Discussion
The function \(f(z)\) of complex variable \(z\) given below,
\[ f(z) = \frac{z^{2} - 5z + 4}{z^{3} + 4z - z^{2} - 4} \]
has singular points at \(z\) =
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.
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