Chapter 14: Problem 24

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

Short Answer

Expert verified

The residue at $z = 0$ is 2.

Step by step solution

Expand the numerator using Taylor Series

Use the Taylor series expansion of $\text{cos} $ around $z = 0$. The Taylor series of $\text{cos}(2z)$ is \[ \text{cos}(2z) = 1 - \frac{(2z)^2}{2!} + \frac{(2z)^4}{4!} - \frac{(2z)^6}{6!} + \text{higher-order terms} \] which simplifies to \[ 1 - 2^2 \frac{z^2}{2} + 2^4 \frac{z^4}{4!} - \text{higher-order terms} = 1 - 2z^2 + \frac{2^4 z^4}{24} - \text{higher-order terms} \]

Substitute the expansion into the function

Replace $ \text{cos}(2z) $ in the original function $\frac{1-\text{cos}(2z)}{z^3}$ with its Taylor series expansion: \[ \frac{1 - \big(1 - 2z^2 + \frac{2^4 z^4}{24} - \text{higher-order terms}\big)}{z^3} \]

Simplify the expression

Simplify the numerator: \[ 1 - (1 - 2z^2 + \frac{2^4 z^4}{24} - \text{higher-order terms}) = 2z^2 - \frac{2^4 z^4}{24} + \text{higher-order terms} \] Hence, the function becomes: \[ \frac{2z^2 - \frac{2^4 z^4}{24} + \text{higher-order terms}}{z^3} = \frac{2z^2}{z^3} - \frac{2^4 z^4}{24z^3} + \text{higher-order terms} = \frac{2}{z} - \frac{16z}{24} + \text{higher-order terms} \]

Identify the residue

From the above expression, the term $\frac{2}{z}$ indicates that the residue at $z = 0$ is 2.

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Key Concepts

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

Taylor series expansion

The Taylor series expansion is a way to express functions as infinite sums of terms calculated from the values of their derivatives at a single point.
The Taylor series of a function $ f(z) $ around a point $ z = a $ is given by:
\[ f(z) = f(a) + f'(a)(z-a) + \frac{f''(a)}{2!}(z-a)^2 + \frac{f'''(a)}{3!}(z-a)^3 + \cdots \]
In our example, we expand $ \text{cos}(2z) $ around $ z = 0 $ using its Taylor series:
\[ \text{cos}(2z) = 1 - \frac{(2z)^2}{2!} + \frac{(2z)^4}{4!} - \frac{(2z)^6}{6!} + \text{higher-order terms} \]
This series is useful for simplifying complex functions, especially when dealing with singularities or points where the function behaves unusually.

Residue at a singularity

A residue is a crucial concept in complex analysis, particularly in the evaluation of complex integrals.
A function $ f(z) $ has a singularity at a point $ z = a $ if the function is not analytic (not differentiable) at that point but analytic elsewhere in some neighborhood of $ a $.
The residue of $ f $ at a singularity $ z = a $ is the coefficient of $ \frac{1}{z-a} $ in the Laurent series expansion of $ f $ around $ z = a $.
For the function in our example, we need to find the residue at $ z = 0 $. After simplifying the expression, we look for the term $ \frac{2}{z} $. Thus, the residue is 2.

Simplify expressions

Simplifying expressions involves reducing them to their most basic form to make calculation and interpretation easier.
After expanding $ \text{cos}(2z) $ in our example, we were left with:
\[ \frac{1 - (1 - 2z^2 + \frac{2^4 z^4}{24} - \text{higher-order terms})}{z^3} = \frac{2z^2 - \frac{2^4 z^4}{24} + \text{ higher-order terms}}{z^3} \]
Further simplification gives us:
\[ \frac{2z^2}{z^3} - \frac{2^4 z^4}{24z^3} + \text{ higher-order terms} = \frac{2}{z} - \frac{16z}{24} + \text{ higher-order terms} \]
This simplification is critical for identifying the residue correctly.

Complex functions

Complex functions are functions of a complex variable, involving both real and imaginary parts. They are fundamentally different from real functions due to their nature and behavior.
The study of complex functions often involves concepts like analyticity, singularities, and residues.
In our case, the function $ \frac{1 - \text{cos}(2z)}{z^3} $ was analyzed. Understanding how to manage such functions requires techniques such as Taylor series expansion and simplification of expressions to detect key features like residues at singular points.