Sir aapki class chahe paid course ho ya RU-vid ki class hr class me kuch nya hi milta hai seekhne ko ❤ thank you so much sir 🙏 for thus amazing class Mere Bahut se doubts the Taylor aur Laurent series ko lekr sb short out ho gye Aur hmko Cauchy integral formula exact nhi pta tha use kaise krte hain Laurent me Mera Sara doubt clear ho gya sir Thank you so much Sir ❤
Sir aapne bola tha sinz/z^2 ke example mai ki hum usme direct cauchy integral formula mhi lga sakte hai kyunki function analytic hai annular region mai to fir humne jo a_0,a_1 nikala tha usme fir kyu lagaya cauchy integral formula? Plz sir bta dijiye bhut confusion ho rha hai isme
@@SaloniJain-p1uyou didn’t notice one thing ☹️ when you are finding a0 and a1 then function is f(z)=sinz / z^2 and when you are solving using CIF then the analytic function in the contour is not f(z). Watch it again with slow speed and find your mistakes.
Sir at 1:20, Q4, what is the correct option? Since limit z tending to ♾️ exists finitely, it means f(z)/z^n=a, a constant which implies that f(z)=az^n, similarly g(z) = b z^n. So which means option 3 is correct. Kindly correct me if i am wrong. Thank u
@@SBTechMathhow is that possible here in this case ? Like if we take h(z)= f(z)/z^n, then limit z tending to infinity h(z) =1, and hence exists finitely, which means h(z) has removable singularity at ♾️, which further implies that h(z) is a constant because only constant functions have removable singularity at infinity. So taking h(z) = constant, we get f(z)= az^n=z^n. How can we here get, less degree terms ?
Sir in last question...f(z) has two poles 1 and 2 If we expand by laurants expansion....why there comes infinite terms of 1/z Pole me to only finite terms hote hai of 1/z???
Nice class sir. Sir i have a doubt, does analytic continuation is possible for the function 1/z-1 beyond the boundary |z|=1?? does there exist series expansion for this function if boundary contains singularity?
