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The function is entire and it is also bounded since $|e^{iz}| = 1$ for all $z$ due to Euler's formula. By Liouville's theorem, it should be constant, but it's not; this does not lead to a contradiction as Liouville's theorem does not apply to functions with a constant modulus.

The function is entire as a polynomial, but it is not bounded since as $|z| \to \infty$, $|z^3 + 2z + 1|$ also goes to infinity. This does not contradict Liouville's theorem because the theorem states that only bounded and entire functions must be constant.

The function $\tan(z)$ is not entire due to poles at $z = (2n+1)\frac{\pi}{2}$ for $n \in \mathbb{Z}$. It is also not bounded as it takes arbitrarily large values near its poles.

This function is not entire due to a singularity at $z = 0$. It is not bounded because the value of $cosh(\frac{1}{z})$ can become arbitrarily large as $z$ approaches zero. Liouville's theorem does not apply to functions with singularities.

The function is not entire as it contains a branch cut and is additionally not defined for $z=0$. Hence, it is not bounded over the entire complex plane and does not contradict Liouville's theorem.

The function is entire but not bounded, as $|z^2|$ can grow without bound when $|z| \to \infty$. Therefore, it does not contradict Liouville's theorem, which states that an entire and bounded function must be constant.

The function has a removable singularity at $z = 0$ and is entire after the singularity is removed. However, it is not bounded since $|\frac{sin(z)}{z}|$ can become arbitrarily large for large imaginary values of $z$. Liouville's theorem does not apply here because the function is not both entire and bounded.

The function is not bounded since the modulus $|z|$ can become arbitrarily large. It is entire, but its unboundedness confirms Liouville's theorem.

This function is entire and it is also bounded since $0 < |\frac{1}{1 + e^z}| \leq 1$. However, by Liouville's theorem, an entire and bounded function must be constant, which this function is not. This presents no contradiction as the function actually does vanish for some $z$ in the complex plane.

The function is not bounded as $|e^z|$ can grow without bound. According to Liouville's theorem, it cannot be entire and bounded.

The function is entire because it is composed of entire functions (polynomials and the exponential function), but it is unbounded due to the exponential term. As $|z| \to \infty$, $|e^z|$ increases rapidly, which means the function does not satisfy the boundedness condition for Liouville's theorem.

The function is not bounded since $|e^{z^2}|$ can become arbitrarily large as $|z| \to \infty$. It is entire, but not bounded and hence does not contradict Liouville's theorem.

This function is not entire due to a singularity at $z = i$. It is also unbounded as it approaches infinity when $z$ approaches $i$. Such a function does not disobey Liouville's theorem, which applies to functions that are both entire and bounded.

This is an entire function due to the removable singularity at $z = 0$. However, it is not bounded because $|(e^z - 1)/z|$ can become arbitrarily large, especially along lines in the complex plane where the imaginary part of $z$ is large.

This function is not entire as it is multi-valued and has a branch point at $z = 0$. It is bounded for $|z| = 1$ but unbounded as $|z|$ approaches infinity. The function does not contradict Liouville's theorem since it is not entire.

The function is entire as both $sin(z)$ and $e^z$ are entire functions. However, it is not bounded because the modulus $|e^{\sin(z)}|$ can be made arbitrarily large for certain values of $z$ in the complex plane. This does not conflict with Liouville's theorem which asserts that an entire and bounded function must be constant.

This function is not entire as it has singularities where $z^2 + 1 = 0$. However, it is bounded on the entire plane due to the denominator growing faster than the numerator.

The function is not bounded because $|cos(z)|$ can be made arbitrarily large for some imaginary parts of $z$. It is entire but not bounded.

The function has singularities at $z = 1$ and $z = -1$, so it is not entire and therefore does not need to be bounded, according to Liouville's theorem.

This is an entire function that is bounded in the entire complex plane, approaching zero as $|z| \to \infty$. Liouville's theorem tells us that it must be a constant function, but in this case, the function is not constant. This doesn't contradict Liouville because boundedness implies constant only if the function is also non-vanishing.

The function is not entire due to poles at $z = i$ and $z = -i$. However, on the complex plane without these points, the function is bounded.

The function is not entire as ar{z} is not differentiable everywhere in the complex plane. However, it is bounded since $|e^{\bar{z}}| = |e^{Re(z)}|$ is bounded. Liouville's theorem does not apply as the function is not entire.

This function is not entire as it has a singularity at $z = 0$. The function is not bounded because it approaches infinity as $z$ approaches zero.

Function $\sinh(z)$ is not bounded because $|\sinh(z)|$ can grow without bound, particularly along certain lines in the complex plane. It is entire but not bounded.

The function is not entire, having singularities at $z = (2n+1)\frac{\pi}{2}$ for $n \in \mathbb{Z}$. Additionally, it is not bounded because it approaches infinity at its singularities. This does not contradict Liouville's theorem, which requires entireness.

The function is not bounded as $|sin(z)|$ can be made arbitrarily large for some imaginary parts of $z$. It is entire but not bounded.

The function is not bounded as $|z^n|$ increases without bound as $|z| \to \infty$. This polynomial is entire but not bounded, satisfying Liouville's theorem.

Similarly to $\sinh(z)$, function $\cosh(z)$ is not bounded as it grows without bound for large values of $z$. It is entire but unbounded thus does not contradict Liouville's theorem.