Closed path complex integral with non anaytic function gives me zero

Question

When i calculate

$$\oint_C \left(\overline z\right)^2 \mathbb dz$$

Along $|z|=1$, it gives me $0$.

Then i remembered with Cauchy's Theorem,

" If we have an analytic function and want to integrate it along closed path, then we have $0$ as the result".

Then i tried to check the analiticity with Cauchy-Riemann Equation, and it's not satisfied.

Well, randomly i change the radius to $123$, and it still gives me $0$

Could i conclude

$$\oint_C (\overline z)^2 \mathbb dz$$

Is always $0$, when it's evaluated along $|z|=R$ ? Where $R$ is any radius?

NB : But it fails to be $0$ when around the $|z-1|=1$

But why?

What is the best reason?

Answer 1

EDIT: this answer is wrong because the part derivative term in the integral is not conjugated.

Since the curve is symmetric about the real axis, the integral is equivalent to $\int_C z^2 dz$. Thus by Cauchy's theorem, it is zero.

Answer 2

Note that on $|z|=1$, $\bar z=\frac1z$. Hence, we have

$$\oint_{|z|=1}(\bar z)^2\,dz=\oint_{|z|=1}\frac1{z^2}\,dz$$

Inasmuch as the residue of $\frac1{z^2}$ is $0$, we conclude that

$$\oint_{|z|=1}(\bar z)^2\,dz=0$$

And we are done.