convex set.

(b) We have
$$\{|\operatorname{Im}(z)| \geq 1\}=\{|y| \geq 1\}=\{y \geq 1\} \cup\{y \leq-1\}$$

Since $f(x, y)=y$ is continuous on $\mathbb{R}^{2}$, both $\{y \geq 1\}$ and $\{y \leq-1\}$ are closed and hence $\{|\operatorname{Im}(z)| \geq 1\}$ is closed. It is not open since the only sets that are both open and closed in $\mathbb{C}$ are $\emptyset$ and $\mathbb{C}$. Since $z_{n}=n+2 i \in\{|\operatorname{Im}(z)| \geq 1\}$ for all $n \in \mathbb{Z}$ and
$$\lim _{n \rightarrow \infty}\left|z_{n}\right|=\lim _{n \rightarrow \infty} \sqrt{n^{2}+4}=\infty,$$
the set is unbounded. The set is not connected. Otherwise, let $p=2 i$ and $q=-2 i$. There is a polygonal path
$$\overline{p_{0} p_{1}} \cup \overline{p_{1} p_{2}} \cup \ldots \cup \overline{p_{n-1} p_{n}}$$
with $p_{0}=p, p_{n}=q$ and $p_{k} \in\{|\operatorname{Im}(z)| \geq 1\}$ for all $0 \leq k \leq n .$
Let $0 \leq m \leq n$ be the largest integer such that $p_{m} \in\{y \geq 1\}$. Then $p_{m+1} \in\{y \leq-1\}$. So $\operatorname{Im}\left(p_{m}\right) \geq 1>0$ and $\operatorname{Im}\left(p_{m+1}\right) \leq-1<0$. It follows that there is a point $p \in$ $\overline{p_{m} p_{m+1}}$ such that $\operatorname{Im}(p)=0$. This is a contradiction since $\overline{p_{m} p_{m+1}} \subset\{|\operatorname{Im}(z)| \geq 1\}$
but $p \notin\{|\operatorname{Im}(z)| \geq 1\}$. Therefore the set is not connected.

Solution. (c) Since $-2 \in\{1 \leq|z+3|<2\}$ and $\{|z+2|<r\} \not \subset\{1 \leq|z+3|<2\}$
for all $r>0,\{1 \leq|z+3|<2\}$ is not open. Similarly, $-1$ is a point lying on its complement
$$\{1 \leq|z+3|<2\}^{c}=\{|z+3| \geq 2\} \cup\{|z+3|<1\}$$
and $\{|z+1|<r\} \not \subset\{1 \leq|z+3|<2\}^{c}$ for all $r>0$. Hence $\{1 \leq|z+3|<2\}^{c}$ is not open and $\{1 \leq|z+3|<2\}$ is not closed. In summary, $\{1 \leq|z+3|<2\}$ is neither open nor closed. Since
$$|z|=|z+3-3| \leq|z+3|+|-3|<5$$
for all $|z+3|<2,\{1 \leq|z+3|<2\} \subset\{|z|<5\}$ and hence it is bounded. The set is connected. To see this, we let $p_{1}=-3 / 2, p_{2}=-3+3 i / 2, p_{3}=-9 / 2$ and $p_{4}=-3-3 i / 2$. All these points lie on the circle $\{|z+3|=3 / 2\}$ and hence lie
$$\text { in }\{1 \leq|z+3|<2\} \text { . }$$
It is easy to check that for every point $p \in\{1 \leq|z+3|<2\}, \overline{p p_{k}} \subset\{1 \leq|z+3|<2\}$
for at least one $p_{k} \in\left\{p_{1}, p_{2}, p_{3}, p_{4}\right\}$. So the set is connected.
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