Let $C=\partial D_1(\mathbf i/2)$, compute $\int_C\frac{dz}{z^2+1}$

Question

Let $C=\partial D_1(\mathbf i/2)$, compute $\int_C\frac{dz}{z^2+1}$

$C=\partial D_1(\mathbf i/2)$ is the boundary of the disc with center $\mathbf i/2$ and radius $1$, then $\mathbf i$ is contained in the disc and $\mathbf -i$ is not inside, therefore

$\displaystyle\int_C\frac{dz}{z^2+1}=\frac{1}{2i}\int_C\left(\frac{1}{z-\mathbf i}-\frac{1}{z-(-\mathbf i)}\right)dz$

using Cauchy Integral Formula the integral of the second fraction is $0$

but for the first one which of the following is correct ?

Cauchy formula $\displaystyle f(z)=\int_C\frac{f(\zeta)}{\zeta-z}d\zeta\implies f\equiv 1$ hence the result is $1$ or;

$\displaystyle \int_C\frac{1}{z-\mathbf i}dz=\int_0^{2\pi}\frac{1}{3}e^{-\mathbf it}3\mathbf ie^{\mathbf it}dt=2\pi\mathbf i$, so the result is $\pi$

Answer 1

The second answer is correct, but you mistated the Cauchy formula in the first answer (which is why you have it wrong) -- it should be $$f(z) = \color{red}{\frac{1}{2\pi i}} \int_C \frac{f(\zeta)}{\zeta - z}\, d\zeta$$