I think the standard way to prove that function has an essential singularity is to show that on an arbitrarily small interval around 0, it takes all values between-1 and 1, meaning there is no way to continuously extend the function to 0 ( so it isn't a removable singularity, it's clearly not a pole because sine is bounded on the reals ).
22.
:It's a removable singularity, and as one professor said, " At this point we assume all removable singularities are removed . " In other words, while f ( z ) is technically undefined at z = a, since it is identical to \ frac { z-b } { z-c } elsewhere and can be analytically continued through it, you can essentially work with the analytic continuation instead of the function as you defined it.
23.
:It's a removable singularity, and as one professor said, " At this point we assume all removable singularities are removed . " In other words, while f ( z ) is technically undefined at z = a, since it is identical to \ frac { z-b } { z-c } elsewhere and can be analytically continued through it, you can essentially work with the analytic continuation instead of the function as you defined it.
24.
Formally, if U \ subset \ mathbb C is an open subset of the complex plane \ mathbb C, a \ in U a point of U, and f : U \ setminus \ { a \ } \ rightarrow \ mathbb C is a holomorphic function, then a is called a "'removable singularity "'for f if there exists a holomorphic function g : U \ rightarrow \ mathbb C which coincides with f on U \ setminus \ { a \ }.
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