Hartogs triangle

A non-trivial example of domain of holomorphy that has someinterestingnon-obvious properties is the Hartogs triangle which is the set

Since it is a Reinhardt domain it can be representedby plotting it on the plane $|z|\times |w|$as follows.

Figure 1: Hartogs triangle

It is obvious then where the name comes from. To see that this is a domain of holomorphy, then given a boundary point we wish to exhibit a holomorphic function on the whole Hartogs triangle which does not extend beyond that point. First note that on the top boundary $z$ is anything and $w=e^{i\theta }$ for some $\theta$, so$f(z,w)=\frac{1}{w-e^{i\theta }}$ will not extend beyond $(z,e^{i\theta })$.Now for the diagonal boundary this is where $|z|=|w|$,that is $z=e^{i\theta }w$ for some $\theta$, so$f(z,w)=\frac{1}{z-e^{i\theta }w}$ will do not extend beyond $(e^{i\theta }w,w)$.

One of the many properties of this domain is that if $U$ is the Hartogstriangle, then it is a domain of holomorphy, but if we take a sufficentlysmall neighbourhood $V$ of $\overline{U}$ (the closure of $U$),then any function holomorphic on $V$ is holomorphic on the polydisc$D^2(0,1)$ (just fill in everything below the triangle in Figure 1). So if $V$ does not include all of $D^2(0,1)$ then it is not a domain ofholomorphy. This is because a Reinhardt domain that contains zero (the point$(0,0)$) is a domainof holomorphy if and only if it is a logarithmically convex set and any neighbourhood of $\overline{U}$ does contain zero while $U$ itself does not.