Line integral around singular point (Residue theorem)

Question

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?

Answer 1

How about this:

f[z_] := 1/(z - I/2)
intF[f_, c_, r_] := 
 NIntegrate[(f[z] /. z -> c + r Exp[I t]) I r Exp[I t], {t, -Pi, Pi}]
intF[f, I/2, 1/2]

(* 0. + 6.28319 I *)