Question

Find the Laurent series for the following functions about the indicated points; hence find the residue of the function at the point. (Be sure you have the Laurent series which converges near the point.) $$\frac{\cosh z}{z^{2}}, z=0$$

Short answer

The residue at \( z = 0 \) is \( 0 \).

Step-by-step solution

Write the function in terms of its Taylor series

The given function is \( \frac{\cosh z}{z^{2}} \). First, recall the Taylor series expansion of \( \cosh z \), which is \(\cosh z = \sum_{n=0}^{\infty} \frac{z^{2n}}{(2n)!}.\)

Divide the Taylor series by \( z^2 \)

Next, divide each term in the Taylor series expansion of \( \cosh z \) by \( z^2 \) to get the Laurent series: \(\frac{\cosh z}{z^2} = \frac{1}{z^2} + \frac{z^2}{2!z^2} + \frac{z^4}{4!z^2} + \cdots = \frac{1}{z^2} + \frac{1}{2} + \frac{z^2}{24} + \cdots \).

Identify the residue

To find the residue at \( z = 0 \), look for the coefficient of the \( \frac{1}{z} \) term in the Laurent series. However, there is no \( \frac{1}{z} \) term present in the series expansion. Therefore, the residue at \( z = 0 \) is \( 0 \).

Key concepts

Complex Analysis

Complex analysis is a field of mathematics that studies complex numbers and functions of complex variables. It is essential in understanding various areas of both pure and applied mathematics. Complex analysis extends concepts like differentiation and integration to the complex plane.

• **Complex Numbers**: These are numbers in the form of \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit, satisfying \( i^2 = -1 \).
• **Complex Functions**: Functions that take complex numbers as inputs and produce complex numbers as outputs. One essential function in complex analysis is \( \text{cosh}(z) \), the hyperbolic cosine of a complex number.
• **Analytic Functions**: Complex functions that are differentiable everywhere in their domain. A typical example is the Laurent series, which represents an analytic function either as a power series or a combination of positive and negative powers.

Residue

The concept of a residue is central to complex analysis, particularly in evaluating complex integrals. The residue of a function at a singularity is the coefficient of the \( \frac{1}{z} \) term in its Laurent series expansion. Residues are crucial in the famous residue theorem, which simplifies the computation of integrals of analytic functions over closed curves.

• **Simple Poles**: These are singularities where a function behaves like \( \frac{1}{z-a} \), with \( a \) being the point where the singularity occurs.
• **Calculating Residue**: To find the residue of a function at a point, look at the Laurent series around that point. The residue is the coefficient of the \( \frac{1}{z} \) term.
• **Example**: In the given function \( \frac{\text{cosh}(z)}{z^2} \), the Laurent series shows no \( \frac{1}{z} \) term, indicating that the residue at \( z=0 \) is \( 0 \).

Understanding residues helps in solving complex integrals and identifying the nature of singularities in functions.

Taylor Series

The Taylor series represents functions as infinite sums of terms calculated from the values of their derivatives at a single point. For a function \( f(z) \) that is infinitely differentiable at \( z=a \), the Taylor series is given by: \[ f(z) = \frac{f(a)}{0!} + \frac{f'(a)}{1!}(z-a) + \frac{f''(a)}{2!}(z-a)^2 + \frac{f^{(3)}(a)}{3!}(z-a)^3 + \frac{f^{(4)}(a)}{4!}(z-a)^4 + \] The series converges to the function within its radius of convergence.

• **Application**: Taylor series is used to approximate functions locally around a point and to simplify complex functions for integration and differentiation.
• **Example**: The Taylor series expansion of \( \text{cosh}(z) \) is \( \text{cosh}(z) = \frac{1}{0!}z^0 + \frac{1}{2!}z^2 + \frac{1}{4!}z^4 + \frac{1}{6!}z^6 + \frac{1}{8!}z^8 + \)
• **Converting to Laurent Series**: By dividing the Taylor series terms of \( \text{cosh}(z) \) by \( z^2 \), we obtain the Laurent series: \( \frac{\text{cosh}(z)}{z^2} = \frac{1}{z^2} + \frac{1}{2!} + \frac{z^2}{4!} + \frac{z^4}{6!} + \frac{z^6}{8!} + \)
Taylor series play a fundamental role in understanding and working with complex functions.