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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

Evaluate the integral \( \oint_C \frac{1}{1 + z^2} \, dz \) where \( |z + i| = 1 \).

Detailed solution to evaluate the contour integral \( \oint_C \frac{1}{1 + z^2} \, dz \) using the residue theorem. The problem includes simplification of the function and identification of poles within the contour \( |z + i| = 1 \). Suitable for undergraduate students studying complex analysis.

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