Re: Complex analysis problem
Setting a = 0 gives f(z) / z^(n+1) given the line integral region of y (gamma). If a function is analytic in the complex plane then all of its derivatives exist by default in the region that the function is itself analytic.
This means if f(z) is analytic in a region then its derivatives are also analytic in the same region (i.e. the f^n(a)).
So you need to break up f(z)/z^(n+1) into f(z)/z and 1/z^(n+1) and show that F(z) is analytic.
You have already shown that f^n(0) exists for all values of n (since 0 is in the region |z| < 1) so you need to use F(z) * 1/z^n in partial fraction decomposition within the integral.