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Why does $f(z)=\cos(z^2)$ not contradict Liouville's theorem?

Is the best approach to put $\cos(z^2)$ into its Taylor expansion?

How can I visualize $\cos(z^2)$?

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    Just pick a trajectory which gives cosh, which happens along the imaginary line for cos. – 2017-02-27
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    See the question that is posed at http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics. – 2017-02-27

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Because in $\mathbb{C}$, $\cos$ and $\sin$ are not bounded functions like they are in $\mathbb{R}$. In particular, $\cos(ix)=\cosh(x)$, so $\cos$ grows exponentially on the imaginary axis.

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    Well, they are nice, they just aren't bounded (indeed, the content of Liouville's theorem is that if a function is nice enough and is not completely boring, then it must be unbounded). – 2017-02-27
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    @WillR They are nice in the sense of regularity, but their unboundedness makes them "not nice" for certain uses (e.g. in contour integration). – 2017-02-27
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    @WillR: agree, the constants have all the nice properties we need and yet are boring enough. :) – 2017-02-27