Question

f(x)=sin ^-1{1/z} has singularity?​

Answer 1

Answer:

f(z) = sin(1/z), z = 0 has the Laurent expansion f(z − z0)=1/z − 1/z33! + 1/z55! ... It has an isolated essential singularity at z0 = 0. Theorem: For f differentiable in 0 < |z − z0| < R, the statements a)z0 is a removable singularity; b) limz→z0 f(z) < ∞; c) f is bounded in a neighbourhood of z0; are equivalents