Originally Posted by

**foolsgold** Unfortunately I'm not very good with singularities, so any help with this proof would be very much apprectiated. Kudos in advance!!!!!

*S (a,r) *denotes a positively orientated circular contour of radius *r* centered at *a*

*D'(a,r)* denotes a disk such that {z: 0< |z-a|< r}

Let f be an analytic function on the domain D'(0,2) and suppose that for all n=0,1,2...

S(0,1)**∫ **(z^n)f(z)dz = 0

Show that f has a removable singularity at z=0

(The s(0.1) is meant to be below the integral sign)