Evaluate $\int_{C}(z-i) \,dz$ where $C$ is the parabolic segment: $z(t) = t + it^2, −1 \le t \le 1$

Question

Evaluate $\int_{C}(z-i) \,dz$ where $C$ is the parabolic segment: $$z(t) = t + it^2, −1 \le t \le 1$$ by integrating along the straight line from $−1+i$ to $1+i$ and applying the Closed Curve Theorem. How should I apply closed curve theorem to such problem?

Answer 1

Note that$$\int_{-1}^1\bigl((z+i)-i\bigr)\,\mathrm dz=\int_{-1}^1z\,\mathrm dz=0.$$So, the integral that you're after is equal to the integral along the closed path which goes from $-1+i$ to $1+i$ along that parabolic arc followed by going in a straigh line from $1+i$ to $1-i$. But this last integral is $0$! So, the integral that you're after is equal to $0$ too.

Answer 2

A possible intention of the exercise is to use the fact that $z-i$ is holomorphic in $\mathbb{C}$ and hence its complex integrals are path-independent.

Hence, you just can calculate the path integral using the complex anti-derivative of $z-i$ and plugging in the endpoints of the path:

$$\int_C (z-i)\;dz = \left[\frac{z^2}{2}-iz\right]_{-1+i}^{1+i}$$ $$=\frac{(1+i)^2}{2} -i(1+i) -\left(\frac{(-1+i)^2}{2} -i(-1+i) \right) = 0$$