Problem 32 - A First Course in Complex Analysis with Applications

Question

Expand the given function in Taylor series centered at each of the indicated points. Give the radius of convergence \(R\) of each series. Sketch the region within which both series converge. $$ f(z)=\frac{1}{z}, z_{0}=1+i, z_{0}=3 $$

Step-by-Step Solution

Answer

Taylor series around \( z = 1+i \) converges within \( |z - (1+i)| < \sqrt{2} \) and around \( z = 3 \) within \( |z - 3| < 3 \).

Step 1: Define the Taylor Series

The Taylor series of a function \( f(z) \) centered at a point \( z_0 \) is given by \[ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!} (z - z_0)^n \]. For \( f(z) = \frac{1}{z}, \) the derivatives \( f^{(n)}(z) \) are calculated.

Step 2: Calculate Derivatives of f(z)

The first derivative of \( f(z) = \frac{1}{z} \) is \(-\frac{1}{z^2}\), the second derivative is \(\frac{2}{z^3}\), the third derivative is \(-\frac{6}{z^4}\), and generally, \( f^{(n)}(z) = \frac{(-1)^n n!}{z^{n+1}} \).

Step 3: Taylor Series at z_0 = 1 + i

To find the Taylor series centered at \( z_0 = 1 + i \), calculate each term using \( f^{(n)}(1+i) = \frac{(-1)^n n!}{(1+i)^{n+1}} \) and substitute into the series formula. Simplify for practical terms.

Step 4: Taylor Series at z_0 = 3

For \( z_0 = 3 \), calculate the terms using \( f^{(n)}(3) = \frac{(-1)^n n!}{3^{n+1}} \). Substitute to find series terms at \( z_0 = 3 \) and simplify.

Step 5: Determine Radius of Convergence R

For \( f(z) = \frac{1}{z} \), find all singularities, which in this case is at \( z = 0 \). Use \( R = |z_0 - 0| \) to find the radius of convergence for each Taylor series: \( R_{1+i} = \sqrt{2} \) and \( R_3 = 3 \).

Step 6: Sketch Region of Convergence

Sketch a circle centered at \( z_0 = 1 + i \) with radius \( \sqrt{2} \), and another centered at \( z_0 = 3 \) with radius \( 3 \). These circles represent the regions of convergence for each series.

Key Concepts

Radius of Convergence

The radius of convergence is an essential concept when expanding functions into Taylor series, especially in the context of complex analysis. It determines the extent to which a Taylor series accurately represents a function. For a function centered at a point \( z_0 \), the series converges within a circle of radius \( R \) on the complex plane.
Finding the radius of convergence involves identifying singularities of the function. For instance, the function \( f(z) = \frac{1}{z} \) has a singularity at \( z = 0 \). This means it cannot be represented by a Taylor series centered at any point that is within this singular region.
To calculate the radius of convergence for a series centered at \( z_0 = 1 + i \), the formula \( R = |z_0 - 0| \) yields \( \sqrt{2} \), which is the distance from \( 1 + i \) to the singularity at the origin. Similarly, for a center at \( z_0 = 3 \), the radius of convergence is \( 3 \) since \( R = |3 - 0| \). These distances define the power of the series to converge accurately to \( f(z) \) within these circles on the complex plane.

Complex Analysis

Complex Analysis is a fascinating field that examines functions of complex numbers. Using complex analysis, we can understand and visualize how functions behave not only on the real number line but within the entire complex plane.
In this exercise, we're diving into expanding \( f(z) = \frac{1}{z} \) into a Taylor series around different center points \( z_0 \). The choice of center point is crucial, as it determines the region where the series will effectively converge.
Working with Taylor series in complex analysis involves taking multiple derivatives and calculating complex powers, both of which are essential for expanding functions into infinite series. Each term of the series contains derivatives evaluated at a particular center, which, combined with a power of \( (z - z_0) \), builds up the series representation for points in the complex plane. The series captures local behavior around \( z_0 \) and converges within a specific radius, offering fascinating insights into the nature of complex functions.

Singularities

Singularities play a crucial role in understanding the limitations and behavior of functions in complex analysis. A singularity of a function is a point where the function fails to be defined or doesn't behave well, such as going to infinity. In the case of \( f(z) = \frac{1}{z} \), there is a clear singularity at \( z = 0 \) because dividing by zero is undefined.
Identifying singularities helps in determining the region within the complex plane where a Taylor series can converge to the function. For example, a Taylor series expansion at a point \( z_0 \) can only represent the function accurately up to the closest singularity, as indicated by the radius of convergence.
Sometimes, these singular points can be classified further into essential singularities, poles, or branch points, depending on the nature of the function's behavior around them. This understanding is part of what makes complex analysis a powerful tool to explore advanced mathematical concepts.