Question

Find the disk of convergence of the series \(\sum(z-2 i)^{n} / n\).

Short answer

The disk of convergence is \(|z-2i|<1\).

Step-by-step solution

Identify the General Term

The given series is \(\sum\ \frac{(z−2i)^n}{n}\). Consider the general term \frac{(z−2i)^n}{n}.\

Apply the Root Test

To find the radius of convergence, use the root test. According to the root test, the radius of convergence is given by \(\frac{1}{\limsup_{n\to\infty} \sqrt[n]{|a_n|}}.\)

Determine \sqrt[n]{|a_n|}

Here, \(a_n=\frac{(z−2i)^n}{n}\) so \(\sqrt[n]{|a_n|} = \sqrt[n]{\frac{|z−2i|^n}{n}} = \frac{|z−2i|}{\sqrt[n]{n}}.\)

Evaluate \limsup_{n\to\infty} \sqrt[n]{|a_n|}

Since \(\sqrt[n]{n}\to1\) as \(n\to\infty\), it follows that \(\lim_{n\to\infty} \sqrt[n]{|a_n|} = |z - 2i|.\)

Calculate the Radius of Convergence

The radius of convergence is given by \(\frac{1}{\lim_{n\to\infty} \sqrt[n]{|a_n|}} = \frac{1}{|z-2i|}\), provided that this limit exists and is finite.

Interpret the Result

Therefore, the series \sum\ \frac{(z−2i)^n}{n} converges when \(|z-2i|<1\). Thus, the disk of convergence is centered at \(2i\) with radius 1.

Key concepts

Radius of Convergence

The radius of convergence determines where a complex series converges. For a series \( \sum a_n(z - z_0)^n \), the radius of convergence is the distance from the center \( z_0 \) within which the series converges.
Imagine dropping a stone in a pond. The ripples represent the radius around the point (the center). If you're inside the ripples, the series converges; outside, it diverges.
In our problem, the series is \( \sum \frac{(z-2i)^n}{n} \). We need to determine the range of \( z \) values that keep the series within its 'convergent circle'. Using root test helps to find this range.

Root Test

The root test helps us determine the radius of convergence for a series. For a series \( \sum a_n \), we use the formula \( \frac{1}{\limsup_{n\to\infty} \sqrt[n]{|a_n|}} \). It involves evaluating the n-th root of the absolute value of terms in the series.
Let's break down this process for the series \( \sum \frac{(z-2i)^n}{n} \):

• Identify the general term \( a_n = \frac{(z-2i)^n}{n} \).
• Find \( \sqrt[n]{|a_n|} \), which simplifies to \( \frac{|z-2i|}{\sqrt[n]{n}} \).
• As \( n \) grows, \( \sqrt[n]{n} \) approaches 1, leading to \( \lim_{n\to\infty} \sqrt[n]{|a_n|} = |z-2i| \).

As a result, the radius of convergence is given by \( \frac{1}{|z-2i|} \). Hence, for convergence, the value \( |z - 2i| \) must be less than 1. This describes a disk with center at \( 2i \) and radius 1.

Complex Series

A complex series is a sum of terms involving complex numbers. Complex numbers include both real and imaginary parts. By working with series like \( \sum\frac{(z - 2i)^n}{n} \), we can understand how these sums behave in the complex plane.
Key points to consider with complex series:

• The center of convergence: Here, it's \( 2i \), a purely imaginary number.
• Your general term describes how each term is positioned relative to this center.

Understanding the behavior of such series helps in various fields like engineering, physics, and computer science, where they model real-world phenomena involving complex variables.
For this problem, focusing on how \( |z-2i| \) behaves gives insights into where the series will converge, providing a 'safe' area for working within the complex plane.