My attempt: I was trying to apply identity theorem that if the zeros of the function do have any limit point and it will be a zero function but setting $g(z)=f(z)-1$ will not help me. Can anyone help me in that!!
2026-03-25 13:57:04.1774447024
$f$ is an entire function s.t $|f(z)|=1$ $\forall z \in \Bbb R$. prove that $f$ has no zeros in $\Bbb C$
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Since $f$ is entire, so is the function $g$ defined by $g(z)=\overline{f(\overline z)}.$ The hypothesis gives $g(z)=1/f(z)$ for $z\in\mathbb R,$ which forces $g=1/f$ by the identity principle. So $f$ is invertible.