wdnaeng
commented
8 months ago Also, a request for clarification on the third example. If $S_3$ is the closed deleted neighborhood, how is it that its interior contains $z_0$ (is equal to the open neighborhood)?
Should it maybe read $\text{Int } S3=B{\frac{1}{2}}\left(2+\sqrt{3}i\right)\setminus \lbrace2+\sqrt{3}i\rbrace$ ?
My understanding of the previous section is that $\text{Int}$, $\text{Ext}$, and $\partial$ have no overlap, but you write that $$\partial S_3=\lbrace z:|z-\left(2+\sqrt{3}i\right)|=\frac{1}{2}\rbrace \cup \lbrace2+\sqrt{3}i\rbrace$$ which contains $z_0$.
Am I misunderstanding something fundamental?
jcponce
The last two examples of the classification section state that $$S2=\overline{B}{\frac{7}{8}}\left(-1-\sqrt{2}i\right)$$ and
$$S3=\overline{B}{\frac{1}{2}}\left(2+\sqrt{3}i\right)\setminus \lbrace2+\sqrt{3}i\rbrace $$
where $$\text{Int } S2=B{\frac{7}{8}}\left(-1-\sqrt{2}\right)$$ and $$\text{Int } S3=B{\frac{1}{2}}\left(2+\sqrt{3}\right)$$
Am I misunderstanding the math or are we missing an i in the Int S statements?
Loving the course so far as someone that learned just enough z-plane to be dangerous in signal processing in college but had forgotten a lot of it since ;)