Singularities and Residues Questions and Answers

Answer: D

The correct statement is the residue of \(f\) at \(z = 0\) is 1/6.

Near 0, \(\sin z = z(1 - z^{2}/6 + \cdots)\), so \(z^{2}\sin z\) has a zero of order 3 and \(f\) has a pole of order 3, not 1 or 2. Expanding,

\[ \begin{aligned} \frac{1}{z^{2}\sin z} &= \frac{1}{z^{3}}\left(1 + \frac{z^{2}}{6} + \cdots\right) \\ &= \frac{1}{z^{3}} + \frac{1}{6z} + \cdots \end{aligned} \]

The coefficient of \(1/z\) is 1/6, so the residue is 1/6, not 1/3.

Answer: B

The integral equals \(\frac{\pi i}{60}\).

The only singularity inside \(|z| = 1\) is the essential singularity at 0, so the integral is \(2\pi i\) times the coefficient of \(1/z\) in the Laurent series:

\[ \begin{aligned} z^{4}\sin\frac{1}{z} &= z^{4}\left(\frac{1}{z} - \frac{1}{3!\,z^{3}} + \frac{1}{5!\,z^{5}} - \cdots\right) \\ &= z^{3} - \frac{z}{6} + \frac{1}{120 z} - \cdots \end{aligned} \]

\[ \begin{aligned} \oint z^{4}\sin\frac{1}{z}\, dz &= 2\pi i \cdot \frac{1}{120} \\ &= \frac{\pi i}{60} \end{aligned} \]

The option \(\pi i/30\) doubles the residue (using 60 instead of 120), and \(60\pi i\) and \(12\pi i\) put the factorial in the numerator.

Answer: C

The residue of \(\cot z\) at \(\pi\) is 1.

\(\cot z = \cos z / \sin z\). At \(z = \pi\), \(\sin z\) has a simple zero and \(\cos \pi = -1 \neq 0\), so the pole is simple and

\[ \begin{aligned} \text{Res}_{z = \pi} \frac{\cos z}{\sin z} &= \frac{\cos \pi}{(\sin z)'|_{z = \pi}} \\ &= \frac{\cos \pi}{\cos \pi} = 1 \end{aligned} \]

The same holds at every \(k\pi\): \(\cot z\) has residue 1 at each multiple of \(\pi\), which is why \(\pi \cot \pi z\) is used to sum series over the integers.

The residue is not 0 (the pole is genuine), and \(\pi\) or \(2\pi\) confuse the location of the pole with its residue.

Answer: D

The correct statement is f(G) is dense in C.

\(e^{1/z}\) has an essential singularity at 0, and G is a punctured disk around it. By the Casorati-Weierstrass theorem, the image of every punctured neighbourhood of an essential singularity is dense in C. Picard's great theorem says more: the image omits at most one value. Here 0 is omitted, since \(e^{w} \neq 0\), so \(f(G) = \mathbf{C} \setminus \{0\}\).

That makes \(f(G) = \mathbf{C}\) false, because 0 is never attained. The set \(\{0 \lt |z| \lt e^{10}\}\) is too small: on G, \(w = 1/z\) covers \(|w| \gt 10\), and \(e^{w}\) for such \(w\) takes arbitrarily large moduli. \(f(G) = G\) is also wrong for the same reason.

Answer: B

f is not one-one but g is one-one.

For \(g(z) = e^{z}\): \(e^{z_{1}} = e^{z_{2}}\) forces \(z_{1} - z_{2} = 2\pi ik\). In A, \(|z_{1} - z_{2}| \lt 1 \lt 2\pi\), so \(k = 0\) and \(z_{1} = z_{2}\). So g is one-one on A.

For \(f(z) = e^{1/z}\): as z ranges over A, \(w = 1/z\) ranges over all of \(|w| \gt 2\). That region contains points such as \(w = 3\) and \(w = 3 + 2\pi i\), which have the same exponential. Their preimages \(z = 1/3\) and \(z = 1/(3 + 2\pi i)\) both lie in A and give equal values of f, so f is not one-one.

Answer: C

The residue is 1.

Expand the numerator:

\[ \begin{aligned} \frac{e^{z} - 1}{z^{2}} &= \frac{z + \frac{z^{2}}{2} + \frac{z^{3}}{6} + \ldots}{z^{2}} \\ &= \frac{1}{z} + \frac{1}{2} + \frac{z}{6} + \ldots \end{aligned} \]

The residue is the coefficient of \(1/z\), which is 1. The value \(\frac{1}{2}\) is the constant term, not the residue, and 0 or 2 come from expanding the exponential incorrectly.

Answer: A

The function has a removable singularity at z = 0.

Both numerator and denominator vanish at 0, each to first order:

\[ \begin{aligned} \sin z &= z - \tfrac{z^{3}}{6} + \ldots \\ 1 - e^{z} &= -z - \tfrac{z^{2}}{2} - \ldots \end{aligned} \]

so \(f(z) = \frac{1 - z^{2}/6 + \ldots}{-1 - z/2 - \ldots} \to -1\) as \(z \to 0\). A finite limit means the singularity is removable; defining \(f(0) = -1\) makes f analytic there.

A pole would need the denominator to vanish to higher order than the numerator, and an essential singularity would need infinitely many negative powers. The poles of this function are at \(z = 2\pi i k\), \(k \neq 0\), not at the origin.

Answer: C

The integral equals \(4i\).

The integrand \(\frac{\sin z}{(z - \pi/2)z}\) has singularities at \(\pi/2\) and 0. The contour is the circle of radius 1 about \(\pi/2 \approx 1.571\), which contains \(\pi/2\) but not 0 (its distance from \(\pi/2\) is 1.571, more than 1).

The residue at the simple pole \(\pi/2\) is \(\frac{\sin(\pi/2)}{\pi/2} = \frac{2}{\pi}\), so by the residue theorem

\[ \int_{\gamma} = 2\pi i \cdot \frac{2}{\pi} = 4i \]

The value 0 would need no singularity inside; \(2\pi i\) treats the residue as 1, forgetting the factor \(\frac{1}{z}\) evaluated at \(\pi/2\); and \(-2\pi i\) adds a sign error as well.

Answer: C

The true statement is \(\overline{f(A)} = \mathbf{C}\).

\(\sin\frac{1}{z - 1}\) has an essential singularity at \(z = 1\): its Laurent series about 1 has infinitely many negative powers. A is a punctured disk about that point, and the Casorati-Weierstrass theorem says that the image of any punctured neighbourhood of an essential singularity is dense in C.

A dense image is neither bounded nor confined to a half-plane, which rules out the first two options. f is analytic at every point of A (the only bad point, \(z = 1\), is excluded), so 'f is not analytic on A' is false.