Complex Integral Analysis

This problem requires the evaluation of a complex contour integral. We need to compute the value of the integral:

\(\mathop \oint \limits_c \frac{{2z + 5}}{{\left( {z - \frac{1}{2}} \right)\left( {{z^2} - 4z + 5} \right)}}dz\)

The integral is taken over the contour \(c\), which is the circle defined by \(\left| z \right| = 1\), and the direction is anti-clockwise. We will use Cauchy's Residue Theorem to solve this.

Identifying Poles of the Integrand

The first step is to locate the singularities (poles) of the function \(f(z) = \frac{{2z + 5}}{{\left( {z - \frac{1}{2}} \right)\left( {{z^2} - 4z + 5} \right)}}\). Poles are the values of \(z\) for which the denominator equals zero.

The denominator is \(\left( {z - \frac{1}{2}} \right)\left( {{z^2} - 4z + 5} \right)\). Setting it to zero:

1. \( z - \frac{1}{2} = 0 \) gives \( z = \frac{1}{2} \). This is a simple pole.
2. \( z^2 - 4z + 5 = 0 \). We solve this quadratic equation using the quadratic formula: \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

\(z = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)}\)

\(z = \frac{4 \pm \sqrt{16 - 20}}{2}\)

\(z = \frac{4 \pm \sqrt{-4}}{2}\)

\(z = \frac{4 \pm 2i}{2}\)

This gives two more poles: \( z = 2 + i \) and \( z = 2 - i \). These are also simple poles.

The poles of the function are \( \frac{1}{2} \), \( 2 + i \), and \( 2 - i \).

Contour Analysis: Poles Inside \( \left| z \right| = 1 \)

We need to determine which of these poles lie inside the contour \( c \), which is the circle \(\left| z \right| = 1\). This means we check if \( \left| z \right| < 1 \).

• Pole \( z = \frac{1}{2} \): The magnitude is \( \left| \frac{1}{2} \right| = \frac{1}{2} \). Since \( \frac{1}{2} < 1 \), this pole is inside the contour.
• Pole \( z = 2 + i \): The magnitude is \( \left| 2 + i \right| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \). Since \( \sqrt{5} \approx 2.236 > 1 \), this pole is outside the contour.
• Pole \( z = 2 - i \): The magnitude is \( \left| 2 - i \right| = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \). Since \( \sqrt{5} \approx 2.236 > 1 \), this pole is also outside the contour.

Only the pole at \( z = \frac{1}{2} \) lies within the specified contour \(\left| z \right| = 1\).

Residue Calculation at \( z = \frac{1}{2} \)

Cauchy's Residue Theorem states that the integral value is \( 2\pi i \) multiplied by the sum of the residues of the poles enclosed by the contour. Here, only \( z = \frac{1}{2} \) is inside.

The residue at a simple pole \(z_0\) for a function \(f(z)\) can be calculated using the formula:

\( \text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0) f(z) \)

We calculate the residue at \( z_0 = \frac{1}{2} \):

\( \text{Res}(f, \frac{1}{2}) = \lim_{z \to \frac{1}{2}} \left( z - \frac{1}{2} \right) \frac{{2z + 5}}{{\left( {z - \frac{1}{2}} \right)\left( {{z^2} - 4z + 5} \right)}} \)

Cancelling the \( (z - \frac{1}{2}) \) term:

\( \text{Res}(f, \frac{1}{2}) = \lim_{z \to \frac{1}{2}} \frac{{2z + 5}}{{{z^2} - 4z + 5}} \)

Now, substitute \( z = \frac{1}{2} \) into the expression:

\( \text{Res}(f, \frac{1}{2}) = \frac{{2(\frac{1}{2}) + 5}}{{{(\frac{1}{2}})^2 - 4(\frac{1}{2}) + 5}} \)

\( \text{Res}(f, \frac{1}{2}) = \frac{{1 + 5}}{{\frac{1}{4} - 2 + 5}} \)

\( \text{Res}(f, \frac{1}{2}) = \frac{6}{{\frac{1}{4} + 3}} \)

Combine the terms in the denominator:

\( \text{Res}(f, \frac{1}{2}) = \frac{6}{{\frac{1 + 12}{4}}} = \frac{6}{{\frac{13}{4}}} \)

Calculate the final value:

\( \text{Res}(f, \frac{1}{2}) = 6 \times \frac{4}{13} = \frac{24}{13} \)

Integral Value Determination

Apply Cauchy's Residue Theorem using the calculated residue:

\( \mathop \oint \limits_c f(z) dz = 2\pi i \times (\text{Sum of residues inside } c) \)

Since only the residue at \( z = \frac{1}{2} \) is needed:

\( \mathop \oint \limits_c f(z) dz = 2\pi i \times \text{Res}(f, \frac{1}{2}) \)

\( \mathop \oint \limits_c f(z) dz = 2\pi i \times \frac{24}{13} \)

\( \mathop \oint \limits_c f(z) dz = \frac{48\pi i}{13} \)

Thus, the value of the contour integral is \(\frac{48\pi i}{13}\).