Residue Calculation for \( \frac{{\cos z}}{z} \) at \( z = 0 \)

The problem asks us to find the residue of the complex function \( f(z) = \frac{{\cos z}}{z} \) at the point \( z = 0 \). In complex analysis, the residue of a function at an isolated singularity is a fundamental concept, particularly useful in evaluating complex integrals using the Residue Theorem.

Understanding the Singularity at \( z = 0 \)

First, we need to identify the type of singularity at \( z = 0 \). A singularity exists where the function is undefined or becomes infinite. For the given function \( f(z) = \frac{{\cos z}}{z} \), the denominator becomes zero at \( z = 0 \). To classify this singularity:

• The numerator is \( \cos z \). When we evaluate it at \( z = 0 \), we get \( \cos(0) = 1 \), which is a non-zero value.
• The denominator is \( z \). This term has a single root (a zero of order 1) at \( z = 0 \).

Since the numerator is non-zero at \( z = 0 \) and the denominator has a simple zero at \( z = 0 \), the function \( f(z) \) possesses a simple pole at \( z = 0 \).

Method 1: Limit Formula for a Simple Pole

For a function \( f(z) \) that has a simple pole at \( z_0 \), the residue at that point can be calculated using the following limit formula:

$$ \text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0) f(z) $$

In this specific case, \( f(z) = \frac{{\cos z}}{z} \) and the pole is at \( z_0 = 0 \). Applying the formula:

$$ \text{Res}\left(\frac{{\cos z}}{z}, 0\right) = \lim_{z \to 0} (z - 0) \left(\frac{{\cos z}}{z}\right) $$

Simplify the expression inside the limit:

$$ = \lim_{z \to 0} z \left(\frac{{\cos z}}{z}\right) $$

The \( z \) terms in the numerator and denominator cancel each other out:

$$ = \lim_{z \to 0} \cos z $$

Now, substitute \( z = 0 \) into the remaining expression:

$$ = \cos(0) $$

$$ = 1 $$

Thus, the residue of \( \frac{{\cos z}}{z} \) at \( z = 0 \) is \( 1 \).

Method 2: Laurent Series Expansion

Another approach to find the residue involves expanding the function into its Laurent series around the singularity. The residue is defined as the coefficient of the \( \frac{1}{(z - z_0)} \) term in this series expansion.

Let's recall the well-known Maclaurin series expansion for \( \cos z \) around \( z = 0 \):

$$ \cos z = 1 - \frac{{z^2}}{{2!}} + \frac{{z^4}}{{4!}} - \frac{{z^6}}{{6!}} + \dots $$

Now, to obtain the Laurent series for \( f(z) = \frac{{\cos z}}{z} \), we divide the series for \( \cos z \) by \( z \):

$$ f(z) = \frac{{\cos z}}{z} = \frac{1}{z} \left( 1 - \frac{{z^2}}{{2!}} + \frac{{z^4}}{{4!}} - \frac{{z^6}}{{6!}} + \dots \right) $$

Distribute the \( \frac{1}{z} \) term across the series:

$$ f(z) = \frac{1}{z} - \frac{z}{2!} + \frac{z^3}{4!} - \frac{z^5}{6!} + \dots $$

The Laurent series is generally of the form \( \dots + c_{-2}(z-z_0)^{-2} + c_{-1}(z-z_0)^{-1} + c_0 + c_1(z-z_0) + \dots \). The residue is precisely the coefficient \( c_{-1} \), which corresponds to the term with \( \frac{1}{(z - z_0)} \).

In our expanded series for \( f(z) \), the term containing \( \frac{1}{z} \) (which is \( \frac{1}{(z-0)} \)) is simply \( \frac{1}{z} \), and its coefficient is \( 1 \). All subsequent terms involve positive powers of \( z \).

Thus, based on the Laurent series expansion, the residue of \( \frac{{\cos z}}{z} \) at \( z = 0 \) is \( 1 \).

Summary of Residue Calculation

Both methods, the limit formula for a simple pole and the Laurent series expansion, consistently show that the residue of \( f(z) = \frac{{\cos z}}{z} \) at \( z = 0 \) is \( 1 \).