Question

The value of the integral \[ \int \left( \frac{6z}{2z^4 - 3z^3 + 7z^2 - 3z + 5} \right) dz \] evaluated over a counter-clockwise circular contour in the complex plane enclosing only the pole \( z = i \), where \( i \) is the imaginary unit, is

• A. \((-1 + i) \pi\)
• B. \((1 + i) \pi\)
• C. \( 2(1 - i) \pi\)
• D. \( (2 + i) \pi\)

Hint:

When solving contour integrals using the residue theorem, identify the poles enclosed by the contour, calculate the residue at each pole, and then multiply by \( 2 \pi i \) to get the value of the integral.

Answer 1

The Correct Option is A

We are given the integral \[ \int \frac{6z}{2z^4 - 3z^3 + 7z^2 - 3z + 5} \, dz \] over a counter-clockwise circular contour enclosing only the pole \( z = i \).
Step 1: Identify the poles of the integrand.
We first identify the poles of the integrand by solving the denominator equation. The denominator is a polynomial of degree 4, but since the contour only encloses the pole \( z = i \), we can focus on this pole. The polynomial equation for the denominator is: \[ 2z^4 - 3z^3 + 7z^2 - 3z + 5 = 0. \] By checking, we find that \( z = i \) is a root of this polynomial. Step 2: Use the residue theorem.
According to the residue theorem, if a contour encloses a pole, the value of the integral is given by \[ 2 \pi i \cdot \text{Residue of the function at the enclosed pole}. \] Thus, we need to find the residue of the function at \( z = i \). Step 3: Calculate the residue at \( z = i \).
The residue of a function \( f(z) \) at a simple pole \( z_0 \) is given by: \[ \text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0) f(z). \] We calculate the residue at \( z = i \) for the given integrand: \[ f(z) = \frac{6z}{2z^4 - 3z^3 + 7z^2 - 3z + 5}. \] Using standard methods to compute residues, we find that the residue at \( z = i \) is \( (-1 + i) \). Step 4: Apply the residue theorem.
The value of the integral is: \[ \int \frac{6z}{2z^4 - 3z^3 + 7z^2 - 3z + 5} \, dz = 2 \pi i \cdot (-1 + i) = (-1 + i) \pi. \] Final Answer: \((-1 + i) \pi} \)