Question

If \( C \) is the unit circle in the complex plane with its center at the origin, then the value of \( n \) in the equation given below is (rounded off to 1 decimal place). \[ \int_C \frac{z^3}{(z^2 + 4)(z^2 - 4)} \, dz = 2 \pi i n \]

Hint:

In contour integration, use the residue theorem to calculate integrals around closed contours by summing the residues of the poles inside the contour.

Answer 1

We are asked to find the value of \( n \) using the residue theorem.

Step 1: Factor the denominator
We begin by factoring the denominator:
\[ (z^2 + 4)(z^2 - 4) = (z + 2i)(z - 2i)(z + 2)(z - 2) \]
So, the integrand becomes:
\[ f(z) = \frac{z^3}{(z + 2i)(z - 2i)(z + 2)(z - 2)} \]

Step 2: Identify the singularities
We now identify the singularities of the function within the unit circle. The singularities occur at \( z = 2i, -2i, 2, -2 \).
\( z = 2i \) and \( z = -2i \) are inside the unit circle.
\( z = 2 \) and \( z = -2 \) are outside the unit circle.
We will only consider the residues at \( z = 2i \) and \( z = -2i \), as these are inside the unit circle.

Step 3: Find the residues at \( z = 2i \) and \( z = -2i \)
We can calculate the residues using the formula for simple poles:
\[ {Residue of } f(z) { at } z = 2i: {Res}(f, 2i) = \lim_{z \to 2i} (z - 2i) f(z) \]
After calculating, we find that:
\[ {Residue at } z = 2i { is } \frac{8i}{16} = \frac{1}{2} \]
Similarly, we find the residue at \( z = -2i \):
\[ {Residue at } z = -2i { is } \frac{-8i}{16} = -\frac{1}{2} \]

Step 4: Apply the residue theorem
The residue theorem tells us that the integral around a closed contour \( C \) is \( 2 \pi i \) times the sum of the residues inside the contour:
\[ \int_C f(z) \, dz = 2 \pi i \left( {Res}(f, 2i) + {Res}(f, -2i) \right) \]
Substituting the calculated residues:
\[ \int_C f(z) \, dz = 2 \pi i \left( \frac{1}{2} + \left( -\frac{1}{2} \right) \right) = 0 \]

Step 5: Conclusion
Since the sum of the residues is 0, we conclude that:
\[ n = 0 \]