Are you sure you're not being asked to show that f(z) = cot(z) is ANALYTIC for all z? f can be ANY function (by definition, a function is defined by the user)...
So I understand how to show that there is no solutions, but my lecturer said before that, I must show that for allSuppose that a function that is analytic in some arbitrary region Ω in the complex plane containing the interval [1,1.2]. Assume for all . Show that has no solutions in Ω.
I'm not sure how to explain it. There are some theorems that say if (for z on some interval) has an accumulation point a the origin then for all z in the complex plane, but this equation has no accumulation point at the origin.
So how do I go about showing f(z) = cot(z) for all z, instead of just the small interval given?
Nope, the question was copy/pasted exactly as is, with a few LaTex added in.
At first I was thinking since f is analytic then it can be represented as a power series. And by the uniqueness theorem for the power series, I can argue that it must be cot(z) everywhere, but the only thing missing is the accumulation point at the origin.