Path integral of $1/z$ along $e^{it}$

Question

I tried to calculate the (complex) path integral of $1/z$ along $y: [0, \pi] \rightarrow \mathbb{C}, y(t) = e^{it}$, and got the result $i\pi$.

Is this correct?

Answer 1

Yes, it is correct, because

$$\int \limits _y \frac 1 z \ \Bbb d z= \int \limits _0 ^\pi \frac 1 {\Bbb e ^{\Bbb i t}} (\Bbb d \Bbb e ^{\Bbb i t}) = \int \limits _0 ^\pi \frac 1 {\Bbb e ^{\Bbb i t}} \Bbb i \ \Bbb e ^{\Bbb i t} \Bbb d t = \int \limits _0 ^\pi \Bbb i \ \Bbb d t = \Bbb i \pi .$$