Question

In Problems 9 and 10, use Laurent series to find the residues of the given functions at the origin. $$\frac{\sin z^2}{z^7}$$

Short answer

The residue at the origin is $\boxed{{\displaystyle -\frac{1}{6}}}$.

Step-by-step solution

Express the Function

Start by writing the given function $$f(z)=\frac{\sin z^2}{z^7}$$

Expand the Sine Function

Recall the Taylor series for sine: $\sin x=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}$. In our case, set $x=z^2$ to get:$$\sin z^2=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(z^2{)}^{2n+1}}{(2n+1)!}$$

Substitute and Simplify

Substitute this series back into the function $f(z)$:$$\frac{\sin z^2}{z^7}=\frac{1}{z^7}\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{z}^{4n+2}}{(2n+1)!}$$Now, this can be written as:$$\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{z}^{4n-5}}{(2n+1)!}$$

Identify the Residue

The residue at the origin is the coefficient of $\frac{1}{z}$ in the Laurent series. We need the term where the exponent of $z$ is $-1$. Set $4n-5=-1$, solving gives $n=1$. Substitute to find:$$\frac{(-1{)}^1{z}^{4\cdot 1-5}}{(2\cdot 1+1)!}=\frac{-z^{-1}}{3!}=-\frac{1}{6z}$$

Write the Residue

The coefficient of $\frac{1}{z}$ in the term is $-\frac{1}{6}$. So, the residue of the function at the origin is $$\boxed{{\displaystyle -\frac{1}{6}}}$$.

Key concepts

Laurent series

The Laurent series is an essential concept in complex analysis. It allows us to represent complex functions as power series, but with terms potentially having negative exponents. Unlike a Taylor series, which only has positive powers of $z$, the Laurent series can handle functions with essential singularities. This means we can expand a function $f(z)$ around a point $z_0$ in a form that includes terms like $a_{-1}/(z-z_0)$, where $a_{-1}$ is known as the residue.

The Laurent series is extremely useful when dealing with functions that have poles. In this exercise, the Laurent series helps us find the residue of $\frac{\sin z^2}{z^7}$ at the origin.

Complex analysis

Complex analysis is the study of functions that operate on complex numbers. Unlike in real analysis, the functions of complex analysis can have properties such as being smooth and differentiable in a more robust sense.

Key topics in complex analysis include:

• Analytic functions: Functions that are locally given by a convergent power series.
• Complex integration: Techniques to integrate functions along paths in the complex plane.
• Singularities and residues: Points where functions fail to be analytic and methods to study them.

In our particular problem, complex analysis provides the groundwork for understanding Laurent series and consequently the residues.

Residue theorem

The residue theorem is a central result in complex analysis. It assists in evaluating complex integrals by relating them to the sum of residues within a closed contour.

According to the residue theorem, if a function $f(z)$ is analytic inside and on some closed contour $C$ except for isolated singularities, then
$${\oint }_{C}f(z)dz=2\pi i\sum \text{Res}(f,z_k)$$
Here, \text{Res}(f, z_k)\ denotes the residue of $f$ at the singularities $z_k$ inside $C$.

In this problem, we work on finding the residue at the origin. This residue is crucial for applying the residue theorem to evaluate integrals over contours encompassing the origin.

Taylor series

The Taylor series lets us expand a differentiable function around a point into an infinite sum of powers of the variable. For $f(z)$, the Taylor series about $z_0$ would look like:
$$f(z)=\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-z_0{)}^n$$
Here, $a_n$ are the coefficients given by $a_n=f^{(n)}(z_0)/n!$.

Though the Taylor series is excellent for analytic functions without singularities, it falls short for functions with isolated singularities. For example, in $\frac{\sin z^2}{z^7}$, the Taylor series for $\sin z^2$ is not sufficient because of the $z^{-7}$ term. Thus, we utilize the Laurent series. However, understanding the Taylor series as an extension of the basic power series is fundamental before learning the more advanced Laurent series.