Here is my problem :
compute $$\int_{\mathcal{C}} \frac{1}{z-2i} \,\mathrm{d}z,$$ and $$\int_{\mathcal{C}} \frac{1}{z-e^{i\pi/4}} \,\mathrm{d}z,$$
where,
$C(t) = t$, $t\in[0,1]$
$C(t) = 1 + (t-1)i$, $t \in [1,2)$,
$C(t) = 3 - t + i$, $t \in [2,3]$
$C(t) = i(4-t)$, $t \in [3,4]$
I find that it will be complicated if I try to calculate directly. But, for those cases, I think that I can use Caushy-Gorsat Theorem. So, the answers for these integral will be equal to 0. Is is correct answer?
For the 1st one it is my impression that CG will work, because 2i is not inside the interior of curve, but the 2nd one e^(ipi/4) = 1+i, which lies in the interior of the curve.