# Is $\sin(z^3)$ bounded? [closed]

I know $\sin x$ is bounded function if $x$ is real. but what about if $x$ is a complex number? moreover, what about $\sin(z^3)$, where $z$ is a complex number? I tried this by Taylor's series but was not successful.

## closed as off-topic by Jack D'Aurizio, Watson, Daniel W. Farlow, Davide Giraudo, user1551May 12 '16 at 14:18

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• A complex analytic function is bounded if and only if it is a constant function. – BigbearZzz May 12 '16 at 11:24

Since $\sin z$ and $z^3$ are analytic, $\sin z^3$ is also analytic. By Liouville's theorem, a bounded function analytic on the whole $\mathbb{C}$ is constant. Since $\sin z^3$ is not constant, it is unbounded.
• Thank you....what about if function is not analytic e.g $\frac{sin(z)}{cos(z)+1}$ – Arun Sharma May 12 '16 at 11:43
• @ArunSharma For this specific case, note that $\cos z$ can be arbitrarily close to $-1$, so the denominator can be made arbitrarily small, and the whole function is unbounded. In general, Liouville's theorem holds for meromorphic functions, too (and your function is meromorphic): math.stackexchange.com/questions/901490/… – lisyarus May 12 '16 at 11:51
• @ArunSharma: it is even worse: such a function has a simple pole at $z=\pi$. – Jack D'Aurizio May 12 '16 at 11:52
• @ArunSharma: The zeros of $\cos(z) + 1$, namely $z = (2k + 1)\pi$ for integer $k$, are also zeros of $\sin z$, so $g(z) = \frac{\sin z}{\cos z + 1}$ has an analytic extension by the Riemann extension theorem, and may therefore be regarded as a non-constant entire function. (If that happened not to be the case, the function would instead have isolated singularities in the complex plane, and not even be locally bounded.) – Andrew D. Hwang May 12 '16 at 11:57
• @AndrewD.Hwang Actually, since $\sin z \approx z$ and $\cos z \approx 1 - \frac{z^2}{2}$, the whole fraction will be $\approx \frac{1}{z}$, without analytic extension. – lisyarus May 12 '16 at 12:54