Solution. Consider \(g(z) = f(z)e^{-z}\), which is entire. The hypothesis says \[\left |g(z)\right | = \frac {\left |f(z)\right |}{\left |e^{z}\right |} \leq 1 \qquad \text {for all } z .\] A bounded entire function is constant, by Liouville, so \(g \equiv c\) with \(\left |c\right | \leq 1\) and \[\boxed {f(z) = c\,e^{z},\qquad \left |c\right | \leq 1 .}\] Conversely every such \(f\) satisfies the hypothesis, so these are exactly the functions asked for. \(\blacksquare \)

Dividing by \(e^{z}\) is legitimate precisely because \(e^{z}\) never vanishes; that is the only property of the exponential used, so the same argument classifies all entire \(f\) with \(\left |f\right | \leq \left |h\right |\) for any zero-free entire \(h\).

Solution. The hypothesis is now only assumed outside the unit disc, so \(g = fe^{-z}\) is bounded there but not obviously on the disc itself. That gap costs nothing.

\(g\) is entire, and \(\left |g\right | \leq 1\) on \(\left |z\right | > 1\). On the closed unit disc \(\overline {D}\), \(g\) is continuous on a compact set, hence bounded there too, say by \(K\). So \(\left |g\right | \leq \max (1,K)\) on the whole plane, and Liouville again gives \(g \equiv c\).

Now the constant must satisfy \(\left |c\right | \leq 1\), because \(\left |g\right | \leq 1\) somewhere — anywhere outside the disc. Hence \[\boxed {f(z) = c\,e^{z},\qquad \left |c\right | \leq 1 ,}\] the same answer as before. \(\blacksquare \)

The lesson is worth stating: a growth hypothesis on an entire function only ever needs to hold near infinity. Any compact set contributes a finite bound for free, by continuity.

Solution. This is the general growth-implies-polynomial statement with \(n = 5\); the argument uses Cauchy’s estimates.

For \(\rho > R\) and any \(k\), the Cauchy inequality on \(\left |z\right | = \rho \) gives \[\left |f^{(k)}(0)\right | \leq \frac {k!\,\max _{\left |z\right |=\rho }\left |f\right |}{\rho ^{k}} < \frac {k!\,M\rho ^{5}}{\rho ^{k}} = k!\,M\,\rho ^{\,5-k}.\] If \(k \geq 6\) the exponent \(5-k\) is negative, so letting \(\rho \to \infty \) forces \[f^{(k)}(0) = 0 \qquad (k \geq 6).\] Since \(f\) is entire it equals its Taylor series about \(0\) everywhere, and all coefficients beyond the fifth vanish: \[\boxed {f \text { is a polynomial of degree at most } 5 .}\] \(\blacksquare \)

Every problem of this family in this chapter is the same computation with a different exponent, and it is worth seeing once that the exponent is the only thing that changes. Liouville itself is the case \(n = 0\).

Solution. Statement. A bounded entire function is constant. That is, if \(f\) is analytic on all of \(\mathbb {C}\) and there is \(M\) with \(\left |f(z)\right | \leq M\) for all \(z\), then \(f\) is constant.

Proof. Fix \(z_0 \in \mathbb {C}\) and \(\rho > 0\). Cauchy’s integral formula for the derivative on the circle \(\left |z-z_0\right | = \rho \) gives \[f'(z_0) = \frac {1}{2\pi i}\oint _{\left |z-z_0\right |=\rho } \frac {f(z)}{(z-z_0)^{2}}\,dz ,\] so by the estimation inequality \[\left |f'(z_0)\right | \leq \frac {1}{2\pi }\cdot \frac {M}{\rho ^{2}}\cdot 2\pi \rho = \frac {M}{\rho }.\] This holds for every \(\rho > 0\), and \(f\) is entire so no upper limit on \(\rho \) applies. Letting \(\rho \to \infty \) gives \(f'(z_0) = 0\).

Since \(z_0\) was arbitrary, \(f' \equiv 0\) on the connected set \(\mathbb {C}\), so \(f\) is constant. \(\blacksquare \)

Everything rests on being able to send \(\rho \) to infinity, which is exactly what being entire buys. On a disc the same estimate gives Cauchy’s inequality, which is useful but not a rigidity statement.

Solution. Boundedness is not assumed — only a one-sided growth condition on the real part, so Liouville cannot be applied to \(f\) itself. Apply it to \(e^{f}\) instead.

Suppose first the stronger-looking claim: \(\operatorname {Re}f\) is bounded above. Then \(\left |e^{f}\right | = e^{\operatorname {Re}f}\) is bounded, \(e^{f}\) is entire, so \(e^{f}\) is constant by Liouville; as \(e^{f}\) never vanishes, \(f\) is constant.

For the hypothesis as given, use the Borel–Carathéodory inequality: for \(0 < r < \rho \), \[\max _{\left |z\right |=r}\left |f(z)\right | \leq \frac {2r}{\rho -r}\max _{\left |z\right |=\rho }\operatorname {Re}f(z) + \frac {\rho +r}{\rho -r}\left |f(0)\right | .\] Write \(A(\rho ) = \max _{\left |z\right |=\rho }\operatorname {Re}f\). The hypothesis \(z^{-1}\operatorname {Re}f(z) \rightarrow 0\) says \(A(\rho ) = o(\rho )\). Taking \(\rho = 2r\), \[\max _{\left |z\right |=r}\left |f\right | \leq 2A(2r) + 3\left |f(0)\right | = o(r) .\] So \(f\) is entire with \(\left |f(z)\right | = o(\left |z\right |)\), and by the growth-implies-polynomial result with \(n = 1\) it is a polynomial of degree at most \(1\); the \(o(r)\) rate kills the linear term as well. Hence \[\boxed {f \text { is constant.}} \qquad \blacksquare \]

The pattern — convert a condition on \(\operatorname {Re}f\) into one on \(\left |e^{f}\right |\), or invoke Borel–Carathéodory — is the standard way to handle one-sided hypotheses, and recurs in the harmonic-function problems.

Solution. By Cauchy’s inequality on the circle \(\left |z\right | = \rho \) with \(\rho \geq R\), \[\left |f^{(k)}(0)\right | \leq \frac {k!}{\rho ^{k}}\max _{\left |z\right |=\rho }\left |f\right | \leq \frac {k!}{\rho ^{k}}\,M\rho ^{\,n} = k!\,M\,\rho ^{\,n-k}.\] For \(k > n\) the exponent is negative, so letting \(\rho \rightarrow \infty \) gives \(f^{(k)}(0) = 0\) for every \(k > n\).

Because \(f\) is entire its Taylor series about \(0\) converges to \(f\) on the whole plane, and all terms beyond degree \(n\) vanish: \[\boxed {f(z) = \sum ^{n}_{k=0}\frac {f^{(k)}(0)}{k!}z^{k},}\] a polynomial of degree at most \(n\). \(\blacksquare \)

Taking \(n = 0\) recovers Liouville’s theorem, so this is the natural generalisation: polynomial growth of order \(n\) forces polynomial form of degree \(n\). Note the hypothesis is only needed for \(\left |z\right | \geq R\); on the compact disc \(\left |z\right | \leq R\) continuity supplies a bound for free.

Solution. Here the exponent is fractional and less than \(1\), which is the point.

By the previous problem’s computation, for \(\rho > R\) and any \(k \geq 1\), \[\left |f^{(k)}(0)\right | \leq k!\,M\,\rho ^{\,\alpha -k}.\] Since \(0 < \alpha < 1\), already \(k = 1\) gives exponent \(\alpha - 1 < 0\), so \[\left |f'(0)\right | \leq M\rho ^{\,\alpha -1} \longrightarrow 0 \qquad (\rho \to \infty ),\] whence \(f'(0) = 0\), and likewise \(f^{(k)}(0) = 0\) for every \(k \geq 1\). Only the constant term survives: \[\boxed {f \text { is constant.}} \qquad \blacksquare \]

So any sub-linear growth rate, however slow, is as restrictive as boundedness. The threshold is exactly \(\alpha = 1\): at \(\alpha = 1\) one gets degree \(\leq 1\), and \(f(z) = z\) shows that this cannot be improved.

Solution. The relation \(f(z) = f(1/z)\) transfers the behaviour of \(f\) near infinity to its behaviour near \(0\), where \(f\) is analytic and therefore tame.

\(f\) is continuous at \(0\), so it is bounded on some disc \(\left |z\right | \leq \delta \), say by \(K\). For \(\left |w\right | \geq 1/\delta \) we have \(\left |1/w\right | \leq \delta \), and the hypothesis gives \[\left |f(w)\right | = \left |f\!\left (\tfrac 1w\right )\right | \leq K .\] So \(f\) is bounded outside the disc \(\left |w\right | < 1/\delta \). Inside that disc — a compact set — \(f\) is bounded by continuity. Hence \(f\) is bounded on \(\mathbb {C}\), and Liouville makes it \[\boxed {\text {constant.}} \qquad \blacksquare \]

Equivalently: the substitution \(z \mapsto 1/z\) turns the point at infinity into the origin, and the hypothesis says \(f\) has a removable singularity there rather than the pole or essential singularity a non-constant entire function must have.

Solution. Since \(\left |f(z)\right | \leq \left |z\right |^{n}\) for all \(z\), taking \(z = 0\) gives \(f(0) = 0\); more, the bound forces \(f\) to vanish to order at least \(n\) at the origin. To see this, note \(g(z) = f(z)/z^{n}\) is analytic on \(\mathbb {C}\setminus \{0\}\) and satisfies \(\left |g\right | \leq 1\) there, so \(0\) is a removable singularity and \(g\) extends to an entire function with \(\left |g\right | \leq 1\) on all of \(\mathbb {C}\).

By Liouville, \(g \equiv c\) with \(\left |c\right | \leq 1\), and therefore \[\boxed {f(z) = c\,z^{n},\qquad \left |c\right | \leq 1 .} \qquad \blacksquare \]

Compare the earlier problems in this chapter: a bound \(M\left |z\right |^{n}\) holding only near infinity gives a polynomial of degree at most \(n\), whereas this bound, holding everywhere including at \(0\), pins the polynomial down to the single monomial \(cz^{n}\). The extra information at the origin is what removes the lower-order terms.

Solution. The hypothesis \(\left |w\right | \leq \left |p\right |\) forces \(w\) to vanish wherever \(p\) does, and to at least the same order: if \(p\) has a zero of order \(m\) at \(a\) then near \(a\), \(\left |w(z)\right | \leq C\left |z-a\right |^{m}\), so \(w\) has a zero of order at least \(m\) there.

Consequently \(g = w/p\), defined on \(\mathbb {C}\) minus the finitely many zeros of \(p\), has removable singularities at each of them, and extends to an entire function. On the extended domain \[\left |g(z)\right | = \frac {\left |w(z)\right |}{\left |p(z)\right |} \leq 1 ,\] and by continuity the bound persists at the removed points. So \(g\) is a bounded entire function, hence constant by Liouville, say \(g \equiv c\) with \(\left |c\right | \leq 1\). Therefore \[\boxed {w(z) = c\,p(z),\qquad \left |c\right | \leq 1 .} \qquad \blacksquare \]

The pattern is the same as the \(ce^{z}\) and \(cz^{n}\) problems: dividing by the dominating function turns a growth hypothesis into a boundedness hypothesis, and Liouville converts that into rigidity. What differs is only how the zeros of the divisor are handled.

Solution. (a). Liouville’s Theorem. If \(f\) is entire and there is \(M\) with \(\left |f(z)\right | \leq M\) for all \(z \in \mathbb {C}\), then \(f\) is constant.

(b). Write \(x = \operatorname {Re}z\) and observe that \[\left |e^{z}\right | = e^{\operatorname {Re}z} = e^{x},\] so the hypothesis \(\left |f(z)\right | \leq e^{x}\) says exactly \[\left |f(z)\right | \leq \left |e^{z}\right | \qquad \text {throughout } \mathbb {C}.\] Put \(g = fe^{-z}\), entire because \(e^{-z}\) is, with \(\left |g\right | \leq 1\) everywhere. By part (a), \(g\) is constant and \[\boxed {f(z) = c\,e^{z} \qquad \text {with } \left |c\right | \leq 1 .}\]

The step to notice is the first one: \(e^{x}\) looks like a real bound depending on only half the variable, and recognising it as \(\left |e^{z}\right |\) turns the problem into the standard one. Every such bound of the form \(e^{\operatorname {Re}(az)}\) works the same way.

Solution. Map the half-plane hypothesis to a disc hypothesis, then apply Schwarz.

Since \(\operatorname {Re}f \geq 0\) and \(f\) is non-constant, \(f\) maps \(D\) into the closed right half plane; by the open mapping theorem the image is open, so in fact \(\operatorname {Re}f > 0\). The Cayley transform \[\varphi (w) = \frac {w-1}{w+1}\] carries the right half plane onto the unit disc and sends \(1 \mapsto 0\). So \[g = \varphi \circ f = \frac {f-1}{f+1}\] is analytic \(D \rightarrow D\) with \(g(0) = 0\), because \(f(0) = 1\).

Schwarz’s lemma gives \(\left |g(z)\right | \leq \left |z\right |\). Inverting \(\varphi \), \(f = \frac {1+g}{1-g}\), so \[\left |f(z)\right | = \frac {\left |1+g(z)\right |}{\left |1-g(z)\right |} \leq \frac {1+\left |g(z)\right |}{1-\left |g(z)\right |} \leq \frac {1+\left |z\right |}{1-\left |z\right |},\] the last step because \(t \mapsto \frac {1+t}{1-t}\) is increasing on \([0,1)\) and \(\left |g(z)\right | \leq \left |z\right |\). \(\blacksquare \)

\[\boxed {\left |f(z)\right | \leq \frac {1+\left |z\right |}{1-\left |z\right |}}\]

The same computation with the reverse triangle inequality gives the companion lower bound \(\left |f(z)\right | \geq \frac {1-\left |z\right |}{1+\left |z\right |}\); together they are the Herglotz bounds for functions with positive real part.

Solution. Since \(f(0) = f'(0) = 0\), the Taylor series of \(f\) begins at \(z^{2}\), so \[f(z) = z^{2}g(z)\] with \(g\) analytic on \(D\).

\(\left |g\right | \leq 1\). Fix \(\left |z\right | = r < 1\). Then \[\left |g(z)\right | = \frac {\left |f(z)\right |}{r^{2}} \leq \frac {1}{r^{2}} .\] By the maximum modulus principle, this bound on the circle \(\left |z\right | = r\) holds throughout \(\left |z\right | \leq r\). Now fix \(z\) and let \(r \uparrow 1\): the right-hand side tends to \(1\), so \(\left |g(z)\right | \leq 1\) for every \(z \in D\).

The bound. Differentiating \(f = z^{2}g\) twice and setting \(z = 0\) gives \(f''(0) = 2g(0)\), so \[\left |f''(0)\right | = 2\left |g(0)\right | \leq \boxed {2 .}\]

Equality. If \(\left |f''(0)\right | = 2\) then \(\left |g(0)\right | = 1\), so the analytic function \(g\) attains the maximum of its modulus at an interior point of \(D\). By the maximum modulus principle \(g\) is constant, say \(g \equiv c\) with \(\left |c\right | = 1\), and \[f(z) = c\,z^{2},\qquad \left |c\right | = 1 . \qquad \blacksquare \]

This is the order-\(2\) Schwarz lemma. The same argument with \(f\) vanishing to order \(k\) gives \(\left |f^{(k)}(0)\right | \leq k!\) with equality only for \(cz^{k}\).

Solution. Statement. Let \(f\) be analytic on a region \(G\). If \(\left |f\right |\) attains a local maximum at a point of \(G\), then \(f\) is constant on \(G\). Equivalently, if \(G\) is bounded and \(f\) is continuous on \(\overline {G}\) and analytic on \(G\), then \[\max _{\overline {G}}\left |f\right | = \max _{\partial G}\left |f\right | .\]

Proof. Suppose \(\left |f\right |\) has a local maximum at \(z_0 \in G\), so \(\left |f(z)\right | \leq \left |f(z_0)\right |\) on some disc \(B(z_0,\rho ) \subseteq G\). For \(0 < r < \rho \) the mean value property — Cauchy’s formula on the circle — gives \[f(z_0) = \frac {1}{2\pi }\int ^{2\pi }_{0} f\!\left (z_0+re^{i\theta }\right )d\theta ,\] so taking moduli, \[\left |f(z_0)\right | \leq \frac {1}{2\pi }\int ^{2\pi }_{0} \left |f\!\left (z_0+re^{i\theta }\right )\right | d\theta \leq \left |f(z_0)\right | .\] The two ends agree, so the middle inequality is equality; since the integrand is continuous and \(\leq \left |f(z_0)\right |\), it must equal \(\left |f(z_0)\right |\) identically. So \(\left |f\right |\) is constant on every circle about \(z_0\) of radius \(< \rho \), hence on the disc.

An analytic function of constant modulus on a region is constant: if \(\left |f\right |^{2} = f\overline {f} = c\) with \(c \neq 0\) then \(\overline {f} = c/f\) is analytic, so both \(f\) and \(\overline {f}\) are analytic and \(f\) is constant; and if \(c = 0\) then \(f \equiv 0\).

So \(f\) is constant on \(B(z_0,\rho )\), and by the Identity Theorem constant on the region \(G\). \(\blacksquare \)

The boundary form follows because \(\left |f\right |\) is continuous on the compact \(\overline {G}\) and so attains its maximum somewhere; by the above, an interior maximum forces \(f\) constant, in which case the boundary value is the same.

Solution. Suppose \(f : D \rightarrow D\) is analytic with two distinct fixed points \(a\) and \(b\).

Move one fixed point to the origin by a disc automorphism. Let \[\varphi _a(z) = \frac {z-a}{1-\overline {a}z},\] which maps \(D\) onto \(D\) bijectively with \(\varphi _a(a) = 0\), and set \[g = \varphi _a \circ f \circ \varphi _a^{-1} : D \rightarrow D .\] Then \(g(0) = \varphi _a(f(a)) = \varphi _a(a) = 0\), so Schwarz’s lemma applies: \[\left |g(w)\right | \leq \left |w\right | \qquad (w \in D).\]

Now let \(c = \varphi _a(b) \neq 0\), since \(\varphi _a\) is injective and \(b \neq a\). Because \(b\) is fixed by \(f\), \[g(c) = \varphi _a\big (f(b)\big ) = \varphi _a(b) = c ,\] so \(\left |g(c)\right | = \left |c\right |\) — equality in the Schwarz bound at a non-zero point.

The equality case of Schwarz’s lemma forces \(g(w) = \lambda w\) with \(\left |\lambda \right | = 1\); and \(g(c) = c\) with \(c \neq 0\) gives \(\lambda = 1\). Hence \(g = \operatorname {id}\), and conjugating back, \[f = \varphi _a^{-1}\circ g\circ \varphi _a = \operatorname {id},\] that is \(f(z) = z\) for all \(z \in D\). \(\blacksquare \)

So a disc self-map is either the identity or has at most one fixed point — the rigidity coming entirely from the equality clause of Schwarz, which is the part of that lemma students most often skip.

Solution. If \(C = 0\) then \(\left |f\right | = 0\) on \(\partial G\), and by the maximum modulus principle \(\left |f\right | \leq 0\) throughout \(\overline {G}\), so \(f \equiv 0\) — which is constant. Assume therefore \(C > 0\).

Suppose \(f\) has no zero in \(G\). Then \(1/f\) is analytic on \(G\) and continuous on \(\overline {G}\), since \(\left |f\right | = C > 0\) on the boundary and \(f \neq 0\) inside. Apply the maximum modulus principle to both \(f\) and \(1/f\): \[\max _{\overline {G}}\left |f\right | = \max _{\partial G}\left |f\right | = C, \qquad \max _{\overline {G}}\frac {1}{\left |f\right |} = \max _{\partial G}\frac {1}{\left |f\right |} = \frac {1}{C}.\] The second says \(\left |f\right | \geq C\) throughout, the first says \(\left |f\right | \leq C\); hence \(\left |f\right | \equiv C\) on \(\overline {G}\).

An analytic function of constant modulus on a region is constant. So \(f\) is constant. \(\blacksquare \)

\[\boxed {\text {Either } f \text { is constant, or } f \text { has a zero in } G .}\]

Boundedness of \(G\) is needed so that \(\overline {G}\) is compact and the maximum principle in its boundary form applies; on an unbounded region the statement fails, as \(f(z) = e^{z}\) on a strip shows.

Solution. No. The Schwarz–Pick lemma says that for any analytic \(f : D \rightarrow D\) and any \(z \in D\), \[\frac {\left |f'(z)\right |}{1-\left |f(z)\right |^{2}} \ \leq \ \frac {1}{1-\left |z\right |^{2}} .\] (It is Schwarz’s lemma conjugated by the automorphisms \(\varphi _a\), exactly as in the fixed-point problem above.)

Evaluate the two sides with the prescribed data at \(z = \tfrac 12\): \[1 - \left |z\right |^{2} = 1 - \tfrac 14 = \tfrac 34, \qquad 1 - \left |f(z)\right |^{2} = 1 - \tfrac {9}{16} = \tfrac {7}{16}.\] The inequality therefore requires \[\left |f'\!\left (\tfrac 12\right )\right | \ \leq \ \frac {7/16}{3/4} = \frac {7}{12} \approx 0.5833 .\] But the prescription demands \[f'\!\left (\tfrac 12\right ) = \frac {2}{3} \approx 0.6667 > \frac {7}{12}.\]

The required derivative exceeds the maximum permitted, so \[\boxed {\text {no such } f \text { exists.}} \qquad \blacksquare \]

Note how little freedom there is: once \(f(\frac 12)\) is fixed at \(\frac 34\), the derivative there is confined to a disc of radius \(\frac 7{12}\), and the bound is attained only by the automorphism carrying \(\frac 12\) to \(\frac 34\).

Solution. (a). \(M(r)\) is continuous and non-decreasing.

Non-decreasing. For \(r_1 < r_2 < R\), the closed disc \(\left |z\right | \leq r_1\) lies inside \(\left |z\right | \leq r_2\), and by the maximum modulus principle the maximum of \(\left |f\right |\) over the larger closed disc is attained on its boundary circle. Hence \[M(r_1) = \max _{\left |z\right |\leq r_1}\left |f\right | \leq \max _{\left |z\right |\leq r_2}\left |f\right | = M(r_2).\] (Note \(M(r) = \sup _{\left |z\right |=r}\left |f\right |\) equals the maximum over the closed disc, again by the maximum principle — this identification is what makes monotonicity immediate.)

Continuity. \(\left |f\right |\) is continuous on the compact annulus \(r_1 \leq \left |z\right | \leq r_2\), hence uniformly continuous there; a small change in \(r\) moves each boundary point a small distance, so \(M(r)\) changes by little. Formally, given \(\varepsilon \), uniform continuity supplies \(\delta \) with \(\left |\,\left |f(z)\right |-\left |f(w)\right |\,\right | < \varepsilon \) whenever \(\left |z-w\right | < \delta \), and comparing the circles of radii \(r\) and \(r'\) with \(\left |r-r'\right | < \delta \) radially gives \(\left |M(r)-M(r')\right | \leq \varepsilon \).

(b). Strictly increasing when \(f\) is non-constant. Suppose \(M(r_1) = M(r_2)\) for some \(r_1 < r_2\). The maximum of \(\left |f\right |\) over the closed disc \(\left |z\right | \leq r_2\) equals \(M(r_2) = M(r_1)\), and it is attained at a point of the circle \(\left |z\right | = r_1\), which is an interior point of that disc. By the maximum modulus principle \(f\) is then constant.

So for non-constant \(f\) the values \(M(r_1)\) and \(M(r_2)\) must differ, and being non-decreasing they satisfy \(M(r_1) < M(r_2)\): \(M\) is strictly increasing. \(\blacksquare \)

Solution. Statement (Jensen’s formula). Let \(f\) be analytic on \(\left |z\right | \leq R\) with \(f(0) \neq 0\), and let \(a_1,\ldots ,a_N\) be its zeros in \(\left |z\right | < R\), repeated according to multiplicity and none on the boundary circle. Then \[\log \left |f(0)\right | = -\sum ^{N}_{k=1}\log \frac {R}{\left |a_k\right |} + \frac {1}{2\pi }\int ^{2\pi }_{0}\log \left |f\!\left (Re^{i\theta }\right )\right |d\theta .\]

Proof. First suppose \(f\) has no zeros in \(\left |z\right | \leq R\). Then \(\log \left |f\right |\) is harmonic there — it is the real part of a branch of \(\log f\), available because the disc is simply connected and \(f\) is zero-free — so the mean value property gives the formula with the empty sum.

In general, divide out the zeros using Blaschke-type factors adapted to radius \(R\): \[g(z) = f(z)\prod ^{N}_{k=1}\frac {R^{2}-\overline {a_k}z}{R\left (z-a_k\right )} .\] Each factor is analytic and zero-free on \(\left |z\right | \leq R\) except that it cancels one zero of \(f\), so \(g\) is analytic and zero-free on the closed disc. Crucially, on \(\left |z\right | = R\) each factor has modulus \(1\): \[\left |\frac {R^{2}-\overline {a_k}z}{R(z-a_k)}\right | = 1 \qquad \text {when } \left |z\right | = R,\] as one checks by writing \(z = Re^{i\theta }\) and using \(\left |R^{2}-\overline {a}z\right | = R\left |z - a\right |\) there.

Applying the zero-free case to \(g\), \[\log \left |g(0)\right | = \frac {1}{2\pi }\int ^{2\pi }_{0}\log \left |g\!\left (Re^{i\theta }\right )\right |d\theta = \frac {1}{2\pi }\int ^{2\pi }_{0}\log \left |f\!\left (Re^{i\theta }\right )\right |d\theta ,\] the last step because the extra factors have modulus \(1\) on the circle. Meanwhile at the centre \[\left |g(0)\right | = \left |f(0)\right |\prod ^{N}_{k=1}\frac {R}{\left |a_k\right |},\] and taking logarithms rearranges to the stated formula. \(\blacksquare \)

The hinted integral \(\int _0^{2\pi }\log \left |1-e^{i\theta }\right |d\theta = 0\) is what licenses the argument when a zero sits on the boundary circle: the logarithmic singularity is integrable and contributes nothing.

Jensen’s formula is the quantitative statement that a function cannot have many zeros without being small somewhere: each zero contributes a positive term \(\log (R/\left |a_k\right |)\) that the boundary average must pay for.

Solution. Statement (Open Mapping Theorem). A non-constant analytic function on a region \(G\) is an open map: it carries open subsets of \(G\) to open subsets of \(\mathbb {C}\).

Sketch of proof. It suffices to show the image contains a disc about \(f(z_0)\) for each \(z_0 \in G\). Translating, assume \(z_0 = 0\) and \(f(0) = 0\).

Because \(f\) is non-constant and \(G\) is connected, the zeros of \(f\) are isolated, so choose \(\rho > 0\) with \(\overline {B(0,\rho )} \subseteq G\) and \(f(z) \neq 0\) for \(0 < \left |z\right | \leq \rho \). Put \[\delta = \min _{\left |z\right | = \rho }\left |f(z)\right | > 0,\] positive because the circle is compact and \(f\) does not vanish on it.

Now let \(\left |w\right | < \delta /2\). On the circle \(\left |z\right | = \rho \), \[\left |f(z) - w - f(z)\right | = \left |w\right | < \frac {\delta }{2} < \delta \leq \left |f(z)\right | ,\] so by Rouché the functions \(f(z)\) and \(f(z) - w\) have the same number of zeros inside \(\left |z\right | < \rho \). Since \(f\) has at least one (namely \(z = 0\)), so does \(f - w\); that is, \(w \in f\left (B(0,\rho )\right )\).

Hence \(B\left (0,\delta /2\right ) \subseteq f\left (B(0,\rho )\right )\), and the image contains a disc about \(f(0)\). \(\blacksquare \)

The maximum modulus principle is an immediate corollary: an open set in \(\mathbb {C}\) contains points of larger modulus than any given one of its points, so \(\left |f\right |\) can have no interior maximum.

Solution. \(M(r) = \max _{\left |z\right | \leq r}\left |f\right |\), the two definitions agreeing by the maximum modulus principle.

Increasing. This is part (a) of the earlier problem: for \(r_1 < r_2\), \(M(r_1) \leq M(r_2)\), and equality would place the maximum over the larger closed disc at an interior point, forcing \(f\) constant. Since \(f\) is non-constant, \(M\) is strictly increasing.

\(M(r) \rightarrow \infty \). Suppose not. Then \(M\) is increasing and bounded above, say \(M(r) \leq K\) for all \(r\). But \(f\) is entire, so this says \(\left |f(z)\right | \leq K\) for every \(z \in \mathbb {C}\): \(f\) is a bounded entire function and Liouville makes it constant, contrary to hypothesis. Hence \[\boxed {\lim _{r\to \infty } M(r) = \infty .} \qquad \blacksquare \]

So the modulus of a non-constant entire function is unbounded, and grows monotonically in \(r\). It says nothing about the growth rate, which may be arbitrarily slow along some directions — \(e^{z}\) decays to \(0\) along the negative real axis while \(M(r) = e^{r}\).

Solution. This is the strict-inequality version of the Herglotz bound, with the equality case asked for as well.

As before, \(\varphi (w) = \frac {w-1}{w+1}\) carries the right half plane onto \(D\) with \(\varphi (1) = 0\), so \[g = \frac {f-1}{f+1} : \Delta \rightarrow \Delta ,\qquad g(0) = 0 ,\] is analytic (the denominator does not vanish, as \(\operatorname {Re}f > 0\)). By Schwarz’s lemma \(\left |g(z)\right | \leq \left |z\right |\), and inverting, \[\left |f(z)\right | = \left |\frac {1+g(z)}{1-g(z)}\right | \leq \frac {1+\left |g(z)\right |}{1-\left |g(z)\right |} \leq \frac {1+\left |z\right |}{1-\left |z\right |}.\]

The equality case. Suppose equality holds at some \(z_0 \neq 0\) in \(\Delta \). Both inequalities above must then be equalities. The second forces \(\left |g(z_0)\right | = \left |z_0\right |\), which is equality in Schwarz’s lemma at a non-zero point; hence \[g(z) = \lambda z, \qquad \left |\lambda \right | = 1 ,\] and consequently \[f(z) = \frac {1+\lambda z}{1 - \lambda z}\] for some unimodular \(\lambda \). These are precisely the extremal functions: each is a Möbius map of the disc onto the right half plane with \(f(0) = 1\).

Conversely equality in the first inequality, \(\left |1+g\right | = 1+\left |g\right |\), requires \(g(z_0)\) to be a positive multiple of \(1\), i.e. real and positive, which pins \(\lambda z_0 > 0\) and fixes \(\lambda = \left |z_0\right |/z_0\). \(\blacksquare \)

Solution. Answer: \(P(z) = z^{n}\), with \(M_P = 1\).

\(M_P \geq 1\) always. Write \(P(z) = z^{n} + a_1z^{n-1} + \cdots + a_n = \sum ^{n}_{k=0}c_k z^{k}\) with \(c_n = 1\). On the unit circle, Parseval’s identity for Fourier coefficients gives \[\frac {1}{2\pi }\int ^{2\pi }_{0}\left |P\!\left (e^{i\theta }\right )\right |^{2}d\theta = \sum ^{n}_{k=0}\left |c_k\right |^{2} \geq \left |c_n\right |^{2} = 1 .\] The maximum of a function is at least its root-mean-square, so \[M_P^{2} \geq \frac {1}{2\pi }\int ^{2\pi }_{0}\left |P\right |^{2}d\theta \geq 1, \qquad M_P \geq 1 .\] (By the maximum modulus principle the maximum over \(\left |z\right | \leq 1\) is attained on the circle, so working there loses nothing.)

Equality forces \(P = z^{n}\). If \(M_P = 1\) then both inequalities above are equalities. The second gives \(\sum ^{n-1}_{k=0}\left |c_k\right |^{2} = 0\), so every lower coefficient vanishes and \(P(z) = z^{n}\). That polynomial does have \(M_P = \max _{\left |z\right |\leq 1}\left |z\right |^{n} = 1\), so the minimum is attained. \(\blacksquare \)

An equivalent one-line argument: the leading coefficient is \(1 = \frac {1}{2\pi i}\oint _{\left |z\right |=1} P(z)z^{-n-1}dz\), and estimating gives \(1 \leq M_P\) at once — though the Parseval route is what delivers the equality case.

Numerically: perturbing \(z^{3}\) to \(z^{3}+0.3z^{2}\) raises the maximum to \(1.3\), and to \(z^{3}+0.5z\) raises it to \(1.5\). Every perturbation costs.

Solution. (a). Schwarz’s Lemma. Let \(f\) be analytic on the unit disc \(D\) with \(f(0) = 0\) and \(\left |f(z)\right | \leq 1\) on \(D\). Then \[\left |f(z)\right | \leq \left |z\right | \ \text { for all } z \in D, \qquad \text {and}\qquad \left |f'(0)\right | \leq 1 .\] If equality holds in the first at any \(z \neq 0\), or in the second, then \(f(z) = \lambda z\) with \(\left |\lambda \right | = 1\).

(b). With \(f(0) = f'(0) = 0\), write \(f(z) = z^{2}h(z)\) with \(h\) analytic on \(D\). For \(\left |z\right | = r < 1\), \[\left |h(z)\right | = \frac {\left |f(z)\right |}{r^{2}} \leq \frac {1}{r^{2}},\] and by the maximum modulus principle this holds on \(\left |z\right | \leq r\); letting \(r \uparrow 1\) gives \(\left |h\right | \leq 1\) throughout \(D\). Hence \[\left |f(z)\right | = \left |z\right |^{2}\left |h(z)\right | \leq \left |z\right |^{2} .\]

Suppose \(\left |f(z_0)\right | = \left |z_0\right |^{2}\) for some \(z_0 \neq 0\). Then \(\left |h(z_0)\right | = 1\), so \(\left |h\right |\) attains its maximum at an interior point and \(h\) is constant of modulus \(1\) by the maximum modulus principle: \[f(z) = e^{i\alpha }z^{2}.\] Therefore, unless \(f\) is of that form, the inequality is strict: \[\boxed {\left |f(z)\right | < \left |z\right |^{2} \quad \text {on } D \qquad \text {unless } f(z) = e^{i\alpha }z^{2}. } \qquad \blacksquare \]

This is the order-\(2\) case of the general principle: vanishing to order \(k\) at the origin improves the Schwarz bound from \(\left |z\right |\) to \(\left |z\right |^{k}\), with the same rigid equality case.