Liouville's theorem

Theorem

Liouville's theorem

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Statement

Let \( f : \mathbb{C} \to \mathbb{C} \) be entire, i.e. holomorphic on all of \( \mathbb{C} \). Suppose \( f \) is bounded, meaning there exists \( M \gt 0 \) such that \( |f(z)| \le M \) for all \( z \in \mathbb{C} \). Then \( f \) is constant: there exists \( c \in \mathbb{C} \) with \( f(z) = c \) for all \( z \in \mathbb{C} \).

Why it matters

Liouville's theorem is the single fact that most sharply separates complex differentiability from real differentiability. A real function such as \( \sin x \) is smooth and bounded on all of \( \mathbb{R} \) without being remotely constant; the theorem says the complex analogue of this situation is impossible. This rigidity — "holomorphic plus bounded forces trivial" — is the engine behind the Fundamental Theorem of Algebra, behind Picard's theorems on the range of entire functions, and behind countless "no nontrivial bounded solution exists" arguments across analysis.

It also exemplifies a recurring theme in complex analysis: local information (a convergent power series, via Cauchy's integral formula) combines with global information (a bound valid everywhere) to produce an extremely strong global conclusion. Nothing this strong is true in real analysis or in several real variables without additional hypotheses.

Hypotheses

Entire (holomorphic on all of \( \mathbb{C} \)).\( f(z) = \tan z \) is bounded on no vertical strip but is holomorphic only away from its poles; more to the point, \( f(z) = 1/z \) is bounded and holomorphic on \( \mathbb{C}\setminus\{0\} \) but is not entire (it has a singularity at \( 0 \)), and it is certainly not constant. Entirety — holomorphy at every point of the plane, with no exceptional points — is essential so that the Cauchy estimates below apply on circles of every radius centred at every point. Bounded on the whole plane.\( f(z) = z \) is entire but unbounded, and it is not constant. \( f(z) = e^z \) is entire, unbounded (though bounded on the imaginary axis and on left half-planes), and not constant. Boundedness must hold on the entire plane \( \mathbb{C} \), not merely on some proper subregion; a function bounded only on a disc or a half-plane need not be constant. The domain must be all of \( \mathbb{C} \), not merely a domain conformally very close to it.The open unit disc \( \mathbb{D} \) is conformally equivalent to \( \mathbb{C} \) as a topological space is not the relevant issue — rather, \( \mathbb{D} \) is not the whole plane, and bounded non-constant holomorphic functions on \( \mathbb{D} \) abound (e.g. \( f(z) = z \) itself, bounded by \( 1 \) on \( \mathbb{D} \)). Liouville's theorem is a statement about entire functions specifically; boundedness on a proper subdomain carries no such force, however large that subdomain is.

Proof

\text{Since } f \text{ is entire, for every } z_0 \in \mathbb{C} \text{ and every } R \gt 0, f \text{ is holomorphic on and inside the circle } |z-z_0| = R.

Restatement of the hypothesis that \( f \) is holomorphic on all of \( \mathbb{C} \); in particular it is holomorphic on the closed disc \( \overline{D}(z_0,R) \) for every choice of \( z_0 \) and \( R \). A

f(w) = \sum_{n=0}^{\infty} a_n (w-z_0)^n, \qquad a_n = \frac{f^{(n)}(z_0)}{n!} = \frac{1}{2\pi i}\int_{|w-z_0|=R} \frac{f(w)}{(w-z_0)^{n+1}}\, dw,

By Cauchy's integral formula for derivatives (itself derived from the Cauchy integral formula \( f(z) = \frac{1}{2\pi i}\int_{|w-z_0|=R} \frac{f(w)}{w-z}\,dw \) applied on the circle of radius \( R \), together with the Cauchy–Goursat theorem justifying differentiation under the integral sign), \( f \) is holomorphic on a disc, hence analytic there, hence equal to its Taylor series about \( z_0 \) with the stated integral formula for the coefficients \( a_n = f^{(n)}(z_0)/n! \). B

|a_n| = \left| \frac{1}{2\pi i}\int_{|w-z_0|=R} \frac{f(w)}{(w-z_0)^{n+1}}\, dw \right| \le \frac{1}{2\pi} \cdot \sup_{|w-z_0|=R} \frac{|f(w)|}{R^{n+1}} \cdot 2\pi R = \frac{\sup_{|w-z_0|=R}|f(w)|}{R^{n}}.

The standard ML-inequality (estimation lemma) for contour integrals: \( \left|\int_\gamma g(w)\,dw\right| \le \left(\sup_{w\in\gamma}|g(w)|\right)\cdot \operatorname{length}(\gamma) \), applied to \( g(w) = f(w)/(w-z_0)^{n+1} \) on the circle \( \gamma \) of radius \( R \), which has length \( 2\pi R \) and on which \( |w - z_0|^{n+1} = R^{n+1} \) identically. This is the Cauchy estimate for the \( n \)-th Taylor coefficient. C

\sup_{|w-z_0|=R}|f(w)| \le \sup_{w \in \mathbb{C}} |f(w)| \le M \quad \text{for every } R \gt 0, \text{ since the circle } |w-z_0|=R \text{ is a subset of } \mathbb{C}.

Direct use of the boundedness hypothesis \( |f(z)| \le M \) for all \( z \in \mathbb{C} \): the supremum of \( |f| \) over any subset of \( \mathbb{C} \), in particular over any circle, is at most the global bound \( M \). A

\text{Combining Steps 3 and 4: } |a_n| \le \frac{M}{R^n} \quad \text{for every } R \gt 0 \text{ and every fixed } n \ge 1.

Substitution of the bound from Step 4 into the Cauchy estimate of Step 3; valid for every radius \( R \), because Step 1 guarantees the disc of radius \( R \) about \( z_0 \) lies in the domain of holomorphy of \( f \) for every \( R \), so the argument places no upper limit on \( R \). B

\text{Fix } n \ge 1. \text{ Let } R \to \infty \text{ in } |a_n| \le M/R^n. \text{ Since the left side is a fixed nonnegative number independent of } R, \text{ we get } |a_n| \le \lim_{R\to\infty} \frac{M}{R^n} = 0, \text{ so } a_n = 0.

Squeeze/monotonicity of limits: \( a_n \) does not depend on \( R \) (Step 2 fixes \( z_0 \) and \( n \); the value of the coefficient \( a_n \) is independent of which circle was used to compute it, by uniqueness of Taylor coefficients), while the upper bound \( M/R^n \to 0 \) as \( R \to \infty \) for each fixed \( n \ge 1 \). A nonnegative constant bounded above by a sequence tending to \( 0 \) must itself be \( 0 \). This is the crux step where boundedness on the whole plane is used essentially: it licenses letting \( R \) grow without bound. B

\text{Therefore } a_n = 0 \text{ for every } n \ge 1, \text{ and the Taylor series of Step 2 collapses to } f(w) = a_0 \text{ for all } w \text{ in every disc } D(z_0, R).

Direct consequence of Step 6 applied to every \( n \ge 1 \), substituted back into the series representation of Step 2. Since \( z_0 \in \mathbb{C} \) was arbitrary in Step 1, this holds with \( a_0 = f(z_0) \) at every point of \( \mathbb{C} \). A

\text{Set } c := f(0). \text{ Then } f(z) = c \text{ for every } z \in \mathbb{C}.

Step 7 shows every point \( z_0 \) is a centre about which \( f \) is locally equal to the constant \( f(z_0) \); in particular \( a_0 = f(z_0) \) from Step 2 with \( n=0 \) gives \( f(w) = f(z_0) \) for \( w \) near \( z_0 \), and since \( z_0 \) ranges over all of \( \mathbb{C} \) this pins \( f \) to the single value \( c = f(0) \) everywhere, i.e. \( f \) is constant, as claimed. A

Result

f \text{ entire and bounded on } \mathbb{C} \;\Longrightarrow\; f \text{ is constant}

Reading. An entire function that never grows unboundedly as \( |z| \to \infty \) — in fact one that is simply bounded, with no growth condition needed at all — has no choice but to be constant. Complex differentiability everywhere on the plane is so rigid a constraint that boundedness alone crushes all the freedom a real-analytic function would retain.

Scope. Applies exactly to functions holomorphic on the whole of \( \mathbb{C} \) (entire functions) that are bounded on the whole of \( \mathbb{C} \). It does not apply to functions with any singularity, however mild (a single removable-looking pole, an essential singularity, a branch point), nor to functions holomorphic only on a proper subset of \( \mathbb{C} \) (a disc, a half-plane, a strip), nor to bounded-only-in-some-direction functions.

Corollaries & converses

Fundamental Theorem of Algebra. Every non-constant polynomial \( p(z) \) with complex coefficients has a root in \( \mathbb{C} \): if \( p \) had no root, \( 1/p(z) \) would be entire, and since \( |p(z)| \to \infty \) as \( |z|\to\infty \) for non-constant \( p \), \( 1/p(z) \to 0 \), making \( 1/p \) bounded and entire, hence constant by Liouville — contradicting that \( p \) is non-constant.
Generalized Liouville (polynomial growth). If \( f \) is entire and \( |f(z)| \le A + B|z|^k \) for constants \( A, B \) and integer \( k \ge 0 \), then \( f \) is a polynomial of degree at most \( k \). (Same Cauchy-estimate argument, keeping the terms \( n \le k \) instead of discarding all of them.)
No entire function has a bounded real part without being constant on the imaginary part suitably normalised — more precisely, if \( \operatorname{Re} f \) is bounded above (or below) for entire \( f \), then \( f \) is constant, by composing with \( e^{-f} \) or \( e^{f} \) to reduce to the bounded case.
Converse fails outright. The converse statement "if \( f \) is constant then any entire extension bounded on \( \mathbb{C} \) exists" is trivially true but vacuous, and more meaningfully, boundedness is not necessary for an interesting rigidity statement: there is no useful converse of the form "constant entire functions are the only bounded ones among some larger class," since the theorem already characterises exactly the bounded entire functions (precisely the constants) — there is nothing to converse.
The theorem does not extend to functions merely bounded on a sequence of circles \( |z| = R_k \to \infty \) rather than on all of \( \mathbb{C} \); this weaker hypothesis is genuinely insufficient (see Fails without).

Fails without

Drop boundedness: \( f(z) = e^z \) is entire, non-constant, and unbounded (as \( \operatorname{Re}(z) \to \infty \), \( |e^z| = e^{\operatorname{Re}(z)} \to \infty \)); it satisfies every other hypothesis of the theorem.
Drop entirety (allow one singularity): \( f(z) = \dfrac{1}{1+z^2} \) is bounded on \( \mathbb{C}\setminus\{\pm i\} \) by \( 1 \) near infinity and remains bounded away from its two poles on any region avoiding small discs about them, yet it is not entire (poles at \( z = \pm i \)) and is manifestly non-constant; more simply \( f(z) = 1/z \) is bounded on \( |z|\ge 1 \) and holomorphic there but not entire and not constant.
Restrict to a proper subdomain: \( f(z) = z \) restricted to the unit disc \( \mathbb{D} \) is holomorphic and bounded (\( |f(z)| \lt 1 \)) on \( \mathbb{D} \), but \( \mathbb{D} \ne \mathbb{C} \), and \( f \) is not constant — the theorem's global-domain hypothesis is doing real work.
Weaken "bounded everywhere" to "bounded on a discrete sequence \( |z|=R_k\to\infty \)": one can construct entire functions bounded on the circles \( |z| = k \) (integers \( k \)) yet unbounded between them and non-constant (e.g. suitably normalised products vanishing to controlled order); the Cauchy-estimate argument needs the bound to hold on every radius \( R \) as \( R \to \infty \) continuously, not merely along a sequence, so this weakening genuinely breaks the proof and the conclusion.

Common errors

Applying Liouville to a function that is bounded only on the real axis or only on a strip (e.g. \( \sin z \) is bounded on \( \mathbb{R} \) but unbounded on \( \mathbb{C} \) since \( |\sin(iy)| = \sinh y \to \infty \)); students conflate "bounded on the reals" with "bounded on the plane."
Forgetting to check entirety and invoking the theorem for functions with removable-looking singularities or branch cuts, e.g. treating \( \sqrt{z} \) or \( \log z \) as entire.
Misapplying the generalized (polynomial-growth) version by using the wrong power \( k \) in the Cauchy estimate, e.g. concluding \( f \) has degree \( k \) when the growth bound only forces degree \( \le k \) (a smaller-degree or even constant polynomial also satisfies the same bound).
Using the Fundamental-Theorem-of-Algebra corollary circularly, e.g. trying to prove Liouville's theorem itself using the Fundamental Theorem of Algebra (the correct logical direction is Liouville \( \Rightarrow \) FTA, not the reverse).
Believing the theorem says "bounded implies constant" for holomorphic functions on any domain; the global hypothesis \( \text{domain} = \mathbb{C} \) is essential and frequently dropped silently in casual restatements.

Discussion

Liouville's theorem, despite its name, was apparently first proved by Cauchy; Joseph Liouville is credited via Cauchy's attribution of a special case (used to derive the Fundamental Theorem of Algebra) in lectures, and the modern textbook name persists partly through historical accident. The theorem is a paradigm example of how much stronger complex differentiability is than real differentiability: a real-valued function of one real variable that is everywhere differentiable and bounded need not be remotely close to constant (e.g. \( \sin x \), or even something with rich oscillation like a bounded but nowhere-monotonic smooth function). The obstruction that saves complex analysis from such richness is that holomorphy forces analyticity (via Cauchy's integral formula), and analyticity on an unbounded domain interacts with a global bound to crush all higher Taylor coefficients simultaneously.

The Cauchy estimate used in Step 3 is itself a foundational tool independent of Liouville's theorem, appearing throughout complex analysis to bound derivatives of holomorphic functions in terms of sup-norms on circles. Liouville's theorem is the \( R \to \infty \) degenerate case of this estimate, exploiting the fact that entire functions admit the estimate on circles of arbitrarily large radius, whereas a function holomorphic only on a bounded domain has no such freedom — the radius is capped by the boundary of the domain, and the estimate degrades as \( R \) approaches that cap rather than improving.

The theorem generalises dramatically. Picard's Little Theorem states that a non-constant entire function omits at most one value of \( \mathbb{C} \) (e.g. \( e^z \) omits only \( 0 \)) — a vastly stronger rigidity statement than Liouville's, which only concerns boundedness (omitting an open set, in effect). Liouville's theorem can be seen as the "linear growth bound \( k=0 \)" instance of the polynomial-growth generalisation, which itself sits inside the broader theory of the order and type of entire functions (Hadamard factorization theory), where growth rates are calibrated far more finely than "bounded vs unbounded."

A conceptually cleaner route to the same result uses the maximum modulus principle together with Cauchy's formula for the derivative directly at a single point \( z_0 \), applied on ever-larger circles: \( f'(z_0) = \frac{1}{2\pi i}\int_{|w-z_0|=R} \frac{f(w)}{(w-z_0)^2}\,dw \), giving \( |f'(z_0)| \le M/R \to 0 \) as \( R \to \infty \), hence \( f'(z_0) = 0 \) at every point, hence \( f \) constant by the real-variable fact that a function with identically zero derivative on a connected open set is constant (which itself requires connectedness of \( \mathbb{C} \), an easily overlooked ingredient). This is a special case (\( n=1 \)) of the argument given above, isolated because it needs nothing about higher Taylor coefficients, only the single fact that the derivative vanishes everywhere — but it still hides the connectedness hypothesis on \( \mathbb{C} \) inside "zero derivative everywhere implies constant," which fails on disconnected domains.

Common misconception: that Liouville's theorem is about functions bounded "at infinity" in the sense of having a limit there. It requires no limit to exist; wild bounded oscillation as \( |z|\to\infty \) (impossible for genuinely entire functions, but this is exactly the point) would still force the conclusion if it could occur at all — the theorem's content is precisely that such oscillation cannot occur for entire functions.

Worked examples

\text{Prove that } f(z) = \sin z \cos z + i\,\text{(anything unbounded on } \mathbb{C}\text{) cannot occur if we require } f \text{ bounded and entire}\text{ — concretely: show } g(z) = \dfrac{1}{2+\cos z} \text{ is NOT entire-and-bounded-hence-nonconstant by locating its failure.}

Reframe as: determine whether \( g \) is entire and bounded, and if both, conclude \( g \) is constant by the theorem; otherwise identify which hypothesis fails. A

\text{Since } |\cos z| \text{ is unbounded on } \mathbb{C} \text{ (e.g. } |\cos(iy)| = \cosh y \to \infty \text{ as } y \to \infty\text{), the denominator } 2 + \cos z \text{ vanishes only if } \cos z = -2, \text{ which } does \text{ occur for some complex } z \text{ (since } \cos \text{ is surjective onto } \mathbb{C} \text{ as an entire non-constant function, by Picard/openness)}.

Complex \( \cos z \) is surjective onto \( \mathbb{C} \setminus \{\text{at most one point}\} \) by Picard's Little Theorem (or, more elementarily, by solving \( \cos z = w \) via \( z = \arccos w \), which has solutions for every \( w \in \mathbb{C} \) since the equation \( \frac{e^{iz}+e^{-iz}}{2} = w \) reduces to a quadratic in \( e^{iz} \) with a nonzero solution for every \( w \)). Hence \( \cos z = -2 \) has solutions, so \( g \) has poles and is not entire. B

\text{Therefore Liouville's theorem does not apply to } g\text{: the entirety hypothesis fails, so no conclusion about constancy follows, consistent with } g \text{ indeed being non-constant.}

Direct application of the Hypotheses discussion: dropping entirety removes the theorem's force entirely; this example confirms the hypothesis is not vacuous — genuine functions fail it. A

g(z) = \frac{1}{2+\cos z} \text{ is not entire (poles where } \cos z = -2\text{), so Liouville is inapplicable and } g \text{ need not be — and is not — constant.}

\text{Prove the Fundamental Theorem of Algebra: every non-constant polynomial } p(z) = a_n z^n + \cdots + a_0, \; a_n \ne 0,\; n\ge 1,\; \text{has a root in } \mathbb{C}.

Proof by contradiction using Liouville's theorem, following the corollary stated above. B

\text{Suppose, for contradiction, } p(z) \ne 0 \text{ for every } z \in \mathbb{C}. \text{ Define } h(z) = \dfrac{1}{p(z)}.

Standard contradiction setup: if \( p \) never vanishes, its reciprocal is defined everywhere on \( \mathbb{C} \). A

h \text{ is entire, since } p \text{ is entire (a polynomial) and, by assumption, } p(z)\ne 0 \text{ for all } z, \text{ so } h = 1/p \text{ is a quotient of entire functions with nonvanishing denominator.}

Quotient rule for holomorphic functions: if \( p, q \) are holomorphic and \( q \) is nowhere zero on a domain, \( p/q \) is holomorphic on that domain. Here the domain is all of \( \mathbb{C} \). A

\text{As } |z| \to \infty,\; |p(z)| = |a_n||z|^n\left|1 + \frac{a_{n-1}}{a_n z} + \cdots + \frac{a_0}{a_n z^n}\right| \to \infty, \text{ since the bracketed factor} \to 1 \text{ and } |z|^n \to \infty.

Elementary estimate: factor out the leading term \( a_n z^n \); each remaining term \( a_k/(a_n z^{\,n-k}) \to 0 \) as \( |z| \to \infty \) for \( k \lt n \), so the bracket tends to \( 1 \), and \( |a_n||z|^n \to \infty \) since \( n \ge 1 \) and \( a_n \ne 0 \). B

\text{Consequently } |h(z)| = \frac{1}{|p(z)|} \to 0 \text{ as } |z|\to\infty; \text{ in particular there exists } R_0 \gt 0 \text{ with } |h(z)| \le 1 \text{ for } |z| \ge R_0.

Direct consequence of Step 4 and the definition of the limit \( |p(z)| \to \infty \): eventually \( |p(z)| \ge 1 \), hence \( |h(z)| \le 1 \), outside some disc of radius \( R_0 \). A

\text{On the closed disc } \overline{D}(0,R_0), \; h \text{ is continuous (indeed holomorphic), hence attains a finite maximum } M_0 \text{ there, by the extreme value theorem.}

Extreme value theorem for continuous functions on a compact set: \( \overline{D}(0,R_0) \) is closed and bounded in \( \mathbb{C} \cong \mathbb{R}^2 \), hence compact by Heine–Borel, and \( |h| \) is continuous, so it attains a maximum \( M_0 \) on this compact set. A

\text{Set } M = \max(1, M_0). \text{ Then } |h(z)| \le M \text{ for every } z \in \mathbb{C}: \text{ for } |z|\ge R_0 \text{ by Step 5, and for } |z| \le R_0 \text{ by Step 6.}

Combining the two regions \( |z| \le R_0 \) and \( |z| \ge R_0 \), which together cover all of \( \mathbb{C} \), gives a single global bound \( M \) for \( |h| \). A

\text{By Liouville's theorem (applied to the entire, bounded function } h\text{), } h \text{ is constant. Hence } p = 1/h \text{ is also constant — contradicting that } p \text{ is non-constant (degree } n \ge 1\text{).}

Direct invocation of the theorem proved above: \( h \) is entire (Step 3) and bounded (Step 7), so \( h \equiv c \) for some constant \( c \). Since \( h \) is nowhere zero (as \( p \) is finite everywhere), \( c \ne 0 \), and \( p(z) = 1/c \) for all \( z \), contradicting \( \deg p = n \ge 1 \). This contradiction refutes the assumption of Step 2. B

\text{Therefore } p \text{ must have a root in } \mathbb{C} \text{ — the Fundamental Theorem of Algebra, derived entirely from Liouville's theorem.}

Problems

Show directly (without quoting the generalized version) that if \( f \) is entire and \( |f(z)| \le M \) for all \( z \), then \( f'' \equiv 0 \) as well as \( f' \equiv 0 \), reproving constancy via second derivatives.
Solution

By Cauchy's estimate applied with \( n=2 \) at an arbitrary point \( z_0 \): \( |f''(z_0)| \le \dfrac{2! \, M}{R^2} \) for every \( R \gt 0 \) (using \( a_2 = f''(z_0)/2! \) and the bound \( |a_2| \le M/R^2 \) from the Cauchy estimate, so \( |f''(z_0)| \le 2M/R^2 \)). Letting \( R \to \infty \) gives \( f''(z_0) = 0 \). Since \( z_0\) was arbitrary, \( f'' \equiv 0 \) on \( \mathbb{C} \). Similarly \( f' \equiv 0 \) by the \( n=1 \) case (as in the Discussion). Either fact alone (via "identically zero derivative on connected open set implies constant") gives \( f \) constant, consistent with the main theorem.
Let \( f \) be entire with \( |f(z)| \le 3 + 4|z|^2 \) for all \( z \). What is the most that can be said about \( f \)? Prove your claim.
Solution

Apply the generalized Liouville corollary with \( k=2 \): writing \( f(w) = \sum a_n w^n \) (Taylor series about \( 0 \)), the Cauchy estimate gives \( |a_n| \le \sup_{|w|=R}|f(w)|/R^n \le (3+4R^2)/R^n \). For \( n \ge 3 \), as \( R\to\infty \), \( (3+4R^2)/R^n = 3/R^n + 4/R^{n-2} \to 0 \), so \( a_n = 0 \) for all \( n \ge 3 \). No further coefficients are forced to vanish by this bound (e.g. \( f(z) = 4z^2 \) satisfies the hypothesis with room to spare). Conclusion: \( f \) is a polynomial of degree at most \( 2 \), i.e. \( f(z) = a_0 + a_1 z + a_2 z^2 \) for some constants \( a_0,a_1,a_2 \in \mathbb{C} \); this is the sharpest general conclusion, and it is achieved (not merely an upper bound) since \( f(z)=4z^2\) is a valid example meeting the hypothesis with equality in growth order.
Does there exist a non-constant entire function \( f \) such that \( |f(z)| \le 5 \) whenever \( z \) lies on the real axis, but \( f \) is otherwise unrestricted? Explain, being careful about what Liouville's theorem does and does not say.
Solution

Yes — for example \( f(z) = \sin z \) satisfies \( |\sin x| \le 1 \le 5 \) for all real \( x\), is entire, and is manifestly non-constant. This does not contradict Liouville's theorem because the theorem requires the bound \( |f(z)|\le M \) to hold for every \( z \in \mathbb{C} \), not just for \( z \) on the real axis; indeed \( \sin z \) is unbounded on \( \mathbb{C} \) since \( |\sin(iy)| = \sinh y \to \infty \). This illustrates precisely the common error of restricting the boundedness check to the real axis.
Use Liouville's theorem to show that if \( f \) is entire and \( f(z) \ne f(z+1) \) is false for no reason relevant here — instead: show that an entire function \( f \) satisfying \( |f(z)| \le |\sin z| \) for all \( z \in \mathbb{C} \) must be identically zero. (Hint: consider where \( \sin z = 0\) and what boundedness of \( f/\sin z \) would require, being careful about removable singularities.)
Solution

The zeros of \( \sin z \) are exactly \( z = k\pi \), \( k \in \mathbb{Z} \), each simple. The hypothesis \( |f(z)| \le |\sin z| \) forces \( f(k\pi) = 0 \) for every integer \( k \), since \( |f(k\pi)| \le |\sin(k\pi)| = 0 \). So \( f \) vanishes at every zero of \( \sin z \), and because these zeros are simple and \( f \) is entire (hence its zeros there are of order at least matching, by the general fact that if \( g \) vanishes to order \( \ge 1 \) wherever \( \sin z \) does with matching or higher order — here both vanish to order exactly \( 1 \) generically, verified since \( f \) vanishing with \( |f| \le |\sin z| \) near each \( k\pi \) forces at least simple-order vanishing), the quotient \( h(z) = f(z)/\sin z \) extends to an entire function via Riemann's removable singularity theorem at each \( k\pi \) (since \( f/\sin z \) is bounded near each \( k\pi\) — indeed \( |f/\sin z|\le 1\) there — a bounded holomorphic function near an isolated singularity extends holomorphically across it). Away from the integer multiples of \( \pi \), \( |h(z)| = |f(z)|/|\sin z| \le 1 \) directly from the hypothesis. So \( h \) is entire (after removable extension) and bounded by \( 1 \) everywhere. By Liouville's theorem, \( h \) is constant, say \( h \equiv c \) with \( |c|\le 1 \). Then \( f(z) = c\sin z \). But we also need \( |f(z)| \le |\sin z| \) to pin down more — actually this alone permits any \( |c| \le 1 \), so the sharper claim "\( f \equiv 0 \)" requires an additional hypothesis not stated; correcting the problem: if instead the strict/generic reading intends \( f \) to also be odd or to vanish faster, the clean forced conclusion from the stated hypothesis alone is \( f(z) = c \sin z \) for some constant \( |c| \le 1 \), not necessarily \( f \equiv 0 \). (Pedagogical note: this problem is a caution against overclaiming — Liouville's theorem here correctly yields "\( f \) is a constant multiple of \( \sin z \)," and the identically-zero conclusion would need the extra hypothesis, e.g. \( f(\pi/2) = 0\), to force \( c = 0\).)
An entire function \( f \) satisfies \( \operatorname{Re}(f(z)) \le 7 \) for all \( z \in \mathbb{C} \). Prove that \( f \) is constant.
Solution

Consider \( g(z) = e^{f(z)} \). Since \( f \) is entire, so is \( g \) (composition of the entire function \( f \) with the entire function \( \exp \)). Moreover \( |g(z)| = |e^{f(z)}| = e^{\operatorname{Re}(f(z))} \le e^{7} \) for all \( z\), using \( \operatorname{Re}(f(z)) \le 7 \) and monotonicity of \( t \mapsto e^t \) on \( \mathbb{R} \). So \( g \) is entire and bounded (by \( M = e^7 \)); by Liouville's theorem, \( g \) is constant, say \( g \equiv c \). Since \( g(z) = e^{f(z)} \) is never zero (the exponential function never vanishes), \( c \ne 0 \). Differentiating \( g(z) = c \) gives \( g'(z) = f'(z) e^{f(z)} \equiv 0 \) (chain rule); since \( e^{f(z)} \ne 0 \) everywhere, \( f'(z) \equiv 0 \) on \( \mathbb{C} \). As \( \mathbb{C} \) is connected and \( f' \equiv 0 \) there, \( f \) is constant (a holomorphic — in particular real-differentiable in each variable — function with identically vanishing derivative on a connected open set is constant, by integrating along paths and using path-connectedness of \( \mathbb{C}\)).