Is there $a,b \in \mathbb C^2$, $b \neq 0$ such that if $t \in \mathbb C, |t|<1$, then $|a+tb|=1$?

I think there shall be not since $|a+tb|=1$ seems to be a closed condition while $t < 1$ is an open condition.

  • $\begingroup$ Is it t < 1 or |t| < 1 ? $\endgroup$ – mathfan27543 Apr 10 '17 at 0:23
  • $\begingroup$ $|t| <1$ and $t$ is in $\mathbb C$ $\endgroup$ – Keith Apr 10 '17 at 0:24
  • $\begingroup$ $|a|=1, b=0$... $\endgroup$ – user251257 Apr 10 '17 at 0:26
  • $\begingroup$ sorry I should add that $b \neq 0$ $\endgroup$ – Keith Apr 10 '17 at 0:27

Assume such $a$ and $b$ existed, then the map $f: t \mapsto a+tb$ would be a holomorphic map from $D^2$ to $\mathbb{C}$, whose image is however $S^1$ and hence not open in $\mathbb{C}$. By the open mapping theorem, $f$ must be constant which contradicts $b \neq 0$, so no, there's no such pair $(a,b)$.


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