Using series you know from Chapter 1, write the power series (about the origin) of the following functions. Use Theorem III to find the disk of convergence of each series. What you are looking for is the point (anywhere in the complex plane) nearest the origin, at which the function does not have a derivative. Then the disk of convergence has center at the origin and extends to that point. The series converges inside the disk. $$\cos z$$

The Maclaurin series is a type of power series that represents a function as an infinite sum of terms. Each term is calculated using the function's derivatives evaluated at zero. It is a special case of the Taylor series centered at zero.
To understand how the Maclaurin series works, consider the function \( f(z) \). The Maclaurin series for \( f(z) \) is given by:
\[ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n \]
Here, \( f^{(n)}(0) \) denotes the \( n \)-th derivative of \( f \) evaluated at zero.
For the cosine function, \( \text{cos}\thinspace z \), the Maclaurin series expansion is:
\[ \text{cos}\thinspace z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!} \]
This infinite series accurately represents \( \text{cos}\thinspace z \) for any complex number \( z \).
It's valuable because it can help approximate complex functions using simpler polynomial terms.

The disk of convergence is the region within which a power series converges. For a power series centered at the origin, this region is a disk in the complex plane.
To find the disk of convergence for \( \text{cos}\thinspace z \), we must identify where the power series stops converging. We look for the nearest point to the origin where the function is not analytic (lacks a derivative).
Since \( \text{cos}\thinspace z \) is entire (analytic everywhere on the complex plane), it does not have a point where it lacks a derivative. Therefore, the disk of convergence for the series of \( \text{cos}\thinspace z \) extends to the whole complex plane.
In general, to find the disk of convergence of any power series, the radius is determined by the distance from the center to the nearest singularity (point where the function is not analytic).

The radius of convergence is a crucial concept for understanding power series. It defines how far from the center (in this case, the origin) the series converges.
For a given power series \( \sum_{n=0}^{\infty} a_n z^n \), the radius of convergence \( R \) is determined by the formula:
\[ \frac{1}{R} = \limsup_{n \to \infty} \sqrt[n]{|a_n|} \]
We apply this concept to our series for \( \text{cos}\thinspace z \):
\[ \text{cos}\thinspace z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!} \]
Here, \( a_n = \frac{(-1)^n}{(2n)!} \). Since \( \text{cos}\thinspace z \) is entire (analytic everywhere), there is no point at which the series fails to converge.
This implies the radius of convergence is infinite, meaning the series converges everywhere on the complex plane.

In complex analysis, entire functions are special. They are functions that are analytic everywhere in the complex plane. That means they have a derivative at every point in \( \mathbb{C} \).
Examples of entire functions include familiar functions like polynomials, the exponential function \( e^z \), sine, and cosine. Their power series converge globally without any restriction.
For the function \( \text{cos}\thinspace z \), we observe that its power series:
\[ \text{cos}\thinspace z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!} \]
is valid everywhere in \( \mathbb{C} \). This happens because \( \text{cos}\thinspace z \) does not have singularities or points where it would not be differentiable.
Understanding entire functions is key as they often simplify complex analysis problems by leveraging their global convergence properties. For any entire function, the radius of convergence for its Maclaurin series is infinite.