Question
Download Solution PDFमान लीजिए f(z) = exp\(\rm\left(z+\frac{1}{z}\right)\), z ∈ ℂ\{0} है। तब f का z = 0 पर अवशेष _______ है।
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFअवधारणा:
f(z) का z = 0 पर अवशेष, f(z) के मैक्लॉरिन श्रेणी प्रसार में \(\frac1z\) के गुणांक के बराबर होता है।
व्याख्या:
f(z) = exp\(\rm\left(z+\frac{1}{z}\right)\)
= \(e^z.e^{\frac1z}\)
= \((1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+...)\)\(.(1+\frac1zz+\frac{1}{z^22!}+\frac{1}{z^33!}+...)\)
इसलिए उपरोक्त व्यंजक में \(\frac1z\) का गुणांक
= \(\frac11+\frac1{2!.1!}+\frac1{3!.2!}+\frac1{4!.3!}+...\)
= \(\sum_{l=0}^{\infty} \frac{1}{l !(l+1) !}\)
इसलिए विकल्प (3) सही है।
Last updated on Jun 5, 2025
-> The NTA has released the CSIR NET 2025 Notification for the June session.
-> The CSIR NET Application Form 2025 can be submitted online by 23rd June 2025
-> The CSIR UGC NET is conducted in five subjects -Chemical Sciences, Earth Sciences, Life Sciences, Mathematical Sciences, and Physical Sciences.
-> Postgraduates in the relevant streams can apply for this exam.
-> Candidates must download and practice questions from the CSIR NET Previous year papers. Attempting the CSIR NET mock tests are also very helpful in preparation.