Question

Evaluate each of the following in \(x+i y\) form, and compare with a computer solution. $$\ln \left(\frac{1-i}{\sqrt{2}}\right)$$

Short answer

-\frac{\pi}{4}i

Step-by-step solution

Convert the Complex Number to Polar Form

Identify the modulus and argument of the complex number \(\frac{1-i}{\sqrt{2}}\). First, find the modulus using \(|z| = \sqrt{x^2 + y^2}\) for the numerator. For \(1 - i\): \(\sqrt{1^2 + (-1)^2} = \sqrt{2}\). Therefore, \(|1-i| = \sqrt{2}\). Now the modulus of \(\frac{1-i}{\sqrt{2}}\) is: \(\frac{\sqrt{2}}{\sqrt{2}} = 1\).

Find the Argument

Calculate the argument \(\theta\) of the numerator \(1 - i\). The argument can be found using \(\tan(\theta) = \frac{y}{x} = \frac{-1}{1} = -1\). So \(\theta = -\frac{\pi}{4}\). For the fraction, the argument is: \(\theta_1 = \-\frac{\pi}{4}\) and \(\theta_2 = 0\) for the denominator \({\sqrt{2}}\). Hence, \(\theta_{result} = -\frac{\pi}{4} - 0 = -\frac{\pi}{4}\).

Apply the Natural Logarithm in Polar Form

Use the property of logarithms for complex numbers in polar form: \(\text{ln}(re^{i\theta}) = \text{ln}(r) + i\theta\). We have \(\text{ln}(1e^{-i\frac{\pi}{4}})\). Since \(\text{ln}(1) = 0\), we get the logarithm of the angle part: \(0 + i(-\frac{\pi}{4}) = -\frac{\pi}{4}i\).

Key concepts

complex numbers

Complex numbers are numbers that have a real part and an imaginary part. They are expressed in the form:

x + iy

Here, x is the real part, and iy is the imaginary part where i is the imaginary unit satisfying \(i^2 = -1\). An example of a complex number is \(1 - i\). We use complex numbers to solve equations that don't have real solutions and in fields such as engineering, physics, and signal processing.

polar form

Converting complex numbers from standard form (x + iy) to polar form can make calculations easier. In polar form, a complex number is represented as

\(z = re^{i\theta}\)

where \(r\) is the modulus (or magnitude) and \(\theta\) is the argument (or angle). The modulus \(r\) is the distance from the origin to the point (x, y) in the complex plane and is found using the formula \( r = \sqrt{x^2 + y^2} \). The argument \(\theta\) represents the angle made with the positive real axis and is calculated using \( \theta = \tan^{-1}(\frac{y}{x}) \). In this exercise, for the complex number \( \frac{1-i}{\sqrt{2}} \), we convert it into polar form to simplify the logarithm calculation.

modulus and argument

The modulus and argument are two key components of a complex number in polar form.

• Modulus: The modulus of a complex number, also known as the magnitude, is the distance between the point representing the complex number and the origin in the complex plane. It is calculated as \( |z| = \sqrt{x^2 + y^2} \). For \( 1 - i \), the modulus is \( \sqrt{2} \).
• Argument: The argument is the angle formed by the complex number with the positive real axis. It is given as \( \theta = \arctan(\frac{y}{x}) \). For \( 1 - i \), the argument is \( -\frac{\pi}{4} \).

In this exercise, we first find the modulus and argument of the complex number \( \frac{1-i}{\sqrt{2}} \) by dealing with its components separately (numerator and denominator).

natural logarithm

The natural logarithm is denoted by ln and is the logarithm to the base \(e\), where \(e\) is an important mathematical constant approximately equal to 2.71828. For a complex number in polar form \( re^{i\theta} \), the natural logarithm is found using:

\( \ln(re^{i\theta}) = \ln(r) + i\theta \)

To find \( \ln( \frac{1-i}{\sqrt{2}} ) \), first convert it to polar form, which simplifies the use of the natural logarithm. In our exercise, once we determined the modulus to be 1 and the argument to be \( -\frac{\pi}{4} \), applying the natural logarithm gave us the result:

\( 0 + i(-\frac{\pi}{4}) = -\frac{\pi}{4}i \)

Euler's formula

Euler's formula is a fundamental equation in complex analysis, given by:

\( e^{i\theta} = \cos(\theta) + i\sin(\theta) \)

This formula links complex exponentiation with trigonometric functions and is essential for converting complex numbers between rectangular form and polar form. In the context of our exercise, when we convert \( 1 - i \) to polar form, we write it as \( \sqrt{2} e^{-i\pi/4} \) by identifying the modulus \( \sqrt{2} \) and the argument \( -\frac{\pi}{4} \). Euler’s formula helps us understand how these trigonometric relationships manifest as complex exponential functions.

Using Euler's formula, we confirmed that \( \frac{1-i}{\sqrt{2}} \) can be depicted as \(1 e^{-i\pi/4} \), setting up for an easy computation with natural logs.