Discussion
Consider the complex function
\[ f(z) = \frac{z^{2}\sin z}{(z - \pi)^{4}} . \]
At \(z = \pi\), which of the following options is(are) CORRECT?
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.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development