We are currently studying Liouville's Theorem and had a few questions which weren't too hard, but I'm kinda struggling with this one. $f$ is entire but $f(z)\notin \left[0,1 \right]$. I tried to play around and prove f is bounded, but nothing seems to work. Would love even a hint :)
2026-03-25 13:56:17.1774446977
Complex Analysis : Prove f is constant if f is entire and $f(z)\notin \left[0,1 \right]$
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Hint: first apply some function $g$ so that $g\circ f$ doesn't take value in some ray going to infinity. Then apply some function $h$ so that $h\circ g\circ f$ doesn't take values n some nonempty open set (try to use a square root somewhere). Finally transform it into a function which is bounded, hence constant.