The function f(z) = sin(z) is holomorphic on all of ℂ (it is entire). Does Liouville's Theorem imply it is constant?
AYes — sin(z) is entire, so by Liouville's Theorem it must be constant
BNo — sin(z) is entire but unbounded on ℂ, so Liouville's Theorem does not apply
CNo — Liouville's Theorem only applies to real-valued functions
DYes — sin(z) is periodic, which Liouville's Theorem classifies as a form of constancy
Liouville's Theorem requires TWO conditions: the function must be entire AND bounded. Sin(z) is indeed entire (holomorphic everywhere on ℂ), but it is NOT bounded on ℂ. While sin(x) is bounded for real x, for complex arguments the exponential growth of sinh causes |sin(z)| → ∞ as Im(z) → ±∞. Since sin(z) fails the boundedness condition, Liouville's Theorem does not apply and does not force it to be constant. Both conditions must hold simultaneously.
Question 2 Multiple Choice
The proof of the Fundamental Theorem of Algebra via Liouville's Theorem assumes a non-constant polynomial p(z) has no roots, then constructs the function 1/p(z). What property does Liouville's Theorem then force on 1/p(z), and why does this give a contradiction?
A1/p(z) must be zero, contradicting that p(z) is non-constant
B1/p(z) must be unbounded, contradicting that it was assumed bounded
C1/p(z) is entire and bounded (since |p(z)| → ∞ for large |z|), so Liouville forces it to be constant — but a constant 1/p means p is constant, contradicting the assumption
D1/p(z) must have a pole, but poles contradict the definition of holomorphic functions
If p(z) has no roots, then 1/p(z) is entire (no poles, since no zeros of p). For large |z|, |p(z)| → ∞ (polynomials grow without bound), so |1/p(z)| → 0 — the function is bounded near infinity. Combined with continuity on the compact disk, 1/p(z) is bounded everywhere. Liouville's Theorem then forces 1/p to be constant, which would make p constant — but p was assumed non-constant. This contradiction proves p must have a root. The entire proof structure is: no root → bounded entire function → Liouville → constant → contradiction.
Question 3 True / False
Liouville's Theorem has a direct analog in real analysis: any bounded smooth function on most of ℝ should be constant.
TTrue
FFalse
Answer: False
This is false, and recognizing it is key to appreciating why Liouville's Theorem is remarkable. In real analysis, sin(x) is a smooth (infinitely differentiable) bounded function on all of ℝ that is not constant. The theorem fails in the real setting because the integral formula argument breaks down: Cauchy's integral formula for derivatives, which allows the bound |f'(z₀)| ≤ M/R → 0, relies on the specific structure of complex holomorphicity and has no real counterpart. Complex differentiability is far stronger than real differentiability.
Question 4 True / False
The key step in the proof of Liouville's Theorem is that the bound on |f'(z₀)| can be made arbitrarily small by taking the integration contour to be a very large circle, and this is only possible because f is entire.
TTrue
FFalse
Answer: True
Exactly. Cauchy's derivative formula gives |f'(z₀)| ≤ M/R, where M bounds |f| and R is the radius of the integration circle. For this bound to be useful, we need R → ∞ — but we can only integrate over arbitrarily large circles if the function is holomorphic everywhere (entire). If f had a singularity anywhere in ℂ, the contour could not be expanded past that point. Boundedness provides the M, but 'entire' is what allows R to grow without limit, forcing M/R → 0.
Question 5 Short Answer
Why does the proof that any bounded entire function is constant fail for bounded smooth real functions like f(x) = sin(x)?
Think about your answer, then reveal below.
Model answer: The proof relies on Cauchy's integral formula for derivatives, which states that f'(z₀) = (1/2πi) ∮ f(z)/(z−z₀)² dz. This formula holds for holomorphic (complex-differentiable) functions and allows the bound |f'(z₀)| ≤ M/R by integrating over a circle of radius R. As R → ∞, M/R → 0, forcing f' = 0. No analogous integral representation for the derivative exists for real smooth functions — real differentiability does not come with the powerful machinery of Cauchy's formula. Sin(x) is smooth and bounded on ℝ, but you cannot expand a real integration contour to infinity and use it to bound the derivative.
Complex holomorphicity is a far stronger condition than real smoothness. A holomorphic function is controlled globally by its values on any contour (via Cauchy's formula), whereas a real smooth function carries no such global rigidity. Liouville's Theorem is essentially a measure of how much more constrained complex-differentiable functions are compared to their real counterparts.