Contour integral – Circle instead of a square

I would like to solve the integral $\int_{\gamma} \frac{1}{z \bar{z}} dz$. Here $\gamma \subset \mathbb{C}$ is a square centered at the origin and where his vertices are parallel to the axes. Could I use the Cauchy theorem in using the fact we will change the contour to a unit circle center to the origin?

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The integrand is not holomorphic, you can’t use Cauchy integral theorem.

However, if you parametric $\gamma$ by $[0,2\pi] \ni \theta \mapsto z(\theta) = r(\theta) e^{i\theta}$ by a suitable chosen real valued function $r(\theta)$.
The integral becomes

$$\int_{\gamma} \frac{dz}{z\bar{z}} = \int_0^{2\pi} \left( \frac{r'(\theta)+ir(\theta)}{r(\theta)^2} \right) e^{i\theta} d\theta$$

Notice $r(\theta+\pi) = r(\theta)$ while $e^{i(\theta+\pi)} = -e^{i\theta}$. the contribution from $[\pi,2\pi]$ in last integral cancel the one from $[0,\pi]$. So the integral is $0$.