Question

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. Check your results by computer. $$\frac{z^2-1}{z^2+1}$$

Short answer

Infinity is a regular point, and the residue at infinity is 0.

Step-by-step solution

- Identify the function

The given function is $\frac{z^2-1}{z^2+1}$. First, we need to analyze the behavior of the function at infinity.

- Transform to analyze infinity

Substitute $w=\frac{1}{z}$ to transform the function's behavior near infinity to the behavior near zero. So, $z=\frac{1}{w}$, and we substitute this into the function.

- Substitute and simplify

Substitute $z=\frac{1}{w}$ into $\frac{z^2-1}{z^2+1}$, transforming it into $\frac{(\frac{1}{w}{)}^2-1}{(\frac{1}{w}{)}^2+1}=\frac{\frac{1}{w^2}-1}{\frac{1}{w^2}+1}=\frac{1-w^2}{1+w^2}$.

- Analyze the transformed function

Now consider the behavior of $\frac{1-w^2}{1+w^2}$ as $w$ approaches 0. The function is regular at $w=0$. Since transforming $w=0$ corresponds to $z=\mathrm{\infty }$, infinity is a regular point for the original function.

- Determine the residue at infinity

To find the residue at infinity, note that the residue at infinity can be found using the residue theorem: $\text{Res}(f,\mathrm{\infty })=-\text{Res}(f,0)$. For $\frac{1-w^2}{1+w^2}$, calculate the residue at 0.

- Calculate residue at zero for transformed function

Expand $\frac{1-w^2}{1+w^2}$ in a Laurent series around $w=0$, and identify the coefficient of $\frac{1}{w}$. For simplicity, observe that it has no $\frac{1}{w}$ term, so the residue at zero is 0.

- Establish residue at infinity

Given the residue at zero is 0, so by the relation $\text{Res}(f,\mathrm{\infty })=-\text{Res}(f,0)$, we find that the residue at infinity is also 0.

- Verify by computation

Using computational tools, re-evaluate the function and residues to double-check the results. Confirm that the computations align with the analytical results.

Key concepts

Singularities

In complex analysis, singularities are points where a function does not behave normally, such as becoming unbounded or not being differentiable. They are classified into different types:

• Removable singularities - Points where a function can be redefined to make it analytic.
• Poles - Points where a function goes to infinity in a predictable way.
• Essential singularities - Points where a function behaves chaotically.

In the given exercise, we are asked to determine the nature of infinity for the function $\frac{z^2-1}{z^2+1}$. By making the substitution $w=\frac{1}{z}$, we transformed the function to analyze its behavior near zero. This helps determine if infinity is a regular point, essential singularity, or pole. Upon analyzing, it turns out infinity is a regular point for the given function.

Residue Theorem

The residue theorem is a powerful tool in complex analysis. It calculates the integral of a function around a closed contour using the sum of residues inside the contour. The theorem states:

$$\frac{1}{2\pi i}{\int }_{C}f(z)dz=\sum \text{Res}(f,z_k)$$ where the $z_k$ are the singular points within the contour $C$.

In the exercise, to find the residue at infinity, we first determined the residue at zero for the transformed function $\frac{1-w^2}{1+w^2}$. Using the relation $\text{Res}(f,\mathrm{\infty })=-\text{Res}(f,0)$, we found the residue at infinity to be zero since the residue at zero was also zero. This application of the residue theorem simplifies complex contour integrals and aids in analyzing functions at infinity.

Laurent Series

The Laurent series is a representation of a complex function that includes both positive and negative powers of $(z-z_0)$. It is given by:

$$f(z)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_n(z-z_0{)}^n$$ where $z_0$ is the point around which the series is expanded.

Laurent series are essential for understanding the behavior of functions at singularities. The coefficient of $\frac{1}{(z-z_0)}$ term is the residue of the function at $z_0$.

In our solution, we expanded the transformed function $\frac{1-w^2}{1+w^2}$ in a Laurent series around $w=0$. We noticed that it did not have a $\frac{1}{w}$ term, leading to a residue of zero at $w=0$. This helped in concluding that the residue at infinity was also zero. Such series expansions provide insights into the local behavior of functions and help identify residues effortlessly.