Complex Analysis and Residues Questions and Answers

Answer: D

The product is \(a \times b \times c = \frac{3}{32}\).

\(U = x^{3}y - xy^{3}\). For \(U + iV\) to be analytic, the Cauchy-Riemann equations \(U_{x} = V_{y}\) and \(U_{y} = -V_{x}\) must hold:

\[ \begin{aligned} 3x^{2}y - y^{3} &= 4by^{3} + 2cx^{2}y \\ x^{3} - 3xy^{2} &= -4ax^{3} - 2cxy^{2} \end{aligned} \]

Matching coefficients: \(2c = 3\), \(4b = -1\) and \(4a = -1\), so \(a = b = -\frac{1}{4}\) and \(c = \frac{3}{2}\). Then

\[ abc = \frac{1}{16} \times \frac{3}{2} = \frac{3}{32} \]

The constant \(k\) drops out of every derivative. The other values do not satisfy both equations; 1/8, for example, would need \(c = 2\).

Answer: B

The correct statement is the order of the pole is 3; it is the only option the key accepts in this multiple-select question.

Put \(w = z - \pi\). Near \(z = \pi\), \(\sin z = -\sin w = -w + \frac{w^{3}}{6} - \ldots\) has a simple zero, which cancels one power of the \((z - \pi)^{4}\) in the denominator, while \(z^{2} \rightarrow \pi^{2} \neq 0\). So

\[ f(z) \approx \frac{-\pi^{2}}{(z - \pi)^{3}} \]

and the pole is of order 3, not 4.

The residue is the coefficient of \(w^{3}\) in \(z^{2}\sin z = (\pi^{2} + 2\pi w + w^{2})(-w + w^{3}/6 - \ldots)\), which is

\[ \text{Res} = \frac{\pi^{2}}{6} - 1 \approx 0.64 \]

so neither \(\pi/6\) nor \(2\pi/3\) is the residue.

Answer: C

The residues are \(-\frac{3}{16}\) and \(\frac{3}{16}\).

At \(z = 0\) (a pole of order 2), the residue is the first derivative of \(\frac{1}{(z-2)^{3}}\) at 0. At \(z = 2\) (order 3), it is half the second derivative of \(\frac{1}{z^{2}}\) at 2:

\[ \begin{aligned} \text{Res}_{0} &= \left.\frac{-3}{(z-2)^{4}}\right|_{0} = -\frac{3}{16} \\ \text{Res}_{2} &= \frac{1}{2}\left.\frac{6}{z^{4}}\right|_{2} = \frac{3}{16} \end{aligned} \]

Check: the function falls off as \(z^{-5}\), so the residues must sum to zero, as they do. The options with \(\frac{3}{8}\) forget the factor \(\frac{1}{2!}\) or the power of 2 in the denominator.

Answer: B

The singular points (poles) are at \(2i\) and \(-2i\).

Factor numerator and denominator. The denominator groups as \(z^{2}(z - 1) + 4(z - 1)\):

\[ \begin{aligned} f(z) &= \frac{(z - 1)(z - 4)}{(z - 1)(z^{2} + 4)} \\ &= \frac{z - 4}{(z - 2i)(z + 2i)}, \quad z \neq 1 \end{aligned} \]

The factor \(z - 1\) cancels, so z = 1 is only a removable singularity: defining \(f(1) = -3/5\) makes the function analytic there. The genuine singularities are the simple poles at \(z = \pm 2i\). The points \(2 \pm i\) are not zeros of the denominator at all, which rules out the other three options.