Complex Integration and Cauchy Theorems Questions and Answers

Answer: C

The limit is analytic, so in particular \(f'\) is analytic.

Weierstrass's theorem says that a uniform limit (indeed a locally uniform limit) of analytic functions on a region is analytic. One proof: for any closed triangle in G, \(\oint f = \lim \oint f_{n} = 0\), and \(f\) is continuous, so Morera's theorem gives analyticity. The derivative of an analytic function is again analytic, and in fact \(f_{n}' \to f'\) locally uniformly.

So the options saying \(f\) is not differentiable, or differentiable but not analytic, are false, and 'f analytic but f' not analytic' is impossible because analytic functions are infinitely differentiable. This is a sharp contrast with real analysis, where a uniform limit of smooth functions can be nowhere differentiable.

Answer: C

The function is \(f(z) = \dfrac{1 + z}{2 - z}\).

Write \(z = 1 - 1/n\), so \(n = 1/(1 - z)\). Substituting in the given values:

\[ \begin{aligned} \frac{2n - 1}{n + 1} &= \frac{\frac{2}{1 - z} - 1}{\frac{1}{1 - z} + 1} \\ &= \frac{2 - (1 - z)}{1 + (1 - z)} = \frac{1 + z}{2 - z} \end{aligned} \]

The points \(1 - 1/n\) lie in G and accumulate at 1, which is also in G. By the identity theorem, an analytic function on the connected set G is determined by its values on such a set, so \(f(z) = (1 + z)/(2 - z)\) everywhere in G (its pole at 2 lies outside G).

Check with \(n = 2\): \(z = 1/2\) gives \(1.5/1.5 = 1\), and \((2n - 1)/(n + 1) = 3/3 = 1\). The option \((2z - 1)/(z + 1)\) gives 0 there, and the other two give values other than 1.

Answer: C

Both (i) and (ii) are true.

(i) If Re \(f \le\) Im \(f\), the entire function \(f\) maps C into the closed half plane \(\{w : \text{Re}\,w - \text{Im}\,w \le 0\}\). Then \(g = e^{(1 + i) f}\) satisfies \(|g| = e^{\text{Re}\,f - \text{Im}\,f} \le 1\), so \(g\) is a bounded entire function, constant by Liouville's theorem, and therefore \(f\) is constant.

(ii) The Schwarz-Pick lemma bounds the derivative of an analytic self-map of the disk: \(|f'(0)| \le 1 - |f(0)|^{2} = 1 - 1/4 = 3/4\). Equality is allowed, and it is attained by the disk automorphism

\[ \begin{aligned} f(z) &= \frac{z + 1/2}{1 + z/2} \\ f(0) &= \frac{1}{2} \\ f'(0) &= 1 - \frac{1}{4} = \frac{3}{4} \end{aligned} \]

So such an \(f\) exists. The trap is to read the Schwarz-Pick bound as strict and call (ii) false.

Answer: D

1/z has a primitive on the half-plane {x ≥ 2}.

A primitive of \(1/z\) is a branch of \(\log z\). Such a branch exists on a domain that does not wind around the origin; on the half-plane \(\text{Re } z \ge 2\) (or any region inside the right half-plane) the principal logarithm works.

The other three regions contain a circle \(|z| = r\) around 0, and

\[ \oint_{|z| = r} \frac{dz}{z} = 2\pi i \neq 0, \]

which is impossible if \(1/z\) had a primitive there, because the integral of a derivative round a closed curve is 0. The disk \(|z| \lt 2\) also contains the singularity at 0 itself.

Answer: C

The false statement is the one about a sequence of distinct complex numbers.

The identity theorem needs the zeros to have a limit point in C. A sequence of distinct points can run off to infinity: \(f(z) = \sin(\pi z)\) vanishes at \(a_{n} = n\) for every integer n, yet \(f \not\equiv 0\).

The other three are true. The zeros in [−1, 1] accumulate inside it; 1/n accumulates at 0; and an uncountable subset of C always has a limit point (otherwise each bounded piece would be finite and the set countable). In each case the identity theorem gives \(f \equiv 0\).

Answer: D

The integral equals 20πi.

The numerator is the derivative of the denominator \(p(z) = 1 + z + \ldots + z^{10}\), so the integrand is \(p'(z)/p(z)\). By the argument principle

\[ \begin{aligned} &\oint_{|z| = R} \frac{p'(z)}{p(z)}\, dz \\ &= 2\pi i \times (\text{number of zeros inside}) \end{aligned} \]

p has degree 10, so it has 10 zeros counted with multiplicity, and all lie inside \(|z| = R\). The integral is \(2\pi i \times 10 = 20\pi i\).

\(10\pi i\) halves the factor \(2\pi i\); 10 forgets it altogether; and 0 would mean no zeros inside.

Answer: A

The true statement is the first: if \(f\left(\frac{1}{2}\right) = \frac{1}{2}\), then \(f(z) = z\) for all \(z \in \Delta\).

Since \(f(0) = 0\) and f maps the open disk into the closed disk (and by the maximum principle, into the open disk unless it is constant), Schwarz's lemma gives \(|f(z)| \le |z|\), with equality at a single non-zero point only if \(f(z) = e^{i\theta} z\). Here \(|f(1/2)| = 1/2\), so \(f(z) = e^{i\theta} z\), and \(f(1/2) = 1/2\) forces \(e^{i\theta} = 1\).

The others fail. \(|f'(0)| = 1\) gives only \(f(z) = e^{i\theta} z\), for example \(f(z) = -z\). Schwarz gives \(|f(z)| \le |z|\), the reverse of option C. And \(f(z) = z^{2}\) has \(f'(0) = 0\) without being constant.

Answer: D

There is an analytic \(g : D \to \mathbf{C}\) with \(f(z) = e^{g(z)}\).

A disk is simply connected. Since f never vanishes, \(\frac{f'}{f}\) is analytic on D and has a primitive h there; then \(\left(f e^{-h}\right)' = 0\), so \(f = c\, e^{h}\) with \(c \neq 0\), and writing \(c = e^{a}\) gives \(g = h + a\). In short, a non-vanishing analytic function on a simply connected domain has an analytic logarithm.

The others do not follow. \(e^{z}\) never vanishes but is not one-one on a disk of radius greater than \(\pi\), and on an unbounded region it need not be bounded. The derivative of an analytic function is always analytic, so 'f ' need not be analytic' is false.