# evaluating contour integrals along a curve

A couple of exercises I'm working on:

$$1)$$ Let $$\gamma$$ be a closed curve lying entirely in the set $$\mathbb{C} \setminus\{z \mid \text{Re} z \leq 0\}$$. Show that $$\int_{\gamma}\frac{1}{z}dz = 0$$.

$$\frac{1}{z}$$ is analytic on and inside $$\gamma$$, so by Cauchy's Theorem we have the result.

Alternatively, using the Fundamental Theorem of Contour Integrals, we have that $$\frac{1}{z}$$ is the derivative of a function that is defined and analytic on $$\mathbb{C} \setminus\{z \mid \text{Re} z \leq 0\}$$, and since $$\gamma$$ is closed, we have that the integral is $$0$$.

2) Evaluate $$\int_{\gamma} \frac{2z+1}{z^2+z}dz$$ for:

$$a)$$ $$\gamma$$ is given by $$|z| = \frac{1}{2}$$

$$\int_{\gamma} \frac{2z+1}{z^2+z}dz = \int_{c_1} \frac{1}{z}dz+\int_{c_2}\frac{1}{z+1}dz$$ (using deformation, $$c_1$$ and $$c_2$$ are circles around each singularity)

Since only one of these is inside the curve, the integral is $$2\pi i$$.

$$b)$$ $$\gamma$$ is given by $$|z|=2$$. Now, using the same process as above, both singularities are contained in $$\gamma$$ so their sum is $$4\pi i$$.

$$c)$$ $$\gamma$$ is the curve $$|z-3i| = 1$$. Here, $$\gamma$$ is just the circle of radius $$1$$ centered at $$3i$$. Neither root is contained in the curve, and each of them are analytic on and inside of $$\gamma$$, so the integral is equal to $$0$$.

Please let me know if anything at all is not correct reasoning or just purely incorrect. I am trying to grasp these concepts :)

For $$2)$$, I don't think you needed a partial fraction decomposition: you could just compute the residues directly. They're both $$1$$.