I must apply Cauchy-Goursat theorem to show that $\displaystyle \int _C f(z)dz=0$ where $\displaystyle C:=\{ z\in \mathbb{C} ; |z|=1 \}$ and $\displaystyle f(z)=\frac{z^2}{z-3}$.

So what I wrote was "f is analytic over $\displaystyle \mathbb{C}-\{ 3 \}$" therefore according to C-G's theorem the line integral is worth 0 since the singularity of f isn't inside the enclosed (by C) domain. Yet I don't feel I've "showed" anything. Do you think it's enough?

So I even tried to directly show that $\displaystyle \int _{|z|=1} \frac{z^2}{z^3-3} dz=0$. I parametrized the unit circle as $\displaystyle \gamma (\theta)=e^{i \theta}$ for $\displaystyle -\pi <\theta <\pi$. I came up with that this integral is worth $\displaystyle i \int _{-\pi} ^{\pi} \frac{e ^{i 3 \theta}}{e^{i \theta}-3} d\theta$. Then I've used Euler's formula and couldn't show anything. (I get a very long and ugly impossible thing). How would you show that the integral is worth 0 using only the definition of a line integral?