We are tasked with evaluating the following contour integral using Cauchy's Integral Formula:

$\int_C \frac{z+1}{z^2 + 2z + 4} \, dz$ where $C$ is the circle $|z + 1 + i| = 2$.

Step 1: Simplify the Denominator

First, recognize that the denominator can be rewritten:

$z^2 + 2z + 4 = (z + 1)^2 + 3$

So the integral becomes:

$\int_C \frac{z+1}{(z+1)^2 + 3} \, dz$

Step 2: Identify Singularities

We need to locate the poles (singularities) of the integrand inside the contour. The denominator $(z+1)^2 + 3$ equals zero when:

$(z+1)^2 = -3$

$z + 1 = \pm \sqrt{3} i$

Thus, the singularities are at $z = -1 + \sqrt{3} i$ and $z = -1 - \sqrt{3} i$.

Step 3: Determine Which Poles Are Inside the Contour

The contour $C$ is the circle $|z + 1 + i| = 2$, which is centered at $z = -1 - i$ with radius 2.

• The singularity $z = -1 + \sqrt{3} i$ can be rewritten as $z = -1 + 1.732i$, which lies inside the contour.
• The singularity $z = -1 - \sqrt{3} i = -1 - 1.732i$ lies outside the contour.

Thus, only $z = -1 + \sqrt{3} i$ is inside the contour.

Step 4: Apply Cauchy's Residue Theorem

To compute the integral, we use Cauchy's Residue Theorem:

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

The residue at $z = -1 + \sqrt{3} i$ is given by:

$\text{Res}\left( \frac{z+1}{(z+1)^2 + 3}, z = -1 + \sqrt{3} i \right)$

Let’s calculate the residue.

At $z = -1 + \sqrt{3} i$, the integrand has a simple pole. The residue is:

$\text{Res}\left( \frac{z+1}{(z+1)^2 + 3}, z = -1 + \sqrt{3} i \right) = \lim_{z \to -1 + \sqrt{3} i} (z - (-1 + \sqrt{3} i)) \cdot \frac{z+1}{(z+1)^2 + 3}$

This simplifies to:

$\text{Res} = \frac{2}{6}$

Thus, the residue is $\frac{1}{3}$.

Step 5: Final Calculation

By Cauchy’s residue theorem, the integral is:

$\int_C \frac{z+1}{z^2 + 2z + 4} \, dz = 2 \pi i \cdot \frac{1}{3} = \frac{2\pi i}{3}$

Final Answer:

$\boxed{\frac{2\pi i}{3}}$

Would you like further clarifications or more details?

Here are some related questions:

1. What are the conditions for applying Cauchy's Integral Formula?
2. How would the integral change if the contour enclosed both singularities?
3. Can we generalize this result to higher powers of $z$ in the denominator?
4. What if the contour were a different shape, like an ellipse?
5. How do residues relate to physical applications like electric fields?

Tip: Always check which singularities are enclosed by the contour before applying residue theorem.