Evaluate the integral∮_C (f(z))dz

Evaluate the integral∮_C (f(z))dz where C is the unit circle enclosing the origin, and f (z) is given as follows: 1/conjugate z

I would not use the Residue Theorem (or any of Cauchy's Theorems for that matter), since the integrand involves a conjugate (and thus is generally not analytic).

Parameterizing C with z = e^(it) with t in [0, 2π], we obtain
∫(t = 0 to 2π) (1/e^(-it)) * (ie^(it) dt)
= ∫(t = 0 to 2π) ie^(2it) dt
= (1/2)e^(2it) {for t = 0 to 2π}
= 0.

I hope this helps!

1/(z-bar) = z / |z|^2 after "rationalizing" which has a single pole at infinity. Hence we get:

∮_C (f(z))dz = (2Pi i) times Sum of the residues of f(z) inside the unit circle = (2Pi i) (0) = 0.

Note: |z| is the modulus of the complex number z = a + bi, which equals sqrt(a^2 + b^2);

z-conjugate is not analytic but 1/z-conjugate is, because it equals z/|z|^2, a polynomial.