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Chapter 6: Problem 22

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

Short Answer

Expert verified
The circle of convergence is \( |z| < \sqrt{2} \).

Step by step solution

01

Identify General Term of the Series

The given power series can be written in the general form as \( a_k z^k \), where \( a_k = \frac{1}{k} \left( \frac{i}{1+i} \right)^k \). This helps us identify \( a_k \) for use in the ratio test to find the circle and radius of convergence.
02

Simplify Complex Coefficient

Simplify the complex coefficient \( \frac{i}{1+i} \). Multiply numerator and denominator by the complex conjugate of the denominator: \( \frac{i(1-i)}{(1+i)(1-i)} = \frac{i-i^2}{1+1} = \frac{i+1}{2} = \frac{1}{2} + \frac{i}{2} \). Thus, the coefficient can be expressed as \( \frac{1}{\sqrt{2}} e^{i\frac{\pi}{4}} \) in polar form.
03

Apply the Ratio Test

To determine the radius of convergence, use the ratio test. Calculate \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \), where \( a_k = \frac{1}{k} \left( \frac{1}{\sqrt{2}} e^{i\frac{\pi}{4}} \right)^k \).
04

Calculate the Limit

Using the ratio of successive terms: \[ \lim_{k \to \infty} \left| \frac{\frac{1}{k+1} \left( \frac{i}{1+i} \right)^{k+1}}{\frac{1}{k} \left( \frac{i}{1+i} \right)^k} \right| = \left| \frac{i}{1+i} \right| \cdot \lim_{k \to \infty} \frac{k}{k+1} \]. As \( k \to \infty \), \( \frac{k}{k+1} \to 1 \). The modulus of the coefficient is \( \frac{1}{\sqrt{2}} \).
05

Find the Radius of Convergence

The radius of convergence \( R \) is given by \( \frac{1}{L} \), where \( L \) is calculated as \( \left| \frac{1}{\sqrt{2}} \right| \). Thus, \( R = \sqrt{2} \). The series converges for \( |z| < \sqrt{2} \).
06

Conclusion

The circle of convergence is centered at the origin with radius \( \sqrt{2} \). This means that the power series converges for all complex numbers \( z \) satisfying \( |z| < \sqrt{2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ratio Test
The Ratio Test is a powerful tool when analyzing the convergence of power series. It involves taking the limit of the absolute value of the ratio of successive terms in a series. For a series \( \sum a_k z^k \), we check
  • Calculate \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \).
  • If the limit \( L < 1 \), the series converges absolutely.
  • If \( L > 1 \), the series diverges.
  • If \( L = 1 \), the test is inconclusive.
In our exercise, applying the Ratio Test helped us to determine that for the series to converge, the expressions related to \( z \) must fall within a specific radius, which is crucial in finding the circle of convergence.
Circle of Convergence
The Circle of Convergence is a concept used to describe where a complex power series converges. For the series \( \sum a_k z^k \), it defines a region in the complex plane. Here's how we determine it:
  • The center is usually the origin \( z = 0 \).
  • Convert the radius \( R \) from the result of the Ratio Test.
  • The series converges when \( |z| < R \), forming a circle.
In the given series, calculated using the Ratio Test, the radius \( R \) is \( \sqrt{2} \). Knowing this means that any point \( z \) satisfying \( |z| < \sqrt{2} \) will ensure the series converges.
Complex Coefficients
Complex Coefficients can complicate series analysis, but they can be managed by simplifying and converting them into more usable forms. For instance, consider a coefficient expressed in the series: \( \frac{i}{1+i} \).
  • Transforming it involves multiplying by the complex conjugate.
  • The simplified form is \( \frac{1}{2} + \frac{i}{2} \).
  • This can then be expressed in polar form: \( \frac{1}{\sqrt{2}} e^{i\frac{\pi}{4}} \).
Changing the form of complex coefficients makes it easier to apply tests like the Ratio Test, as these coefficients often influence the convergence characteristics of the entire series. The polar form is particularly useful as it separates magnitude and direction, simplifying calculations.

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Most popular questions from this chapter

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