Discussion
The bilinear transformation :
\(f(z) = \dfrac{3z + 4}{5z + 6}\) from \(\mathbf{C} - \left\{-\dfrac{6}{5}\right\}\) to \(\mathbf{C}\)
assumes all complex values, except :
Answer: D
The omitted value is 3/5.
Solve \(w = \dfrac{3z + 4}{5z + 6}\) for \(z\): \(5wz + 6w = 3z + 4\), so
\[ z = \frac{4 - 6w}{5w - 3} \]
This gives a valid \(z\) for every \(w\) except \(w = 3/5\). That value is the limit of \(f(z)\) as \(z \to \infty\), the ratio of the leading coefficients, and since \(\infty\) is not in the domain it is never taken.
\(-6/5\) is the excluded point of the domain, not a missing value of the range. 5/3 and \(-4/3\) are both taken, at \(z = -9/4\) and \(z = -36/29\) respectively, as the formula for \(z\) shows.
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