Discussion
\(I = \int_{c} \frac{z^{2}\,dz}{(z^{2}+1)(z^{2}+4)}\), in the z-plane.
Here c is a contour as shown : the closed contour made of the real axis from -R to R and the semicircular arc of radius R about Z = 0 in the upper half-plane, with the points i and 2i marked inside it.
then :
Answer: D
The integral is \(I = \pi/3\).
The integrand has simple poles at \(z = \pm i\) and \(z = \pm 2i\); the contour encloses \(z = i\) and \(z = 2i\). Their residues are:
\[ \begin{aligned} \text{Res}_{z=i} &= \frac{i^{2}}{(2i)(i^{2}+4)} = \frac{-1}{6i} \\ \text{Res}_{z=2i} &= \frac{(2i)^{2}}{(4i)((2i)^{2}+1)} = \frac{1}{3i} \end{aligned} \]
\[ \begin{aligned} I &= 2\pi i\left(\frac{-1}{6i} + \frac{2}{6i}\right) \\ &= \frac{2\pi i}{6i} = \frac{\pi}{3} \end{aligned} \]
The option \(-\pi/3\) is the contribution of the pole at i alone, missing the pole at 2i; \(\pi/6\) is half the correct value; and 0 would hold only if the contour enclosed no pole.
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