Show that [int_{C} frac{d z}{(z-1-i)^{n+1}}=left{begin{array}{cc} 0, & n eq 0 2 pi i, & n=0 end{array} ight.] for (C) the boundary of the square

Show that

\[\int_{C} \frac{d z}{(z-1-i)^{n+1}}=\left\{\begin{array}{cc} 0, & n eq 0 \\ 2 \pi i, & n=0 \end{array}\right.\]

for \(C\) the boundary of the square \(0 \leq x \leq 2,0 \leq y \leq 2\) taken counterclockwise. [Use the fact that contours can be deformed into simpler shapes (like a circle) as long as the integrand is analytic in the region between them. After picking a simpler contour, integrate using parametrization.]