We have function $\frac{1}{x-\frac{i}{2}}$ which has one singular point $x=\frac{i}{2}$ so if we integrate around it we get by Residue theorem $2 \pi i$.

Integrate[1/(x-I/2),{x,-1,1,1+I,-1+I,-1}]
(* 2 Pi I *)

I constructed the rectangular path {x,-1,1,1+I,-1+I,-1} manually.

Now I want to do the same integral but instead use a circle path.

I tried:

Integrate[1/((a+I b)-I/2),{a,b}\[Element]Circle[{0,1/2},1]]
(* 0 *)

But it seems it is not a circle around the singular point as the result is 0. What am I doing wrong?