Problem 12 - A First Course in Complex Analysis with Applications

Question

Show that the given sequence \(\left\\{z_{n}\right\\}\) converges to a complex number \(L\) by computing \(\lim _{n \rightarrow \infty} \operatorname{Re}\left(z_{n}\right)\) and \(\lim _{n \rightarrow \infty} \operatorname{Im}\left(z_{n}\right)\). $$ \left\\{\left(\frac{1+i}{4}\right)^{n}\right\\} $$

Step-by-Step Solution

Answer

The sequence converges to the complex number 0.

Step 1: Express the Sequence

The sequence given is \( z_n = \left(\frac{1+i}{4}\right)^n \). Our goal is to compute the limits of both the real and imaginary parts as \( n \to \infty \).

Step 2: Simplify the Base of the Sequence

First, express \( \frac{1+i}{4} \) in polar form. The modulus is \( |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2} \), and thus dividing by 4 gives \( \frac{\sqrt{2}}{4} = \frac{1}{2\sqrt{2}} \). The argument is \( \frac{\pi}{4} \). So, \( z_n = \left(\frac{1}{2\sqrt{2}} e^{i\frac{\pi}{4}}\right)^n \).

Step 3: Expand the Expression

Expand \( z_n \) using Euler's formula: \( z_n = \left( \frac{1}{2\sqrt{2}} \right)^n \left( \cos(n\frac{\pi}{4}) + i\sin(n\frac{\pi}{4}) \right) \). This breaks \( z_n \) into real and imaginary parts: \( \operatorname{Re}(z_n) = \left( \frac{1}{2\sqrt{2}} \right)^n \cos(n\frac{\pi}{4}) \), and \( \operatorname{Im}(z_n) = \left( \frac{1}{2\sqrt{2}} \right)^n \sin(n\frac{\pi}{4}) \).

Step 4: Compute the Limits

As \( n \to \infty \), \( \left( \frac{1}{2\sqrt{2}} \right)^n \to 0 \) because \( \frac{1}{2\sqrt{2}} \) is less than 1. Therefore, both \( \operatorname{Re}(z_n) \) and \( \operatorname{Im}(z_n) \) tend to zero, as their coefficients \( \cos(n\frac{\pi}{4}) \) and \( \sin(n\frac{\pi}{4}) \) are bound between -1 and 1.

Step 5: Conclude the Convergence

Since both real and imaginary parts of the sequence \( z_n \) tend to zero, we conclude that \( z_n \to 0 \) as \( n \to \infty \). Therefore, the sequence converges to the complex number \( L = 0 \).

Key Concepts

Complex Numbers

Complex numbers extend the idea of one-dimensional numbers into a two-dimensional space by including an imaginary component. A complex number is represented as \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit with the property \(i^2 = -1\).
This representation is powerful because it allows the expression of numbers in terms of real and imaginary axes, thus enabling the description of more complex algebraic operations and concepts.

Another important aspect of complex numbers is their ability to be expressed in polar form. In polar form, a complex number is represented as \(r \times e^{i\theta}\), where \(r\) is the modulus of the complex number, and \(\theta\) is the argument (or angle) of the number. The conversion between standard form \(a + bi\) and polar form \(r \times e^{i\theta}\) is facilitated by the relationships:

• Modulus \(r = \sqrt{a^2 + b^2}\)
• Argument \(\theta = \tan^{-1}{\left( \frac{b}{a} \right)}\)

Polar representation is often useful in simplifying problems involving powers and roots as seen in our sequence problem.

Euler's Formula

Euler's formula establishes a deep relationship between trigonometry and complex numbers. It states that for any real number \(x\), the exponential function can be expressed as:
\ e^{ix} = \cos(x) + i\sin(x) \
This powerful formula allows us to convert complex exponentials into trigonometric functions.

In our sequence problem, Euler's formula is used to express the base of the sequence in polar form, revealing its real and imaginary components without direct reliance on Cartesian coordinates.
By rewriting \( \left( \frac{1+i}{4} \right)^n \) as \( \left( \frac{1}{2\sqrt{2}} \right)^n \times \left( \cos(n\frac{\pi}{4}) + i\sin(n\frac{\pi}{4}) \right) \), we break the sequence into parts we can more easily analyze. Euler's formula essentially bridges the gap between exponential growth functions and periodic trigonometric functions, making it an invaluable tool in dealing with complex numbers and their behaviors in sequences.

Convergence

Convergence is a crucial concept in mathematics, especially in sequences and series. A sequence is said to converge to a limit \(L\) if its terms get arbitrarily close to \(L\) as the sequence progresses, specifically as \(n\) approaches infinity. In simple terms, as you progress along the sequence, the numbers get closer and closer to a particular value, known as the limit.

In the problem of the sequence \( \left\{ \left(\frac{1+i}{4}\right)^n \right\} \), we are tasked to find out if it converges.The convergence is determined by observing the behavior of both the real and imaginary parts separately as \( n \to \infty \).

• The modulus \( \left( \frac{1}{2\sqrt{2}} \right)^n \) tends to zero because it is less than one.
• The cosine and sine components \( \cos(n\frac{\pi}{4}) \) and \( \sin(n\frac{\pi}{4}) \) are bounded, not affecting the limit going to zero.

Therefore, the sequence converges to 0, showing that when both parts of a complex sequence converge to zero, the sequence as a whole also converges to that limit.