[Math] Are the sets $Re(z^2)>0$ and $\left | arg(z) \right |<\pi /4$ domains

complex-analysis

So I was doing some practice problems on complex analysis and got confused on this problem.

I know that a domain is an open connected set, which in other words means a set with every point being interior and the set having "no holes".

  • Now, we can write the first equation $Re(z^2)>0$ as
    $$x^2-y^2>0$$
    which is a hyperbola as in the figure below

enter image description here

and as written on the figure, this is not a domain. (Because there is no neighborhood of the point $0$ that lies entirely inside the set – right?)

  • However, for $\left | arg(z) \right |<\pi /4$, we have the
    $$-\pi/4<arg(z)<\pi /4$$
    which is the region between the rays $arg(z)=\mp \pi /4$

enter image description here

which I don't understand why is it considered a domain? Isn't the same case as above, with the point $0$ not having any neighborhood that lies entirely inside the set, and hence not considered an interior point and therefore making the set not open (and hence not a domain)?

By the way, these figure are taken from the solution manual of Zill's
book: "Advanced Engineering Mathematics, 4th edition"

Best Answer

Your reason why the first set is not a domain is wrong.

$0$ is not a member of the set at all, since $\operatorname{Re}(0^2)=0\not>0$. Therefore it doesn't matter that it has no neighborhood fully within the set.

Instead the first set fails to be a domain because it is not connected.

Whether or not the second set is a domain depends on whether $0$ is a member. As written, the set is ill-defined because the condition $|\operatorname{arg} z|<\pi/4 $ is undefined for $z=0$ because $0$ doesn't have an argument.

If you specify explicitly that $0$ is not in the set, then it is a domain.

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