Complex integral on rectangle

Question

I have to calculate the integral $$\oint_C \frac{z^2+z+1}{(z-i)^2}dz$$ where $C$ is the positively oriented rectangle between $-1, 1, 1+2i, -1+2i$.

My attempt is to use Cauchy's integral formula, but obviously the integrand is not differentiable at $i$ which lies inside $C$ which means that it is not holomorphic. Does this matter? If yes, how do I fix it?
If this does not matter, I simply choose e.g. $z_0=1/2(1+i)$ which lies inside $C$ and calculate its index $I_C(z_0)=\frac{1}{2\pi i}\oint_C \frac{d\xi}{\xi-z_0}$ respective to $C$ and then use $$f(z_0)I_C(z_0)=\frac{1}{2\pi i}\oint_C\frac{f(\xi)}{\xi-z_0}d\xi$$ Am I on the right track?

Answer 1

Note the Cauchy integration is $$ f'(\zeta)=\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-\zeta)^2}dz $$ where $f(z)$ is analytic in $C$ which is a simple closed curve. Now $C$ is the four sides of a square and $z=i$ is inside $C$. Hence $$\oint_C \frac{z^2+z+1}{(z-i)^2}dz=2\pi i (z^2+z+1)'|_{z=i}=2\pi (2i+1)=2\pi+4\pi i. $$ Here

Answer 2

Let us write $w=z-i$, so we have to compute an integral on the "moved square" contour $S$, $$ \oint_D\frac{h^2 +(2i+1)h +i}{h^2}\; dh\ . $$ Now use ttne Residue Theorem, or just integrate on the contour $D$ the functions $1$ (to get zero), $1/h$ (to get $2\pi i$), and $1/h^2$ (with primitive $-1/h$ to get zero).