Solution. The zero set of \(f\) contains \(\{z_n\}\), which has the limit point \(z_0\) lying in \(D\). By the Identity Theorem — an analytic function on a region whose zero set has a limit point in that region vanishes identically — we get \(f \equiv 0\) on \(D\). \(\blacksquare \)

The real counterexample. Define \[g(x) = \begin {cases} e^{-1/x^{2}} & x > 0,\\ 0 & x \leq 0 . \end {cases}\] Every derivative of \(e^{-1/x^{2}}\) is a rational function times \(e^{-1/x^{2}}\) and so tends to \(0\) as \(x \downarrow 0\); hence \(g \in C^{\infty }(\mathbb {R})\), and \(g\) vanishes on the whole interval \((-\infty ,0]\) without vanishing identically.

Why there is no contradiction. \(g\) is infinitely differentiable but not analytic at \(0\): its Taylor series there is identically zero, which does not represent \(g\) on any neighbourhood of \(0\). Complex differentiability at every point of an open set forces analyticity — a Taylor series that converges to the function — and it is that, not smoothness, which the Identity Theorem uses.

Solution. The series is geometric with ratio \(z/2\): \[\sum ^{\infty }_{n=0}\frac {z^{n}}{2^{\,n+1}} = \frac {1}{2}\sum ^{\infty }_{n=0}\left (\frac {z}{2}\right )^{n} = \frac {1}{2}\cdot \frac {1}{1 - \frac z2} = \boxed {\frac {1}{2-z}} \qquad \left (\left |z\right | < 2\right ).\]

The function \(\frac {1}{2-z}\) is analytic on \(\mathbb {C}\setminus \{2\}\), and that is its largest domain of analyticity: \(z = 2\) is a simple pole, so no continuation past it is possible.

The continuation of the element \(\left (\text {series},\ \left |z\right |<2\right )\) to that domain is the function \(\frac {1}{2-z}\) itself, and by uniqueness of analytic continuation it is the only one. The disc \(\left |z\right | < 2\) is simply the largest disc centred at the origin avoiding the pole; it is an artefact of where we chose to expand, not a boundary of the function.

Checked numerically at \(z = 0.5,\ -1.3,\ 1.9\) and \(0.7+i\): series and closed form agree to eight decimals.

Solution. Consider \(g(z) = z^{2}\), which is analytic on the disc and satisfies \(g\!\left (\tfrac 1n\right ) = \tfrac 1{n^{2}}\) for every \(n\).

Then \(f - g\) is analytic on the unit disc and vanishes at every point of \(\left \{\tfrac 1n : n \geq 2\right \}\). That set has limit point \(0\), which lies in the disc, so by the Identity Theorem \(f - g \equiv 0\). Hence \[\boxed {f(z) = z^{2}}\] and it is the only such function. \(\blacksquare \)

The whole force is that the limit point is interior. Had the data been prescribed at points accumulating only at the boundary — say at \(1 - \tfrac 1n\) — no such conclusion would follow, and in fact interpolation at such a sequence is always possible, which is the content of the Weierstrass interpolation problem appearing elsewhere in these papers.

Solution. Suppose such an \(f\) existed. Look at the even and odd indices separately.

Even \(n = 2m\). Here \(f\!\left (\tfrac {1}{2m}\right ) = \tfrac {1}{2m}\), so \(f\) agrees with \(g(z) = z\) on the set \(\left \{\tfrac {1}{2m}\right \}_{m\geq 1}\), which has limit point \(0\) in the disc. By the Identity Theorem \(f(z) = z\) throughout.

Odd \(n = 2m+1\). Here \(f\!\left (\tfrac {1}{2m+1}\right ) = -\tfrac {1}{2m+1}\), so by the same argument on \(\left \{\tfrac {1}{2m+1}\right \}\) — also with limit point \(0\) — we get \(f(z) = -z\) throughout.

But \(z\) and \(-z\) are different functions. The contradiction shows no such \(f\) exists. \(\blacksquare \)

Equivalently in one line: any candidate would have to satisfy \(f(z) = z\) and \(f(z) = -z\) simultaneously, forcing \(f \equiv 0\), which matches neither prescription.

Compare the previous problem. There the data came from a single analytic function and the answer was unique; here the data oscillates between two analytic patterns, and analyticity cannot accommodate that near an interior accumulation point. Prescribing values on a sequence tending to an interior point is enormously restrictive — it either determines the function completely or is impossible.

Solution. Start with the branch of \(z^{1/2}\) near \(z = 1\) normalised by \(1^{1/2} = 1\), and continue along \(\gamma (t) = e^{2\pi i t}\), \(t \in [0,1]\).

Writing the element at parameter \(t\) as \(e^{\frac 12 \log z}\) with the argument carried continuously, the value at the point of the path is \[\gamma (t)^{1/2} = e^{\frac 12 (2\pi i t)} = e^{\pi i t}.\] At \(t = 1\) the path has returned to \(z = 1\), but the element has arrived with \[e^{\pi i} = -1 .\] So the continuation is \(-z^{1/2}\): one circuit reverses the sign.

A second circuit multiplies by \(-1\) again, returning \(+z^{1/2}\). Hence \[\boxed {\text {two circuits are required.}}\]

Comparison with the logarithm. Each circuit adds \(2\pi i\) to \(\log z\), and \(2\pi i \neq 0\) however many times it is added — so \(\log \) never returns to its initial element, and no finite number of circuits closes it up. The square root has a monodromy group of order \(2\); the logarithm’s is infinite. In the same way \(z^{1/m}\) closes after \(m\) circuits, and \(z^{c}\) for irrational \(c\) never does.

Solution. \(D\) is simply connected. \(D = \mathbb {C}\setminus (-\infty ,0]\) is star-shaped about the point \(1\): for any \(z \in D\) the segment from \(1\) to \(z\) stays in \(D\), since the segment can only meet \((-\infty ,0]\) if \(z\) does. A star-shaped set is contractible — shrink along the segments to the centre — so every closed path in \(D\) is homotopic to a constant, which is simple connectedness.

Applying the Monodromy Theorem. Take the function element given by the principal logarithm on a small disc about \(1\). Since \(0 \notin D\), the element can be continued along every path in \(D\): at each point \(a \in D\) the branch extends to the disc of radius \(\left |a\right |\) about \(a\), which avoids the origin. The hypotheses of the Monodromy Theorem are met, so continuation along homotopic paths gives the same result; and as \(D\) is simply connected, any two paths from \(1\) to a given point are homotopic. The continuations therefore fit together into a single-valued analytic branch of \(\log z\) on all of \(D\). \(\blacksquare \)

Where it breaks without the cut. On \(\mathbb {C}\setminus \{0\}\) the continuation along every path still exists, so that hypothesis survives. What fails is simple connectedness: the unit circle is a closed path that is not homotopic to a constant, because contracting it would have to sweep across the origin, which has been removed. The Monodromy Theorem then says nothing, and indeed the conclusion is false — continuing round that circle returns \(\log z + 2\pi i\).

The cut does exactly one job: it destroys the loop.

Solution. The extension. Let \(D = \mathbb {C}\), symmetric about the real axis, with \(D^{+}\) the upper half plane and \(D^{0} = \mathbb {R}\). The hypotheses of the Schwarz Reflection Principle hold: \(f\) is analytic on \(D^{+}\), continuous on \(D^{+}\cup \mathbb {R}\), and real-valued on \(\mathbb {R}\). Hence \(f\) continues analytically to the whole plane, the continuation being \[f(z) = \overline {f\!\left (\overline {z}\right )} \qquad \text {for } \operatorname {Im}z < 0 .\] An analytic function on all of \(\mathbb {C}\) is entire. \(\blacksquare \)

Bounded implies constant. Suppose in addition \(\left |f(z)\right | \leq M\) on the upper half plane and on \(\mathbb {R}\). In the lower half plane the extension satisfies \[\left |f(z)\right | = \left |\overline {f\!\left (\overline {z}\right )}\right | = \left |f\!\left (\overline {z}\right )\right | \leq M,\] because conjugation preserves modulus and \(\overline {z}\) lies in the upper half plane. So the extended \(f\) is a bounded entire function, and Liouville’s theorem makes it constant. \(\blacksquare \)

Note that the bound transfers to the reflected half automatically — reflection cannot enlarge \(\left |f\right |\) — so “bounded on the upper half plane” is enough; boundedness on the whole plane need not be assumed separately.

Solution. Write \(f(z) = \displaystyle \sum ^{\infty }_{n=0} z^{n!}\). The coefficients are \(0\) and \(1\) with infinitely many equal to \(1\), so \(\limsup \left |a_k\right |^{1/k} = 1\) and the radius of convergence is \(1\).

Fix \(m \geq 1\) and let \(\zeta \) be any \(m!\)-th root of unity, so \(\zeta ^{m!} = 1\). For every \(n \geq m\) the factorial \(n!\) is a multiple of \(m!\), hence \[\zeta ^{\,n!} = 1 \qquad (n \geq m).\] Approach \(\zeta \) along the radius, \(z = r\zeta \) with \(r \uparrow 1\). Splitting the series at \(n = m\), \[f(r\zeta ) = \underbrace {\sum ^{m-1}_{n=0} (r\zeta )^{\,n!}}_{\text {bounded, finitely many terms}} \;+\; \sum ^{\infty }_{n=m} r^{\,n!} .\] Every term of the second sum is a positive real, and each individual term \(r^{\,n!} \rightarrow 1\) as \(r \uparrow 1\). So for any \(K\) we may take \(r\) close enough to \(1\) that the first \(K\) of them each exceed \(\tfrac 12\), giving \(\sum _{n\geq m} r^{n!} > K/2\). Hence \[\left |f(r\zeta )\right | \longrightarrow \infty \qquad (r \uparrow 1),\] and \(\zeta \) cannot be a regular point: no analytic continuation exists on any neighbourhood of it.

The \(m!\)-th roots of unity, taken over all \(m\), are dense in the unit circle — already the \(m!\)-th roots alone are spaced \(2\pi /m!\) apart. Every arc of the circle therefore contains a singular point, so no arc admits a continuation and \[\boxed {\left |z\right | = 1 \text { is a natural boundary for } \sum z^{n!}.}\] \(\blacksquare \)

The argument is the one used for \(\sum z^{2^{n}}\) with \(m!\) in place of \(2^{m}\); all it needs is that the exponents are eventually divisible by any fixed one. Numerically the blow-up is slow — at \(r = 0.999999\) the tail has still only climbed to about \(6.7\) — because the terms switch off in enormous jumps. Slow divergence is still divergence.