Problem 1

Question

For the functions defined by the following expressions compute all residues in all singular points. (a) $\frac{1-\cos z}{z^{2}}$, (b) $\frac{z^{3}}{(1+z)^{3}}$, (c) $\frac{1}{\left(z^{2}+1\right)^{3}}$, (d) $\frac{1}{\left(z^{2}+1\right)(z-1)^{2}}$, (e) $\frac{\exp (z)}{(z-1)^{2}}$, (f) $z \exp \left(\frac{1}{1-z}\right)$, (g) $\frac{1}{\left(z^{2}+1\right)(z-\mathrm{i})^{3}}$ (h) $\frac{1}{\exp (z)+1}$ (i) $\frac{1}{\sin \pi z}$.

Step-by-Step Solution

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Answer

Determine singular points, classify singularities, evaluate residues for poles.

1Step 1: Identify the Singular Points

Identify the singular points for each function. Singular points are where the function is not analytic (not defined or not differentiable). For example, for (a) $ \frac{1-\cos z}{z^{2}} $, the singular point is $ z=0 $.

2Step 2: Determine the Type of Singularity

Classify each singular point: whether it is removable, a pole, or essential. For function (a), $ z=0 $ is a removable singularity because the limit exists as $ z \to 0 $. In contrast, for (b), the function has a pole of order 3 at $ z=-1 $.

3Step 3: Calculate the Residues

For non-removable singularities, compute the residues. For a pole of order $ n $, use the formula: $ \text{Residue} = \lim_{z \to z_0} \frac{1}{(n-1)!} \left( \frac{d^{n-1}}{dz^{n-1}} \left[ (z-z_0)^n f(z) \right] \right). $ For example, in (b) at $ z=-1 $, the residue is computed using this formula.

4Step 4: Gather Results

Summarize the residues found in the previous steps. For instance, in (a) the residue is 0 at $ z=0 $ because it's a removable singularity. In (b), the residue at $ z=-1 $ is 0 because the poles of residues require further derivatives to extract non-zero values.

Key Concepts

Residue theoremSingular pointsPole of a functionRemovable singularityEssential singularity

Residue theorem

The residue theorem is a powerful tool in complex analysis. It provides a way to evaluate complex integrals over closed curves by examining the singular points inside the curve.
Specifically, the theorem states that the integral of a function around a closed loop is equal to the sum of the residues at the singular points within that loop, multiplied by $2\pi i$.

Make sure all singular points are inside the contour.
Compute residues for those points.
Apply the formula: $ \oint_C f(z) \, dz = 2\pi i \sum \text{Res}(f, z_k) $.

This method simplifies the computation of complex integrals, especially when dealing with many singularities or complex contours.

Singular points

In complex analysis, singular points are locations in the complex plane where a function is not defined or not differentiable.
Identifying these points is the first step in understanding the behavior of a complex function. These points can be classified into various types:

A point where a function approaches infinity.
A location where a limit exists but the function is undefined.
A point where neither regularity nor a limit exists.

Understanding the nature of singular points helps in the application of the residue theorem and the classification of the singularity type.

Pole of a function

A pole is a type of singularity where a complex function goes to infinity.
Poles are classified by their order, which tells us how "strong" the singularity is at that point. The simplest type of pole is a pole of order one, known as a simple pole.

A pole of order $ n $ means the function behaves like $ \frac{1}{(z-z_0)^n} $ near $ z_0 $.
Residue computation involves derivatives, especially for higher-order poles.

Understanding poles is crucial in analysis, as they directly influence the calculation of residues, which are essential for applying the residue theorem.

Removable singularity

A removable singularity is a point at which a complex function is not initially defined but can be made analytic by extending it.
This occurs when the function approaches a finite limit as one approaches the singularity point, suggesting that if defined properly at that point, the function would be regular.

The function limit exists but is undefined at the point.
After redefining the function, the singularity is "removed" and the function becomes smooth.
Residues are zero at removable singularities.

These are important as they often simplify analysis, allowing the application of regular calculus methods for functions near the singularity.

Essential singularity

Essential singularities are singular points around which a function behaves erratically.
In contrast to poles or removable singularities, an essential singularity is a point where the function exhibits extreme variation and does not converge to any limit.
A classic property of functions near essential singularities is that they take on every possible complex value in any neighborhood of the singularity, with possibly one exception.

Cannot be reduced to a simpler expression like poles.
Residue calculation often involves complex analysis techniques.

These points are fascinating because they exhibit chaotic behavior, and studying them can reveal insights into the nature of complex functions.

Problem 1

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