# Find $\displaystyle{\int_C \left(1+ \frac{2}{z}\right) dz}$ , where $C(\theta)= e^{i\theta}, 0 \le \theta \le \pi$

Find $\displaystyle{\int_C \left(1+ \frac{2}{z}\right) dz}$ along the contour $C(\theta)=e^{i\theta}, 0 \le \theta \le \pi$, that is $C$ is the unit half-circle in the upper plane center at $0$.

I think the solution rely on the choice of an appropriate branch of the logarithmic function and the answer should be $-2+i2\pi$.
I obtain $-2+i2\pi$ when I use the branch $[0,2\pi]$.
But our instructor used the branch $(-\frac{\pi}{2},\frac{7}{2}\pi)$.
Are those two branches appropriate?
Is it ok to use a closed interval for a branch or only open interval are allowable?

My attempt:
A primitive for $1+ \frac{2}{z}$ is $z+2 \ln z$. Choosing the branch $$\ln z = \ln r + i \theta , \theta \in [0,2\pi]$$ by Cauchy theorem $$\int_C \left(1+ \frac{2}{z}\right) dz = z+2 \ln z \bigg|_{z=1}^{z=-1}=-2+i2\pi$$ as $\ln 1= 0 + i 0$ and $\ln -1 = 0 + i \pi$.

-

$$\int_C \left(1 + 2z^{-1}\right)\, dz$$
where $C(\theta) = e^{i\theta}$ and $0 \le \theta \le \pi$. This is just a standard substitution problem, like you were doing since you were 16. Put $z = e^{i\theta}$ so that $dz = ie^{i\theta} \, d\theta.$ You get:
$$\int_C \left(1 + 2z^{-1}\right) \, dz \equiv \int_0^{\pi} (1 + 2e^{-i\theta}) \cdot i e^{i\theta} \, d\theta = i \int_0^{\pi} (e^{i\theta} + 2) \, d\theta = i \left[2\theta-ie^{-\theta} \right]_0^{\pi} = 2(i\pi - 1).$$