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Complex number - Practise set 9
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Prakhar Shrivastava
INDUSTRIAL AND PRODUCTION | IIT B | NITIE | GATE - 2016 ( AIR -03, SCORE - 939/1000) |Tennis | Street plays | Foodie

Unacademy user
Sir ..your way to describe is very nice .short and easy so I like ur all video ..
  1. Engineering Mathematics - Complex numbers Practice set PRAKHAR SHRIVASTAVA

  2. EDUCATOR'S INTRODUCTION Prakhar Shrivastava I am an Industrial and production engineering graduate JEC, Jabalpur GATE 2016 AIR 03 (PI GATE SCORE 939/1000 Cracked written and interview for 1101, IIT B Currently pursuing PGDIE from NITIE, Mumbai My hobbies Tennis, Street plays and big time Foodie Code PrakharShrivastava333 Https:/

  3. INSTRUCTIONS . These questions are not developed by me, various books, materials videos and notes are referred. My coribution is to bring everything under one roof These questions are only for educational purpose and learning You need to practice along with video for better results Iwill not speak for most of the time as I want you to practice and make 2. 3. mistakes so that you can learn in most interactive way 5. Lesson contain both solved and unsolved questions in randomly distributed manner 6. I tried my best but still some images are blurr, so please adjust as it is beneficial for you because we together have to solve as many questions as possible MORE YOU PRACTISE MORE YOU WILL GET PERFECT 7.

  4. 6,27 ExpundI3 n a arent series, valid for; (a) 1 3ljl <3-6)l3.(O0<Iz+ Solution . . (o) Resolving into partial fractio I. ,i iel 3,is en the required Laurent exea siondalid for bothjz/> 1 and Izl ,:::. (b) If Izl > 1, we. have as in part ten ..:; 22 2z 2z* E 12 3 927, [i .:: Then the required Laurent expansion valid for both Iz/> 1 and ld > 3, i.e. 12 > 3, is by subtraction . ....4 13 40

  5. 4.27 Let C is the curvey4 joining points (and (2,3). find the value of (12z2 - 4iz) dz : Metltod L. By Problem 4.17, the integral is independent of the path joining (1.and (2.3). Hence, any path can be chosen. In particular, let us choose the straight line paths from (1, 1) to (2. 1) and then from (2, 1) to (2,:3). i Solution Case 1. Along the path from (1. 1) to (2. 1). y- 1. dy -0 so ihat z*x t iy-xi, di - dt - Then, the integral equals 422+30F igi Case 2. Along the path: frorn (2,1) to (2, 3), x Then, the integral equals 2, dr-O so that z-x + iy-2 + l dz . (12(2 +)4(2 ) dy(4(222176+si hen, adding the required:value (20.+ 0176815638i.