Discussion
The Taylor series for \(\dfrac{z}{z^{2} + 1}\) around 0 is :
Answer: C
The Taylor series is \(z - z^{3} + z^{5} - \ldots\).
For \(|z| \lt 1\), expand \(1/(1 + z^{2})\) as a geometric series and multiply by \(z\):
\[ \begin{aligned} \frac{z}{1 + z^{2}} &= z(1 - z^{2} + z^{4} - \ldots) \\ &= z - z^{3} + z^{5} - \ldots \end{aligned} \]
The radius of convergence is 1, the distance from 0 to the poles at \(\pm i\).
\(z - z^{3}/3! + z^{5}/5! - \ldots\) is \(\sin z\); \(z - z^{3}/3 + z^{5}/5 - \ldots\) is \(\tan^{-1} z\), which is the integral of \(1/(1 + z^{2})\), not \(z/(1 + z^{2})\); and \(z + z^{3} + z^{5} + \ldots\) is \(z/(1 - z^{2})\).
No comment is present. Be the first to comment.
Loading…
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