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Magazine Chapter 6: Problem 16
Expand the given function in a Taylor series centered at the indicated point \(z_{0}\). Give the radius of convergence \(R\) of each series. $$ f(z)=\frac{1}{z}, z_{0}=1+i $$
Short Answer
Expert verified
The Taylor series is \( \frac{1}{1+i} \sum_{n=0}^{\infty} \left( \frac{z-(1+i)}{1+i} \right)^n \) and \( R = \sqrt{2} \).
Step by step solution
01
Identify the Function and Point of Expansion
The function given is \( f(z) = \frac{1}{z} \) and we're asked to expand this function in a Taylor series centered at the point \( z_0 = 1+i \).
02
Rewrite the Function Related to the Center
To expand around \( z_0 = 1+i \), rewrite the function in terms of \( z - z_0 \). Thus, \( f(z) = \frac{1}{z} = \frac{1}{z_0 + (z - z_0)} = \frac{1}{1+i + (z - (1+i))} \).
03
Express as a Power Series
Recognize the expression \( \frac{1}{1+i + (z - (1+i))} \) as a geometric series form \( \frac{1}{a + b} \) if \( |b/a| < 1 \). It transforms to \( \frac{1}{a}(1 - \frac{b}{a} + (\frac{b}{a})^2 - ...) \). Let \( a = 1+i \) and \( b = z - (1+i) \).
04
Determine the Basic Expansion and Substitute Back
Use the basic expansion formula for a geometric series: \( \frac{1}{1-x} = 1 + x + x^2 + \cdots \). Here, substitute \( x = \frac{z-(1+i)}{1+i} \), giving the series: \( \frac{1}{1+i} \cdot \left( 1 + \frac{z-(1+i)}{1+i} + \left(\frac{z-(1+i)}{1+i}\right)^2 + \cdots \right) \).
05
Determine the Radius of Convergence
The geometric series \( \frac{1}{1-x} \) converges when \( |x| < 1 \). Here, \( |\frac{z-(1+i)}{1+i}| < 1 \), meaning that \( |z-(1+i)| < |1+i| \). Compute \( |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2} \). Therefore, the radius of convergence \( R = \sqrt{2} \).
06
Finalize the Taylor Series Expansion
Combine results from previous steps. Thus, the Taylor series for \( f(z) = \frac{1}{z} \) around \( z_0 = 1+i \) is: \( \frac{1}{1+i} \sum_{n=0}^{\infty} \left( \frac{z-(1+i)}{1+i} \right)^n \), and it converges for \( |z-(1+i)| < \sqrt{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Complex Analysis
Complex analysis is a fascinating area of mathematics focused on functions of complex numbers. A complex number is expressed as \( z = x + yi \), where \( x \) and \( y \) are real numbers, and \( i \) is the imaginary unit satisfying \( i^2 = -1 \). When we analyze functions in complex analysis, we explore the unique properties and behaviors of complex-valued functions.
Some key aspects include:
Some key aspects include:
- **Analytic Functions**: These are complex functions that are differentiable at every point in their domain. This property is one of the hallmarks of complex analysis, leading to many important results and applications.
- **Contour Integrals**: Integrals of complex functions taken over a path, or contour, in the complex plane. These types of integrals are central to many complex analysis techniques, like the famous Cauchy Integral Theorem.
- **Taylor and Laurent Series**: These series allow the representation of complex functions as infinite sums. Taylor series are used when functions are analytic at a point, while the Laurent series is more general, also accommodating functions with singularities.
Radius of Convergence
The radius of convergence is a critical concept in series expansions. It tells us the largest distance for which the series converges when expanded around a particular point. Understanding this allows us to determine the region in the complex plane where our power series representation of a function remains valid.
To compute the radius of convergence:
To compute the radius of convergence:
- When working with a power series centered at a point \( z_0 \), written as \( \sum_{n=0}^{\infty} a_n (z-z_0)^n \), the radius of convergence \( R \) can be found using the formula \( R = \frac{1}{\limsup_{n \to \infty} |a_n|^{1/n}} \).
- For geometric series like \( \frac{1}{1-x} \), the series converges when \( |x| < 1 \), illustrating a straightforward method to assess convergence.
- In the exercise, we found \( R = \sqrt{2} \) for the Taylor series expansion of \( f(z) = \frac{1}{z} \) around \( z_0 = 1+i \) using the geometric series property, ensuring convergence when \( |z-(1+i)| < \sqrt{2} \).
Geometric Series
A geometric series is one of the fundamental types of series in mathematics. It takes the form \( a + ar + ar^2 + \cdots \), where \( a \) is the first term and \( r \) the common ratio. Geometric series can be finite or infinite.
For infinite geometric series:
For infinite geometric series:
- The series \( a + ar + ar^2 + \cdots \) converges if the absolute value of the ratio \( |r| < 1 \). The sum of this infinite series is \( \frac{a}{1-r} \).
- This simple convergence criteria make them a powerful tool in calculus and complex analysis, particularly in expanding functions into series.
- In the exercise, we used the geometric series formula to rewrite \( f(z) = \frac{1}{z} \) around \( z_0=1+i \) as a power series, leveraging the geometric series with \( x = \frac{z-(1+i)}{1+i} \).
