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Is the set $\mathbb{C}^2 \setminus\{z_1=0\}$ connected?

My try:

Let points $p=(p_1,p_2)$ i $q=(q_1,q_2)$ such that $p_1=a_1+ib_1=r_1e^{i \phi_1}$ and $q_1=c_1+id_1=r_2e^{i \phi_2}$, $r_i>0$. Then $((tr_1+(1-t)r_2)e^{i(t \phi_1+(1-t) \phi_2},tp_2+(1-t)q_2)$ is path between $p$ i $q$ and doesn't intersect $z_1=0$ because $r_1,r_2>0$ and $t$ or $(1-t)>0$.

Does this make sense?

Can this be generalized for more lines or higher dimensions of complex space?

Anne Bauval
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stranger
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    your proof is fine. For a generalization, see this duplicate: Product of path connected spaces is path connected $\Bbb C^2\setminus{z_1=0}=\Bbb C\times(\Bbb C\setminus{0})$ and $\Bbb C\setminus{0}$ is homeomorphic, via polar coordinates, to $(0,+\infty)\times S^1$). – Anne Bauval May 31 '23 at 21:19
  • What about more lines? For finite set of lines I suppose it is ok, but can't tell why. For infinite maybe not? – stranger May 31 '23 at 21:24
  • Consider a complex line between these two points. It meets ${z_1=0}$ in at most one point. You can safely connect these two pint by a path in the complex line (a plane) avoiding this point. This argument works for a finte set of lines, or any algebraic set $(P(z,w)=0$, because a finite set do not disconnect a 2-plane (complex line).. – Thomas Jun 01 '23 at 10:00

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