Consider the complex function $f(z) = \sin{z}.$ It is analytic and bounded (max $|f(z)| = 1$). However, it's not constant. This seems to disprove Liouville's Theorem. What did I go wrong here?
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13bounded when $z$ is in the real line. When $z$ is pure imaginary, say $z=it$ for real $t,$ it is $ i \sinh t$ – Will Jagy Jan 23 '24 at 22:49
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2$|\sin(i)|=1.17\ldots$ – jjagmath Jan 23 '24 at 22:59
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$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$ – agent_cracker103 Jan 23 '24 at 23:52
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Good catch. This is an important insight. The complex sin is not bounded. – calc ll Jan 24 '24 at 03:36