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

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.

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