Complex Integration and Cauchy Theorems Questions and Answers

Answer: C

One value is \(a = 3 + i\).

The three vertices lie on the unit circle at angles \(90^{\circ}\), \(225^{\circ}\) and \(315^{\circ}\), so the triangle encloses the origin but not \(z = 2\). The term \(1/(z - 2)^{4}\) is analytic inside \(C\) and the constant 4 integrates to zero, so only the pole at 0 contributes:

\[ \begin{aligned} -(a - 2)^{2}\cdot 2\pi i &= 4\pi \\ (a - 2)^{2} &= \frac{4\pi}{-2\pi i} = 2i \\ a - 2 &= \pm(1 + i) \end{aligned} \]

since \((1 + i)^{2} = 2i\). So \(a = 3 + i\) or \(a = 1 - i\), and only \(3 + i\) is offered. \(1 + i\), \(2 + i\) and \(4 + i\) give \((a-2)^{2} = -2i\), \(-1\) and \(3 + 4i\), none equal to \(2i\).

Answer: B

\(f(2) = 5\).

Since \(f\) is entire, \(f''\) is entire too, and it is bounded by \(M\) on the whole plane. By Liouville's theorem \(f''\) is constant, so \(f\) is a polynomial of degree at most 2: \(f(z) = az^{2} + bz + c\).

\[ \begin{aligned} f(0) = 1 &\Rightarrow c = 1 \\ f'(0) = 0 &\Rightarrow b = 0 \\ f(1) = 2 &\Rightarrow a + 1 = 2,\ a = 1 \end{aligned} \]

So \(f(z) = z^{2} + 1\) and \(f(2) = 5\). The value is real because every coefficient is real, which rules out \(2 + 5i\) and \(5 + 2i\); 2 is what you get by taking \(f\) linear and ignoring the quadratic term.

Answer: D

An entire function is bounded if and only if it is a constant.

One direction is trivial: a constant is bounded. The other is Liouville's theorem. If \(|f(z)| \le M\) on C, Cauchy's estimate on a circle of radius R gives \(|f'(z_{0})| \le M/R\) for every R, so \(f' \equiv 0\) and f is constant.

The other options fail. A non-constant polynomial tends to infinity as \(|z| \to \infty\). The reciprocal of a non-constant polynomial has poles, so it is not entire. Polynomials in \(\sin z\) and \(\cos z\) are entire but generally unbounded, since \(|\sin(iy)| = |\sinh y|\) grows without limit.

Answer: D

The keyed condition is f is an analytic function on D.

The 'only if' direction is Morera's theorem: if f is continuous on an open set D and its integral over every closed curve in D is 0, then f has a primitive F locally (integrate along paths from a fixed point), F is analytic, and so its derivative f is analytic too.

The 'if' direction is Cauchy's theorem. Strictly it needs D to be simply connected: on the punctured plane, \(1/z\) is analytic but its integral around the unit circle is \(2\pi i\). The question follows the textbook pairing of Morera and Cauchy, under which analyticity is the characterising property.

The other conditions are too narrow or irrelevant. The zero function and polynomials do have zero integrals, but so does every analytic function such as \(e^{z}\), so neither is necessary. Periodicity is neither necessary nor sufficient: \(z^{2}\) is not periodic yet integrates to zero, while \(\sin(\text{Re}\, z)\) is periodic and continuous but its integral around the unit circle is \(i\int_{0}^{2\pi} \sin(\cos t)\cos t\, dt \ne 0\).

Answer: B

The integral is 0.

At \(z = 0\) the integrand has a removable singularity, because

\[ \begin{aligned} \frac{\sin z}{z} &= 1 - \frac{z^{2}}{3!} \\ &\quad + \frac{z^{4}}{5!} - \ldots \end{aligned} \]

has no negative powers. Defining it as 1 at 0 makes it an entire function, and by Cauchy's theorem the integral of an entire function around the closed curve \(|z| = 1\) is 0. Equivalently, the residue at 0 is the coefficient of \(1/z\), which is 0.

\(2\pi i\) is the answer for \(\cos z/z\) or \(e^{z}/z\), where the numerator is 1 at the origin; \(-2\pi i\) is the same with the orientation reversed, and \(2\pi\) drops the factor \(i\).

Answer: B

The value is 1.

The curve \(\nu(t) = e^{4\pi it}\), \(0 \le t \le 1\), runs round the unit circle twice, so its winding number about 0 is 2. By the residue theorem,

\[ \begin{aligned} &\frac{1}{2\pi i}\int_{\nu} \frac{e^{z}}{z^{3}}\,dz \\ &= n(\nu, 0)\,\mathrm{Res}_{z=0}\frac{e^{z}}{z^{3}} \end{aligned} \]

The residue is the coefficient of \(z^{2}\) in \(e^{z}\), which is \(1/2\). So the value is \(2 \cdot \tfrac{1}{2} = 1\).

Forgetting the double winding gives 1/2 (not offered); 2 is the winding number alone; and \(4\pi i\) is the integral without the factor \(1/(2\pi i)\) and with residue 1.

Answer: A

The answer is \(\mathrm{I}_{1} = \mathrm{I}_{2}\).

Both curves go once anticlockwise around the origin: \(\nu_{1}\) is the unit circle and \(\nu_{2}\) the circle of radius 10. They are homotopic in \(\mathbf{C} - \{0\}\), through the circles of radius \(r\) between 1 and 10, and \(f\) is analytic on the annulus between them. By Cauchy's theorem (the integral over the boundary of the annulus is zero), the two integrals are equal.

The radius does not scale the integral. For \(f(z) = 1/z\), both integrals equal \(2\pi i\), which rules out \(\mathrm{I}_{1} = 10\,\mathrm{I}_{2}\) and \(\mathrm{I}_{1} = \mathrm{I}_{2}/10\), and the relation is a definite one, not 'no relation'.

Answer: D

The impossible statement is \(f(\mathbf{C}) \subseteq \mathbf{R}\).

By the open mapping theorem, a non-constant entire function maps C onto an open set, and no non-empty open subset of the plane lies inside the real line. (Alternatively, if \(f\) were real, \(e^{if}\) would be a bounded entire function, hence constant by Liouville.)

The others occur: \(f(z) = z\) is onto C (A); \(e^{z}\) omits exactly 0 (B), consistent with Picard's theorem; and the image of any non-constant entire function is dense in C (C), by the Casorati-Weierstrass argument.

Answer: A

Only (a) is true.

(a) \(e^{\sin z}\) is a non-constant entire function, so by the maximum modulus principle its modulus has no maximum on C; indeed it is unbounded.

(b) is false. \(\sin(e^{z})\) has zeros: \(e^{z} = \pi\) has the solution \(z = \ln \pi\), and \(\sin \pi = 0\). So \(|\sin(e^{z})|\) attains its minimum value 0. The minimum modulus principle forbids an interior minimum only for functions without zeros.

So 'only (b)', 'both true' and 'both false' are wrong.

Answer: A

The false statement is that \(f\) maps a line onto a circle.

The growth bound \(|f(z)| \le |z|\) for large \(|z|\) and the Cauchy estimates make \(f\) a polynomial of degree at most 1. Being non-constant, \(f(z) = az + b\) with \(a \neq 0\).

Such a map is a rotation, scaling and translation, so it sends lines to lines and circles to circles, never a line onto a circle. The other statements are true for \(az + b\): it is one-one, onto, and \(f'(z) = a \neq 0\) everywhere.