Finding $\int_{C}a\ dz$, where $C$ is the contour $z(\theta) = \rho e^{i\theta}$ for $\theta\in[-\pi,\pi]$.

Question

I have been working with some complex integrations but I would like to know if I am right about this simple integration:

Question : For complex plane: let $C$ be the contour $z(\theta) = \rho e^{i\theta}$ for $(-\pi\leq\theta\leq\pi)$ and let $a$ and $\rho$ be any constant. Find $\displaystyle\int_{C}a\ \mathrm{d}z\ .$

My Attempt

$$\int_{C}a\,\mathrm{d}z = \int\limits_{-\pi}^{\pi} a \rho\ ie^{i\theta}\,\mathrm{d}\theta,$$ (where we make a variable change for $z$ s.t $\mathrm{d}z = \rho\ ie^{i\theta}\ \mathrm{d}\theta$).

$$= a \rho i\int\limits_{-\pi}^{\pi} e^{i\theta}\mathrm{d}\theta\\ = a \rho i \cdot\frac{e^{i\theta}}{i}\bigg|_{-\pi}^{\pi}\\ = 2a \rho $$

does this look right?

Answer 1

It's easy to see that you made a mistake, because a constant function $f(z)=a$ is analytic, and you are integrating around a closed curve. By Cauchy's theorem the integral is then zero.

It was right until you forgot that $e^{iπ}=e^{-iπ}$.