Problem 42 - A First Course in Complex Analysis with Applications

Question

Sketch the region in the complex plane for which \(\sum_{k=0}^{\infty}\left(\frac{z-1}{z+2}\right)^{k}\) converges

Step-by-Step Solution

Answer

The series converges for \(\text{Re}(z) > -\frac{1}{2}\).

Step 1: Identify the Condition for Convergence

The series \(\sum_{k=0}^{\infty}\left(\frac{z-1}{z+2}\right)^{k}\) is a geometric series with a common ratio \(\frac{z-1}{z+2}\). For the series to converge, the absolute value of the common ratio must be less than 1: \(\left|\frac{z-1}{z+2}\right| < 1\).

Step 2: Rearrange the Inequality

We want to find the region in the complex plane, so we need to analyze where the inequality \(\left|\frac{z-1}{z+2}\right| < 1\) holds. Consider the expression under the absolute value: \(\frac{z-1}{z+2}\). This can be rewritten as \(\frac{x+yi-1}{x+yi+2}\), where \(z = x + yi\).

Step 3: Set Up the Inequality for Real Numbers

To solve \(\left|\frac{z-1}{z+2}\right| < 1\), multiply both sides of the inequality by the absolute value of the denominator: \(|z-1| < |z+2|\). This translates to \(\sqrt{(x-1)^2 + y^2} < \sqrt{(x+2)^2 + y^2}\).

Step 4: Remove the Square Roots

Square both sides of the inequality so we can eliminate the square roots: \((x-1)^2 + y^2 < (x+2)^2 + y^2\).

Step 5: Simplify the Resulting Inequality

Cancel \(y^2\) from both sides to obtain: \((x-1)^2 < (x+2)^2\). Expanding both sides gives: \(x^2 - 2x + 1 < x^2 + 4x + 4\).

Step 6: Final Simplification

Subtract \(x^2\) from both sides and simplify to find: \(-2x + 1 < 4x + 4\). Remove the constants and solve for \(x\): \(-2x - 4x < 4 - 1\) \(-6x < 3\) Divide both sides by \(-6\) (remember to flip the inequality): \(x > -\frac{1}{2}\).

Step 7: Sketch the Solution Region

The condition \(x > -\frac{1}{2}\) describes all points to the right of \(x = -\frac{1}{2}\) on the complex plane. Therefore, the region of convergence is the half-plane of complex numbers where the real part is greater than \(-\frac{1}{2}\).

Key Concepts

Complex Plane

The complex plane is a fundamental concept in complex analysis, providing a geometric representation of complex numbers. A complex number is typically expressed as a combination of a real part and an imaginary part:

• Real numbers are mapped onto the horizontal axis, also known as the real axis.
• Imaginary numbers are mapped onto the vertical axis, also known as the imaginary axis.

This two-dimensional plot is similar to the Cartesian coordinate system you might be familiar with from algebra. Here, any complex number, say, \(z = x + yi\), can be represented as a point \( (x, y)\) on the plane.
The exercise asks us to find a specific region on this plane where a given series converges. It means we are looking for a set of points \(z\) (or \(x + yi\)) that satisfies certain conditions.

Geometric Series

A geometric series is one of the simplest forms of series and is vital in mathematics. It involves terms derived from multiplying a starting value by a consistent ratio in succession. The series is given by:\[\sum_{k=0}^{\infty} ar^k\]where \(a\) is the initial term and \(r\) is the common ratio.
To ensure that the series converges, the absolute value of \(r\) must be less than 1. This condition ensures that as \(k\) increases, the terms will get smaller and eventually the sum will stabilize to a certain value.
In the given exercise, the expression \(\frac{z-1}{z+2}\) serves as the common ratio. Thus, for convergence in the complex plane, we must ensure:

• The inequality \(\left|\frac{z-1}{z+2}\right| < 1\) is satisfied.
• This involves analyzing it both as an algebraic expression and a geometric interpretation on the complex plane.

Region of Convergence

The region of convergence is a key concept when it comes to complex series like the one in the exercise. It identifies the set of values within which a given series is guaranteed to converge. For our geometric series, we derived an inequality based on its common ratio:

• The inequality \(\left|\frac{z-1}{z+2}\right| < 1\) simplifies into a geometric region on the complex plane.
• By converting the inequality into a form involving real numbers, we can identify it as \(x > -\frac{1}{2}\).

This means that the series converges for any complex number whose real component is greater than \(-\frac{1}{2}\). Geometrically, this forms a half-plane to the right of the line \(x = -\frac{1}{2}\). Understanding this region is crucial for solving problems involving convergence on the complex plane.