Integrate counterclockwise or as indicated. Show the

details.C
dz
z
2  4
, C: 4x 2  ( y  2)2 -
4

To solve the given integral, let's first clarify the problem:

The integral provided is a complex line integral over a contour \(C\). The integral appears to be:

\[
\oint_C \frac{dz}{z^2 - 4}
\]

where \(C\) is a contour defined by \(4x^2 + (y + 2)^2 = 4\). This represents an ellipse in the complex plane.

### Step 1: Rewrite the Integral

The given integral is:

\[
\oint_C \frac{dz}{z^2 - 4}
\]

The integrand has singularities (poles) at \(z = 2\) and \(z = -2\). To evaluate this integral, we need to know if the contour \(C\) encloses these singularities.

### Step 2: Analyze the Contour \(C\)

The contour \(C\) is defined by the ellipse:

\[
4x^2 + (y + 2)^2 = 4
\]

We rewrite this in standard form:

\[
\frac{x^2}{1} + \frac{(y+2)^2}{4} = 1
\]

This is an ellipse centered at \((0, -2)\) with a semi-major axis of 2 (in the vertical direction) and a semi-minor axis of 1 (in the horizontal direction).

### Step 3: Determine the Enclosed Singularities

The poles of the integrand, \(z = 2\) and \(z = -2\), need to be checked if they lie inside the contour \(C\). The ellipse is centered at \((0, -2)\), and its vertical extent runs from \(y = -4\) to \(y = 0\).

- \(z = 2\) is not inside the ellipse since it lies at \((2, 0)\), outside the horizontal range \([-1, 1]\) of the ellipse.
- \(z = -2\) is inside the ellipse, as it lies at \((-2, 0)\), which is within the horizontal range and well within the vertical range.

### Step 4: Apply the Residue Theorem

Since the only singularity inside the contour \(C\) is at \(z = -2\), we apply the residue theorem. The residue of \( \frac{1}{z^2 - 4} \) at \(z = -2\) is calculated as follows:

\[
\text{Res}\left( \frac{1}{z^2 - 4}, z = -2 \right) = \lim_{z \to -2} (z + 2) \frac{1}{z^2 - 4} = \lim_{z \to -2} \frac{1}{z - 2} = \frac{1}{-4} = -\frac{1}{4}
\]

### Step 5: Compute the Integral

By the residue theorem:

\[
\oint_C \frac{dz}{z^2 - 4} = 2\pi i \times \left( -\frac{1}{4} \right) = -\frac{\pi i}{2}
\]

### Final Answer:

\[
\oint_C \frac{dz}{z^2 - 4} = -\frac{\pi i}{2}
\]

Would you like a more detailed breakdown or have any questions about this solution?

### 5 Related Questions:

1. How do you find the residues of higher-order poles?
2. What are the applications of the residue theorem in physics?
3. How do you parameterize different types of contours in the complex plane?
4. What is the significance of contour orientation in complex integration?
5. How do you apply the residue theorem when multiple singularities are enclosed by a contour?

### Tip:

When performing contour integration, always check whether the singularities are inside the contour and consider the direction of traversal (counterclockwise or clockwise).

Learn how to solve a complex contour integration problem using the residue theorem. The problem involves integrating counterclockwise over a contour defined by a complex equation. Suitable for advanced mathematics enthusiasts.

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