The Holomorphically Convex Hull is defined as $\hat{K}_\Omega= \{z \in \Omega: |f(z)| \leq \sup_K |f|, \forall f\in \mathcal{O}(\Omega)\}$, where $\Omega\underset{open}\subset \mathbb{C}^n$, $K\underset{compact}\subset\Omega$.
Suppose $\Omega\subset\mathbb{C}^2$ is given by $$ \{(z_1,z_2)\in\mathbb{C}^2:|z_1|<1,|z_2|<1\}-\left\{(z_1,z_2)\in\mathbb{C}^2:|z_1|\leq \frac{1}{2},|z_2|\leq\frac{1}{2}\right\} $$ I am asked to find the holomorphically convex hull of the compact set $$ K=\left\{ (z_1,z_2)\in\Omega : z_1=0,|z_2|=\frac{3}{4}\right\} $$
I have no idea on how to compute this. Of course, I can describe $K$ as the image of the curve $$ \gamma(\theta)=\left(0,\dfrac{3}{4}e^{2\pi i\theta}\right),\theta\in\mathbb{R} $$ but how can I find $z$ such that $|f(z)| \leq \sup_{\theta\in\mathbb{R}} |f(\gamma(\theta))|$, for an arbitraty $f$?
Thanks in advance.
First of all, taking $f(z_1,z_2)=z_1$ shows that $\hat K \subseteq \{z_1=0\}$, and taking $f(z_1,z_2)=z_2$ shows that $\hat K\subseteq \{ |z_2| \le3/4\}$. In fact, the hull is exactly $$L = \Omega \cap \{ z_1=0, |z_2|\le 3/4 \}. $$ To prove this, I'm going to assume that you know Hartogs' extension theorem. (Which makes it a lot easier). By Hartogs, every $f\in\mathcal{O}(\Omega)$ extends to a holomorphic function on the bidisc. In particular, the restriction of $f$ to $z_1=0$ is holomorphic on the unit disc. The one-variable maximum modulus principle then implies that $|f(z)|\le \sup_K |f|$ for every $z\in L$ and we are done.
Hence, the exercise shows that $\Omega$ is not holomorphically convex. In fact, it turns out that holomorphic convexity is equivalent to pseudoconvexity which in turn is equivalent to the domain being the domain of existence for some holomorphic function. For contrast, try computing the hull with respect to $\Omega = \{ |z_1|<1, \frac 12 <|z_2|<1 \}$.