Solution: Note that \[ z^2 - 2z + 2 = (z - (1+\iota )) (z - (1-\iota )). \] Let us write $\alpha = 1 + \iota $ and $\beta = 1 - \iota $. We need to evaluate the integral \[ \oint_C \frac{z^2 \mathrm{d}z }{(z-\alpha )(z-\beta )}. \] Since, \begin{align*} f(\alpha ) = \vert \alpha -2 \vert + \vert \alpha +2 \vert = \sqrt{4} + \sqrt{10} \lt 8 \text{ and }\\ f(\beta ) = \vert \beta -2 \vert + \vert \beta +2 \vert = \sqrt{4} + \sqrt{10} \lt 8, \end{align*} so both $\alpha ,\beta $ lies in the interior of the given contour. We will use Cauchy integral formula for evaluating the integral.

Note that \begin{align*} \frac{z^2}{(z - \alpha )(z-\beta )} & = \frac{1}{\alpha -\beta }\left( \frac{z^2}{z-\alpha } - \frac{z^2}{z-\beta } \right) \\ & = \frac{1}{2\iota }\left( \frac{z^2}{z-\alpha } - \frac{z^2}{z-\beta } \right). \end{align*} So, \begin{align*} \oint_C \frac{z^2 \mathrm{d}z }{(z-\alpha )(z-\beta )} & = \frac{1}{2\iota }\oint_{C} \frac{z^2}{z-\alpha }\mathrm{d}z - \frac{1}{2\iota } \oint_{C} \frac{z^2}{z-\beta }\mathrm{d}z \\ & = \frac{1}{2\iota } \left( 2\pi \iota (\alpha ^2) \right) - \frac{1}{2\iota }(2\pi \iota (\beta )^2) \\ & = \pi \left( (1+\iota)^2 - (1-\iota )^2 \right) \\ & = 4\pi \iota. \end{align*} Thus, \[ \textcolor{teal}{\boxed{\oint_C \frac{z^2 d z}{z^2-2 z+2} = 4 \pi \iota}} \]