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Find out whether infinity is a regular point, an essential singularity, or a pole (and if a pole, of what order) for each of the following functions. Using Problem 1 , or Problem 2, or (8.3), find the residue of each function at infinity. $$ \frac{z}{z^{2}+1} $$

Short Answer

Expert verified
It is a pole of order 1 and the residue at infinity is -1.

Step by step solution

01

- Analyze the function at infinity

Given the function \( f(z) = \frac{z}{z^{2}+1} \). To analyze the behavior at infinity, consider the transformation \( w = \frac{1}{z} \), which implies \( z = \frac{1}{w} \). Substitute into the function: \[ f\left( \frac{1}{w} \right) = \frac{\frac{1}{w}}{\left(\frac{1}{w}\right)^2 + 1} = \frac{1/w}{1/w^2 + 1} = \frac{1}{w(w + 1)} \]
02

- Simplify the transformed function

Simplify the expression \( \frac{1}{w(w+1)} \): \[ f\left( \frac{1}{w} \right) = \frac{1}{w^2 + 1} \]
03

- Identify the nature of the singularity

The simplified function \( \frac{1}{w^2 + 1} \) has singularities at \( w = i \) and \( w = -i \), both of which are finite. Since \( z = \frac{1}{w} \), these correspond to singularities at \( z = \infty \). Given that these singularities are of finite order (second order), \( z = \infty \) is a pole of order 1.
04

- Use residue calculation for infinity

To find the residue at infinity, use the formula \( \text{Res}(f(z), \infty) = -\text{Res}(g(z), 0) \) where \( g(z) = \frac{1}{z^2 f(1/z)} \). Simplify it: \[ g(z) = \frac{1}{z^2 \left( \frac{z}{z^2+1} \right)} = \frac{1}{z(z^2+1)} \] Now, at \( z = 0 \), this becomes a simple pole. The residue at 0 can be calculated as: \[ \text{Res}(g(z), 0) = \lim_{z \to 0} z \cdot \frac{1}{z(z^2 + 1)} = 1 \] Therefore, \( \text{Res}(f(z), \infty) = -1 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Infinity Singularities
In complex analysis, understanding the behavior of functions at infinity is essential. Infinity can present different types of singularities: regular points, essential singularities, or poles. To analyze these, we often use a transformation such as changing the variable to its reciprocal. If we denote this transformation by \( w = \frac{1}{z} \), it helps in simplifying the study of the function at infinity. By substituting \( z = \frac{1}{w} \) back into the original function \( f(z) \), we simplify the problem to analyze how the function behaves as \( w \) approaches zero instead of tackling the behavior at infinity directly.
Residue Calculation
Calculating residues is a core part of complex analysis, particularly in evaluating complex integrals. The residue at a particular point of a function is essentially the coefficient of \( \frac{1}{z - z_0} \) in the function's Laurent series expansion around a point \( z_0 \). To find the residue at infinity, we can use the formula: \( \text{Res}(f(z), \infty) = - \text{Res}(g(z), 0) \), where \( g(z) = \frac{1}{z^2 f(1/z)} \). This transformation allows us to shift our focus from infinity to a point we can more easily analyze. In our example, we transformed \( f(z) = \frac{z}{z^{2}+1} \) to \( g(z) = \frac{1}{z(z^2 + 1)} \) and calculated the residue at zero to get the residue at infinity.
Poles in Complex Functions
Poles are singularities where a complex function tends to infinity. The order of a pole is the smallest integer \( n \) for which the limit \( \lim_{{z \to z_0}} (z - z_0)^n f(z) \) exists and is finite. Identifying poles and their orders helps in understanding the behavior and characteristics of complex functions. In our example, the transformed function \( \frac{1}{w^2 + 1} \) showed finite singularities at \( w = i \) and \( w = -i \). This translates back to poles at infinity for our original function. Because these singularities are of second order for \( w \), it means that \( z = \infty \) in the original function is actually a simple pole.
Singularity Analysis
Singularities such as poles, essential singularities, and regular points characterize how complex functions behave in certain regions. A regular point is where the function is analytic. A pole is a point where the function goes to infinity, and an essential singularity is where the function exhibits chaotic behavior. By analyzing the transformed function \( \frac{1}{w^2 + 1} \) and identifying its poles at \( w = i \) and \( w = -i \), we concluded that \( z = \infty \) for the original function \( f(z) = \frac{z}{z^2+1} \) is a pole. This knowledge helps in deeper understanding and proper handling of complex integrals, residue calculus, and other mathematical applications.

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Most popular questions from this chapter

Use Cauchy's theorem or integral formula to evaluate the integrals in Problems 17 to 20 . \(\oint_{c} \frac{e^{3 z} d z}{z-\ln 2}\) if \(C\) is the square with vertices \(\pm 1 \pm i\).

Find the residues of the following functions at the indicated points. Try to select the easiest method. $$ \frac{1-\cos 2 z}{z^{3}} \text { at } z=0 $$

Two long parallel cylinders form a capacitor. (Let their cross sections be the images of \(u=a\) and \(u=-a .\) ) If they are held at potentials \(V_{0}\) and \(-V_{0}\), find the potential \(V(x, y)\) at points between them. Given that the charge (per unit length) on a cylinder is \(q=\) \(V_{0} /(2 a)\), show that the capacitance (per unit length), that is, \(4 /\left(2 V_{0}\right)\), is given by \(1 /(4\) arc cosh \(d / 2 r)\), where \(d\) is the distance between the centers of the two cylinders, and their radii are \(r\).

Evaluate the following integrals by computing residues at infinity. Check your answers by computing residues at all the finite poles. (It is understood that \(\oint\) means in the positive direction.) \(\oint \frac{1-z^{2}}{1+z^{2}} \frac{d z}{z} \quad\) around \(|z|=2\).

For each of the following functions find the first few terms of each of the Laurent series about the origin, that is, one series for each annular ring between singular points. Find the residue of each function at the origin. (Warming: To find the residue, you must use the Laurent series which converges near the origin.) Hints: See Problem 2. Use partial fractions as in equations (4.5) and (4.7). Expand a term \(1 /(z-a)\) in powers of \(z\) to get a series convergent for \(|z|a\). $$ \frac{1}{z(z-1)(z-2)^{2}} $$

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