Problem 24 - Advanced Engineering Mathematics

Question

Find the circle and radius of convergence of the given power series. \(\sum_{k=1}^{\infty} \frac{1}{k^{2}(3+4 i)^{k}}(z+3 i)^{k}\)

Step-by-Step Solution

Answer

The radius of convergence is 5, centered at \(-3i\).

Step 1: Identify the Power Series Form

The power series is given by \( \sum_{k=1}^{\infty} \frac{1}{k^{2}(3+4i)^{k}}(z+3i)^{k} \). It has the general form of \( \sum_{k=0}^{\infty} a_k (z-c)^k \). Here, \( a_k = \frac{1}{k^2 (3+4i)^k} \) and the center \( c = -3i \).

Step 2: Determine Absolute Value of Coefficient Ratio

To find the radius of convergence, use the formula \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \). Since \( a_k = \frac{1}{k^2 (3+4i)^k} \), compute: \( a_{k+1} = \frac{1}{(k+1)^2 (3+4i)^{k+1}} \). Hence, \( \left| \frac{a_{k+1}}{a_k} \right| = \left| \frac{(3+4i)^k}{(3+4i)^{k+1}} \cdot \frac{k^2}{(k+1)^2} \right| = \left| \frac{1}{3+4i} \right| \cdot \left| \frac{k^2}{(k+1)^2} \right| \).

Step 3: Calculate Limit for Unit Contribution

Since \( \left| \frac{1}{3+4i} \right| \) is constant, focus on the limit \( \lim_{k \to \infty} \left| \frac{k^2}{(k+1)^2} \right| = \lim_{k \to \infty} \frac{k^2}{(k+1)^2} = \lim_{k \to \infty} \left( \frac{k}{k+1} \right)^2 = 1 \).

Step 4: Evaluate the Magnitude of the Complex Number

Calculate \( \left| \frac{1}{3 + 4i} \right| \). First, find \( |3 + 4i| = \sqrt{3^2 + 4^2} = 5 \). Therefore, \( \left| \frac{1}{3+4i} \right| = \frac{1}{5} \).

Step 5: Determine the Radius of Convergence

With \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = \frac{1}{5} \), the radius of convergence \( R = \frac{1}{\lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right|} = 5 \).

Step 6: Define the Circle of Convergence

The circle of convergence is centered at \( -3i \) with a radius of 5. Therefore, it includes all points \( z \) such that \( |z + 3i| < 5 \).

Key Concepts

Radius of Convergence

The radius of convergence is a crucial concept when dealing with power series, especially in complex analysis. It essentially tells us how "far" we can travel away from the center of a power series while ensuring that the series converges. In simpler terms, imagine having a circle drawn around the center of a series, and you can explore within this circle, where the series will behave nicely and converge to some value.

To compute this radius, you'll often use the ratio test, which involves examining the absolute value of the ratio of consecutive terms in the series. In the problem provided, the series centered at \(-3i\) has a radius of convergence, \(R\), calculated using the formula \(R = \frac{1}{\lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right|}\).

In our example, we find that \(\left| \frac{1}{3+4i} \right| \) is constant and simplifies to 1/5. This leads to a radius, \(R = 5\), indicating that points within a distance of 5 units from the center will allow the series to converge.

Complex Analysis

Complex analysis is the branch of mathematics dealing with complex numbers and functions of a complex variable. It provides a lens through which we can understand the behavior of functions beyond real numbers, and it's particularly potent in studying phenomena like electrical engineering, fluid dynamics, and other fields.

In our context, the power series \(\sum_{k=1}^{\infty} \frac{1}{k^{2}(3+4 i)^{k}}(z+3 i)^{k}\) exhibits properties typical in complex analysis. The key operation involves complex numbers, such as \(3+4i\), whose magnitude is vital in determining convergence through calculations like \(\left|3+4i\right| = \sqrt{3^2 + 4^2} = 5\).

Understanding complex magnitudes and manipulations is at the heart of solving problems in complex analysis, making series like these fascinating subjects of study.

Power Series

Power series are infinite series of terms in the form \(a_k(z-c)^k\), where each term contributes progressively smaller elements to the sum if the series converges. They resemble polynomials but can have infinitely many terms. What makes power series so interesting is their ability to approximate functions and solve differential equations.

In the problem given, the power series \(\sum_{k=1}^{\infty} \frac{1}{k^{2}(3+4 i)^{k}}(z+3 i)^{k}\) is centered at \(-3i\) with coefficients \(a_k = \frac{1}{k^2 (3+4i)^k}\). These particular coefficients affect the rate of convergence and the size of the radius we discussed previously.

Power series are not only central in complex analysis but also used in a myriad of mathematical applications from physics to engineering. By manipulating and understanding their behavior, we can model real-world phenomena with amazing precision.