Show that arbitrary $(z_1,z_2)$ in the Hartogs' triangle $$\triangle_H = \{(z_1,z_2) \in \Bbb C^2: |z_1| < |z_2| < 1\}$$ can be joined by a path (in $\triangle_H$) to $w = (0,r)$ for some fixed $r\in \mathbb R \cap (0,1)$ in each of the following cases:
- $0 < r \le |z_1| < |z_2| < 1$
- $0\le |z_1| \le r \le |z_2| < 1$
- $0\le |z_1| < |z_2| \le r < 1$
Conclude that $\triangle_H$ is an open connected subset of $\mathbb C^2$.
I don't see why it is required to consider all the three cases above to prove the path connectedness (which implies connectedness) of $\triangle_H$. If we can show the requirement for any one case, we will have a paths $\gamma_1, \gamma_2$ between $(z_1,z_2)$ & $(0,r)$, and $(z_1',z_2')$ & $(0,r)$ respectively, for arbitrary $(z_1,z_2), (z_1',z_2') \in \triangle_H$. Then, we can simply consider $\gamma_1 \ast \overline{\gamma_2}$ to get a path from $(z_1,z_2)$ to $(z_1',z_2')$ in $\triangle_H$. In the first case, I thought of considering $\gamma: I = [0,1]\to \mathbb C^2$ given by $\gamma(t):= ((1-t)z_1, (1-t)z_2 + tr)$ but it doesn't seem to be true that $\gamma(I) \subset \triangle_H$ in this case.
How should I construct the required paths in each of the three cases? Thanks!
Best Answer
These cases don't look particular useful to me. Just use polar coordinates $z_k=r_k e^{i\varphi_k}; k=1,2$. Rotating the complex point doesn't change the condition $|z_1|<|z_2|$. The map $t\mapsto (r_1 e^{i\varphi_1\cdot t},r_2 e^{i\varphi_2\cdot t})$ defines a path from $(r_1,r_2)$ to $(z_1,z_2)$. Two real point in Hartog's Triangle are easily connected by a path.