48.10 ....tell us the valiid reason... function is boundes and entire they why it's not bounded ...the theorem says that entire and bounded function are constant
Sir in first question option d is also correct because you have given open interval.... And at 34:00 time only option a is correct and b and c option is wrong because f(3)= 1/3 is true but 3 doesn't belong in the domain D.. similar for simple pole -3 does not belong in domain D....... Please correct sir.. i am confused..
Sir. I have one doubt in 33:56. In option 3, 3 does not belong to the domain, |z| less than 1. Then how can we choose the given function? Can we take any value for z??
Sir in the gate question in time stamp 39:55 , how limit point of Zn is 1, it is zero for all. Isn't we find limit point of images in co-domain??!! Kindly please take a look and reply, otherwise it gets confusing
32:40 Sir please solve my doubt if function is f( z)=2/(3+z) then how we can say -3 is pole of f while -3 is not in the domain, and second 3 is also does not belong the domain D then how we calculate value of f(3) . By identiy theorem, if we take g(z)= 2 /(3+z) then f(z)= g(z) but g(z) have no singularity in domain, neither pole nor removebale so how -3 is pole of f(z). Kya identity theorem me function exist karne ke bad codomain se singularity check ki jati hai ya domain se , yadi domain se check hoti hai to -3 is not pole , and 3 is also does not belong the domain D fir se only option a is wright aa rha hai . Kya kare?