I'm studying the Cauchy’s integral theorem on circular contour. Is given in the study material that $z = re^{i\theta}$ and $dz = ire^{i\theta}d\theta$. I don't know how to go about using $\mid z \mid$ in this problem. In both examples, I don't know how to solve the integrals to use the $\mid z \mid$ later.
In first case I try:
$$I = \int_{C} \frac{z^{2}}{z-1} dz$$ $$I = \int_{C} \frac{dz}{z-z_{0}} = 2\pi i f(z_{0})\text{.}$$ How $$f(z) = z^{2}\text{ , }z_{0} = 1\text{,}$$ so $$I = 2\pi if(z_{0}) = 2\pi i 1^{2} = 2\pi i $$
In second:
$$I = \int_{C} \frac{e^{z}}{z^{2}}dz$$ $$z^{2} = (z-0)^{2} = z^{2}-2 z (0)+0^{2}=z^{2}$$ Using the general formula, $$I = \int_{C}\frac{e^{z}}{(z-0)^{2}} = 2\pi i \frac{f^{n-1}(a)}{(n-1)!}$$ How $f(z) = e^{z}$, $a=0$ and $f(a)=f(0)=e^{0}=1$,
$$I = \frac{2\pi i f^{(2-1)}(0)}{(2-1)!}=2\pi i \text{.}$$
I still don't know where to use the module. Can anyone help with this?