The integral presented is:

$\oint_C \frac{1}{1 + z^2} \, dz$

where $|z + i| = 1$.

This integral involves a contour $C$ around the point $z = -i$ with a radius of 1, which is a circle centered at $z = -i$.

Solution Outline

The function $\frac{1}{1 + z^2}$ can be simplified using the identity: $1 + z^2 = (z - i)(z + i)$

So, the function becomes: $\frac{1}{1 + z^2} = \frac{1}{(z - i)(z + i)}$

The integrand has poles at $z = i$ and $z = -i$. Since the contour $C$ is defined by $|z + i| = 1$, this contour only encloses the pole at $z = -i$.

Applying the Residue Theorem

To evaluate the integral, we use the residue theorem. The residue theorem states that for a function $f(z)$ with isolated singularities inside a contour $C$:

$\oint_C f(z) \, dz = 2 \pi i \times \text{(sum of residues of } f(z) \text{ inside } C)$

In this case, we need the residue of $\frac{1}{(z - i)(z + i)}$ at $z = -i$.

Calculating the Residue at $z = -i$

The residue of $\frac{1}{(z - i)(z + i)}$ at $z = -i$ is given by:

$\text{Residue} = \lim_{z \to -i} \frac{1}{z - i}$

This residue