Problem 22 - A First Course in Complex Analysis with Applications

Question

Use Cauchy's residue theorem to evaluate the given integral along the indicated contour. $$ \oint_{C} \frac{1}{z^{3}(z-1)^{4}} d z, C:|z-2|=\frac{3}{2} $$

Step-by-Step Solution

Answer

The integral evaluates to zero as there are no poles inside the contour.

Step 1: Identify the Singularities

The given function is \( f(z) = \frac{1}{z^3(z-1)^4} \). The singularities occur where the denominator is zero. Thus, we find singularities at \( z = 0 \) (a pole of order 3) and \( z = 1 \) (a pole of order 4).

Step 2: Check Contour Inclusion

The contour \( C \) is given as \(|z-2|=\frac{3}{2}\). This is a circle centered at 2 with radius \(1.5\). Determine which singularities are inside this contour.\( z = 0 \) is within distance \(2\) from \(2\), which is outside the contour. \( z = 1 \) is within distance \(1\) from \(2\), hence inside \(C\).

Step 3: Compute the Residue at z=1

Since the only singularity inside the contour is \( z=1 \), calculate the residue of \( f(z) \) at \( z=1 \). For a pole of order 4, the residue is determined by:\[ \text{Residue of } f(z) ext{ at } z=1 = \frac{1}{3!} \left. \frac{d^3}{dz^3} \left( \frac{1}{z^3} \right) \right|_{z=1} \]Calculate \( \frac{d^3}{dz^3} \left( \frac{1}{z^3} \right) \) and evaluate at \( z=1 \).

Step 4: Use Cauchy's Residue Theorem

Cauchy's Residue Theorem states that the integral of \( f(z) \) around \( C \) is \( 2\pi i \) times the sum of residues inside \( C \). Since \( z=1 \) is the only pole inside \( C \), the integral is:\[ \oint_C \frac{1}{z^3(z-1)^4} \, dz = 2\pi i \times \text{Residue at } z=1 \].

Key Concepts

Singularities

In complex analysis, singularities are specific points at which a function ceases to be analytic. In simpler terms, they are the points where a given function behaves badly, often due to division by zero. Identifying these singularities is crucial when solving problems involving contour integration because they directly affect the behavior of the function along a path.
To find singularities, look at the denominator of the function. In the given exercise, we start from the function \( f(z) = \frac{1}{z^3(z-1)^4} \). Here, singularities occur where the denominator equals zero.

• For \( z^3 = 0 \), this implies a singularity at \( z = 0 \), specifically a pole of order 3.
• For \((z-1)^4 = 0 \), this gives a singularity at \( z = 1 \), a pole of order 4.

Such categorization helps in determining which singularities are significant to problems requiring residue calculations.

Contour Integration

Contour integration involves integrating a complex function along a path or contour in the complex plane. This path in our case is defined as a circle centered at a point with a specific radius. For this exercise, the contour \( C \) is \( |z-2| = \frac{3}{2} \), a circle centered at 2 with a radius of 1.5.
The contour shape and its location are crucial for determining which singularities fall inside the contour. Only the singularities within the contour need to be considered when calculating certain integrals, such as those involving Cauchy's residue theorem.
By checking each singularity identified:

• \( z = 0 \) is 2 units away from the center (2), hence outside the contour.
• \( z = 1 \) is 1 unit away and thus within the contour.

Understanding this process helps focus on relevant singularities for the succeeding residue calculations.

Residue Calculation

Residue calculation is a method developed from complex analysis used to evaluate integrals, especially with functions having poles. This involves determining the residue at the singularity. Cauchy's residue theorem then uses this residue to compute integrals around closed contours in the complex plane.
In the given exercise, the residue involves calculating it at \( z=1 \), the only pole inside the contour. Since we have a pole of order 4, we use the formula for residues of higher order poles:\[\text{Residue at } z=1 = \frac{1}{3!} \left. \frac{d^3}{dz^3} \left( \frac{1}{z^3} \right) \right|_{z=1} \]This requires:

• Computing the third derivative of \( \frac{1}{z^3} \)
• Evaluating it at \( z=1 \)

Once the residue is found, it is multiplied by \( 2\pi i \), as dictated by Cauchy's residue theorem, to determine the integral of the function around the contour. This streamlined approach makes it efficient to solve complex integrals involving singularities.