Discussion
Let \(f(z)\) be an entire function, then \(f(z)\) is also bounded if and only if :
Answer: D
An entire function is bounded if and only if it is a constant.
One direction is trivial: a constant is bounded. The other is Liouville's theorem. If \(|f(z)| \le M\) on C, Cauchy's estimate on a circle of radius R gives \(|f'(z_{0})| \le M/R\) for every R, so \(f' \equiv 0\) and f is constant.
The other options fail. A non-constant polynomial tends to infinity as \(|z| \to \infty\). The reciprocal of a non-constant polynomial has poles, so it is not entire. Polynomials in \(\sin z\) and \(\cos z\) are entire but generally unbounded, since \(|\sin(iy)| = |\sinh y|\) grows without limit.
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