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 :)


It looks like your answers are correct.

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


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