Solution. (a). The map \(w = \frac {z-a}{z-b}\) is a Möbius transformation sending \(a \mapsto 0\) and \(b \mapsto \infty \). A branch of \(\ln w\) exists wherever \(w\) omits a ray from \(0\) to \(\infty \), so the question is which set of \(z\) we must delete to make that happen.

Parametrise the segment joining \(a\) and \(b\) by \(z = a + t(b-a)\), \(t \in (0,1)\). Then \(z - a = t(b-a)\) and \(z - b = (t-1)(b-a)\), so \[w = \frac {t}{t-1} < 0 \qquad \text {for } t \in (0,1).\] The segment \([a,b]\) therefore maps exactly onto \((-\infty , 0]\), and on \[D = \mathbb {C} \setminus [a,b]\] the principal branch of the logarithm composes with \(w\) to give an analytic \(f\).

This is the answer whenever \(c \notin [a,b]\) — which covers \(c\) off the line through \(a\) and \(b\), and \(c\) on that line but outside the segment.

If instead \(c\) lies on the open segment \((a,b)\), then \(D\) misses it and we must cut elsewhere. Delete the two outward rays, \[\{a + t(b-a) : t \leq 0\} \cup \{a + t(b-a) : t \geq 1\}.\] The same computation gives \(w = \frac {t}{t-1} \in [0,1)\) for \(t \leq 0\) and \(w \in (1,\infty )\) for \(t > 1\), so these rays cover \([0,\infty )\) and a branch of \(\ln \) analytic off the positive axis serves.

(b). \(f\) is analytic on \(D\) and cannot be continued past the cut, since the cut carries the branch points \(a\) and \(b\). The radius of convergence at \(c\) is therefore the distance from \(c\) to the deleted set: \[R = \operatorname {dist}\big (c, [a,b]\big ) \quad \text {in the first case,}\qquad R = \min \big (\left |c-a\right |,\ \left |c-b\right |\big ) \quad \text {in the second.}\]

Solution. It converges for no \(p > 0\).

An infinite product \(\prod u_n\) is said to converge when its partial products tend to a non-zero limit, and a necessary condition for that is \[u_n \longrightarrow 1 .\] Here \(u_n = n^{-p} \rightarrow 0\) for every \(p > 0\), so the condition fails at once.

Directly: the \(N\)-th partial product is \[\prod ^{N}_{n=1} n^{-p} = \big (N!\big )^{-p} \longrightarrow 0 ,\] so the products tend to \(0\), which is divergence to zero rather than convergence. Since convergence fails for every \(p>0\), absolute convergence does too.

The question is worth its place because the non-zero requirement looks like a technicality and is not one: without it every product containing a single zero factor would “converge”, and the correspondence with \(\sum \log u_n\) — the reason products are tractable at all — would break down.

Solution. Write \(f(z) = \sqrt {(z-1)(z+1)}\). The branch points are \(z = \pm 1\), where the radicand vanishes to first order.

A circuit around one branch point multiplies \(f\) by \(-1\); a circuit enclosing both multiplies it by \((-1)^2 = +1\) and returns \(f\) unchanged. So it suffices to prevent loops that separate the two points, and the cut \[[-1, 1]\] does exactly that: on \(\mathbb {C}\setminus [-1,1]\) any closed curve either encloses both branch points or neither, and \(f\) is single-valued and analytic.

(The complementary choice \((-\infty ,-1] \cup [1,\infty )\) works equally well, for the same reason.)

Radius of convergence at \(z = i\). With the cut \([-1,1]\), the nearest point of the boundary to \(i\) is the origin, the foot of the perpendicular from \(i\) to the real segment. Hence \[R = \operatorname {dist}\big (i, [-1,1]\big ) = \left |i - 0\right | = 1 .\]

Solution. (a). The branch points are \(z = \pm i\), and by the computation of Problem 1 the segment joining them — here the piece of the imaginary axis \([-i, i]\) — maps under \(w = \frac {z+i}{z-i}\) onto \((-\infty , 0]\).

But \(z = 0\) lies on that segment, so cutting along \([-i,i]\) would remove the very point we need. Cut the outward rays instead: \[D = \mathbb {C} \setminus \{iy \ :\ \left |y\right | \geq 1\},\] on which \(w\) omits \([0,\infty )\) and a branch of \(\ln \) analytic off the positive real axis makes \(f\) analytic. This \(D\) is a neighbourhood of \(0\), as required.

(b). The nearest boundary points are \(\pm i\), so \[R = \operatorname {dist}\big (0, \{iy : \left |y\right | \geq 1\}\big ) = 1 .\]

Note how the two branch points, both at distance \(1\), act together: the radius is set by whichever piece of the cut comes closest, and here they tie.

Solution. (a). The branch points are \(z = 1\) and \(z = -1\). Putting \(z = -1 + 2t\) with \(t \in (0,1)\) — the segment from \(-1\) to \(1\) — gives \[\frac {z-1}{z+1} = \frac {2t-2}{2t} = \frac {t-1}{t} < 0 ,\] so the segment maps onto the negative reals and the principal branch of the square root is analytic on \[D = \mathbb {C}\setminus [-1,1].\]

(b). As in Problem 3 the nearest boundary point to \(i\) is the origin, so \[R = \operatorname {dist}\big (i,[-1,1]\big ) = 1 .\]

The two functions \(\sqrt {z^2-1}\) and \(\sqrt {\frac {z-1}{z+1}}\) have the same branch points and the same natural cut, and differ only by the factor \(z+1\), which is entire and non-vanishing off \(z=-1\). It is unsurprising that they share a radius of convergence at \(i\).

Solution. (a). The radius of convergence of \(\displaystyle \sum ^{\infty }_{n=0}a_n(z-a)^n\) is the number \(R \in [0,\infty ]\) given by the Cauchy–Hadamard formula \[\frac {1}{R} = \limsup _{n \rightarrow \infty }\left |a_n\right |^{1/n},\] with the conventions \(1/0 = \infty \) and \(1/\infty = 0\). The series converges absolutely for \(\left |z-a\right | < R\) and diverges for \(\left |z-a\right | > R\).

(b). Let \(0 < r < R\) and choose \(\rho \) with \(r < \rho < R\). Since \(\rho < R\) the series converges absolutely at any point at distance \(\rho \) from \(a\), so \[\sum ^{\infty }_{n=0}\left |a_n\right |\rho ^{n} < \infty .\] For \(\left |z - a\right | \leq r\), \[\left |a_n (z-a)^n\right | \leq \left |a_n\right | r^{n} = \left |a_n\right |\rho ^{n}\left (\frac {r}{\rho }\right )^{n} \leq \left |a_n\right |\rho ^{n} =: M_n ,\] a bound independent of \(z\) with \(\sum M_n < \infty \). By the Weierstrass \(M\)-test the series converges uniformly on \(\left |z-a\right | \leq r\). \(\blacksquare \)

The restriction \(r < R\) cannot be dropped: convergence is generally not uniform on the whole open disc. The geometric series on \(\left |z\right |<1\) is the standard witness.

Solution. (a). The branch points are \(z = i\) and \(z = 1\), and by Problem 1 the segment joining them maps under \(w = \frac {z-i}{z-1}\) onto \((-\infty ,0]\).

Is \(0\) on that segment? Points of it are \(i + t(1-i)\) for \(t \in [0,1]\), that is \(t + i(1-t)\). This vanishes only if \(t = 0\) and \(1 - t = 0\) simultaneously, which is impossible. So \(0 \notin [i,1]\) and we may take \[D = \mathbb {C} \setminus [i,1],\] a domain containing a neighbourhood of the origin, on which the principal branch of \(\ln \) makes \(f\) analytic.

(b). The segment from \(i\) to \(1\) lies on the line \(x + y = 1\). The foot of the perpendicular from the origin is \(\left (\tfrac 12,\tfrac 12\right )\), which lies on the segment, so the distance is the perpendicular distance: \[R = \frac {\left |0 + 0 - 1\right |}{\sqrt {1^2+1^2}} = \frac {1}{\sqrt {2}} = \frac {\sqrt {2}}{2} \approx 0.7071 .\]

Checking that the foot lands on the segment matters. Had it fallen outside, the nearest point would have been an endpoint and the answer would have been \(\min (\left |i\right |, \left |1\right |) = 1\) instead.

Solution. Convergence and analyticity. For \(\left |z\right | \leq r < 1\), \[\sum ^{\infty }_{n=1}\left |n z^{n}\right | \leq \sum ^{\infty }_{n=1} n r^{n} = \frac {r}{(1-r)^{2}} < \infty ,\] so \(\sum n z^n\) converges absolutely and uniformly on every closed subdisc. A product \(\prod (1 + c_n)\) converges absolutely and uniformly exactly when \(\sum \left |c_n\right |\) does, so with \(c_n = -nz^n\) the product converges absolutely and uniformly on compact subsets of \(\left |z\right |<1\). By Weierstrass’ theorem on analytic limits its value \(f(z)\) is analytic there.

The zeros. A convergent product vanishes precisely where a factor does, so \(f(z) = 0\) if and only if \[1 - n z^{n} = 0 \qquad \text {for some } n, \qquad \text {that is}\qquad z^{n} = \frac {1}{n}.\] For each \(n\) this has \(n\) roots, equally spaced in argument, all of modulus \[\left |z\right | = n^{-1/n}.\] Now \(n^{-1/n} = e^{-(\ln n)/n} \rightarrow e^{0} = 1\), so these circles of zeros approach the unit circle, and the \(n\)-th of them carries \(n\) zeros spread evenly round it. Given any \(\zeta \) with \(\left |\zeta \right | = 1\) and any \(\varepsilon > 0\), choose \(n\) large enough that both \(1 - n^{-1/n} < \varepsilon /2\) and the angular gap \(2\pi /n < \varepsilon /2\); some \(n\)-th root then lies within \(\varepsilon \) of \(\zeta \). Hence every point of \(\left |z\right | = 1\) is an accumulation point of zeros of \(f\).

The unit circle is consequently a natural boundary: \(f\) cannot be continued past any arc of it, since a continuation would be analytic and vanish on a set with a limit point.

Solution. Let \(P_N = \displaystyle \prod ^{N}_{n=1}(1 + w_n)\) and suppose \(P_N \rightarrow P\) with \(P \neq 0\) — the non-vanishing being part of what convergence of a product means.

Since \(P \neq 0\), the partial products are eventually bounded away from \(0\), so for large \(N\) we may divide: \[1 + w_N = \frac {P_N}{P_{N-1}} \longrightarrow \frac {P}{P} = 1 ,\] whence \(w_N \rightarrow 0\). \(\blacksquare \)

This is the exact analogue of “if \(\sum a_n\) converges then \(a_n \rightarrow 0\)”, and it fails in the same way as a converse: \(w_n \rightarrow 0\) does not make the product converge, as \(w_n = -\tfrac {1}{n+1}\) shows, where the partial products \(\prod \frac {n}{n+1} = \frac {1}{N+1} \rightarrow 0\).

It is also the step where the non-zero convention earns its keep. Without it, \(P = 0\) would be allowed and the division above would be illegitimate.

Solution. (a). Let \(L = \limsup \left |a_n\right |^{1/n} = 1/R\) and fix \(z\) with \(\left |z\right | < R\). Choose \(\rho \) with \(\left |z\right | < \rho < R\). Since \(1/\rho > L\), the definition of \(\limsup \) gives an \(N\) with \[\left |a_n\right |^{1/n} < \frac {1}{\rho }\qquad \text {for all } n \geq N,\] that is \(\left |a_n\right | < \rho ^{-n}\). Hence for \(n \geq N\), \[\left |a_n z^{n}\right | < \left (\frac {\left |z\right |}{\rho }\right )^{n},\] and \(\left |z\right |/\rho < 1\), so the tail is dominated by a convergent geometric series. Therefore \(\sum \left |a_n z^n\right |\) converges. \(\blacksquare \)

(b). For \(\displaystyle \sum ^{\infty }_{n=0} z^{n!}\) the coefficient sequence is \[a_k = \begin {cases} 1 & k = n! \text { for some } n,\\ 0 &\text {otherwise.}\end {cases}\] Infinitely many \(a_k\) equal \(1\), so \(\left |a_k\right |^{1/k} = 1\) along that subsequence and \(0\) elsewhere; hence \(\limsup \left |a_k\right |^{1/k} = 1\) and \[R = 1 .\] The largest disc centred at \(0\) on which the series defines an analytic function is therefore \[\boxed {\left |z\right | < 1 .}\]

It is worth adding that this disc cannot be enlarged in any direction whatever: the series has the unit circle as a natural boundary, by the argument of the analytic continuation chapter applied with \(m!\) in place of \(2^m\). So the answer is not merely ”the largest disc” but the largest open set of any shape.