#csirnet #mathematics #Feb2022 | Cauchy Integral Formula - Complex Integration

Hello friends, this is Mahesh Velikar and you are watching NextGenMath. If you are not subscribed to my channel, then please subscribe to the channel and press the bell icon. In this video, we are going to discuss the problem of Sir Saini's 5th and 18th problem of complex analysis. The question is, let gamma be the positively oriented circle, z belongs to C, modulus of z equal to 3 by 2. Suppose that integration e power i pi z divided by z minus 1 z minus 2i dz equal to 2 pi i c. Then modulus of C equals, we are given 4 options, first option 2, second option 5, third option 1 by 2 and fourth option 1 by 5. It is a problem of Cauchy integral formula. So first we recall the Cauchy integral formula. Let C is a simple closed curve and the function f of z is analytic on a region containing C and its interior. We assume C is oriented counterclockwise. Then for any z0 inside C, integration f of z divided by z minus z0 dz over Cauchy has value 2 pi i f of z0. Note that z0 inside C and f of z is analytic on C and inside C. Now here we have given simple closed curve, modulus z equal to 3 by 2 which is a circle with the center origin and the radius 3 by 2. For the given integration, e power i pi z divided by z minus 1 z minus 2i square. So integrant has two singular points z equal to 1 and z equal to 2i. But 2i is outside the circle so it does not affect the analyticity of the function. While z equal to 1 inside C so it may impact on the analyticity of the function. Therefore this integrant can be written in the form of e power i pi z divided by z minus 2i whole square divided by z minus 1. So here e power i pi z divided by z minus 2i square is analytic function inside and over the curve C. While z minus 1 that is z0 equal to 1 is a singular point. So by the Cauchy integral formula this integration has value 2 pi i f of 1 where f of z equal to e power i pi z divided by z minus 2i whole square. This integration has value 2 pi i c which is given in the data. So from the both side if we eliminate 2 pi i then we have c equal to f of 1. In this function if we put z equal to 1 then we have e power i pi divided by 1 minus 2i whole square. e power i pi has value minus 1 and if we open this bracket then we have 1 minus 4i plus 4i square. i square equal to minus 1 then we have minus 1 divided by minus 3 minus 4i. If we simplify then we have 1 upon 3 plus 4i. Now if we multiply by conjugate of 3 plus 4i then we have 3 minus 4i divided by 25. So 3 by 25 minus 4 by 25i equal to c. We want modulus of c. So if we take modulus of c we have square root a square plus b square a equal to 3 by 25 and b equal to minus 4 by 25. 9 square is 9 by 625 plus 16 by 625. If we simplify then we have 25 divided by 625 which is square root of 1 upon 25 equal to 1 upon 5. Modulus of c has value 1 by 5 which is a option number 4. So option number 4 is the correct answer. Thanks for watching. For more videos like, share, subscribe, press the bell icon of NextGenNet.