We are asked to evaluate the integral of the function

$\int_C \frac{4z^2 + z + 5}{(z+1)^2} \, dz$

where $C$ is the contour $|z| = 2$, which is a circle centered at the origin with radius 2.

This is a complex contour integral, and the function $f(z) = \frac{4z^2 + z + 5}{(z+1)^2}$ has a singularity at $z = -1$. Since $z = -1$ is within the contour $C$ (because $|z| = 2$ includes $z = -1$), we can apply the Residue Theorem to evaluate the integral.

Step 1: Identifying the singularity

The only singularity inside $|z| = 2$ is at $z = -1$. We need to find the residue of the function at this point.

Step 2: Finding the residue at $z = -1$

The function has a pole of order 2 at $z = -1$, since the denominator is $(z+1)^2$. To find the residue for a second-order pole, we use the formula:

$\text{Res}(f, -1) = \lim_{z \to -1} \frac{d}{dz} \left( (z + 1)^2 f(z) \right)$

First, let's multiply $f(z)$ by $(z + 1)^2$:

$f(z) = \frac{4z^2 + z + 5}{(z+1)^2}, \quad \Rightarrow \quad (z + 1)^2 f(z) = 4z^2 + z + 5$

Now, we differentiate $4z^2 + z + 5$ with respect to $z$:

$\frac{d}{dz}(4z^2 + z + 5) = 8z + 1$

Next, we evaluate this derivative at $z = -1$:

$8(-1) + 1 = -8 + 1 = -7$

Thus, the residue at $z = -1$ is $-7$.

Step 3: Applying the Residue Theorem

By the Residue Theorem, the integral of a function around a closed contour that encloses singularities is $2\pi i$ times the sum of the residues of the enclosed singularities. Since there is only one singularity at $z = -1$, we have:

$\int_C \frac{4z^2 + z + 5}{(z+1)^2} \, dz = 2\pi i \cdot (-7) = -14\pi i$

Final Answer:

$\int_C \frac{4z^2 + z + 5}{(z+1)^2} \, dz = -14\pi i$

Would you like any further clarification on this, or a more detailed explanation of any step?

Here are five related questions for you to explore:

1. What is the Residue Theorem and how is it used in contour integrals?
2. How do you compute residues for higher-order poles in complex analysis?
3. What are the conditions for applying the Residue Theorem?
4. How do you evaluate integrals involving simple poles versus higher-order poles?
5. Can contour integrals with no singularities inside the contour be non-zero?

Tip: When dealing with complex integrals, always check for the type of singularities inside the contour to decide the best evaluation method!