Question

Evaluate each of the following in \(x+i y\) form, and compare with a computer solution. $$\cos (2 i \ln i)$$

Short answer

The result is \-1\, or \(-1 + 0i\).

Step-by-step solution

Understand the Problem

We need to evaluate \(\text{cos}(2i \ln i)\) and express the result in the form \((a + bi)\), where \(a\) and \(b\) are real numbers. This involves some understanding of complex logarithms and hyperbolic functions.

Simplify the Logarithm

First, simplify \(\text{ln}(i)\). We know that \(i\) can be written as \(e^{i \(\pi/2\)}\), so \(\text{ln}(i) = i \(\pi/2\)\).

Multiply by 2i

Now multiply by \(2i\): \(2i \text{ln}(i) = 2i(i\pi/2) = -\pi\).

Evaluate the Cosine Function

Now substitute \(-\pi\) into the cosine function: \(\text{cos}(-\pi)\). We know that \(\text{cos}(-x) = \text{cos}(x)\) and \( \text{cos}(\pi) = -1\).

Express in \(x + i y\) Form

The cosine of \(-\pi\) is \(-1\). In the form \((a + bi)\), this is simply \(-1 + 0i\).

Compare with Computer Solution

Using a computer or a calculator, we can confirm \( \text{cos}(2i \ln i) = -1 \).

Key concepts

complex logarithms

Complex logarithms extend the concept of logarithms into the complex plane. While the logarithm of a real number is straightforward, for a complex number, we include both magnitude and angle. The natural logarithm of a complex number z, written as \text{ln}(z)\, is given by: \[ \text{ln}(z) = \text{ln}|z| + i \text{arg}(z) \] Understanding the argument (angle) is key. For example, the complex number \text{i}\ can be represented as \text{e}^{i \frac{\text{\pi}}{2}}\ due to its position on the complex plane at \frac{\text{\pi}}{2}\ radians. Thus, \text{ln}(i) = i \frac{\text{\pi}}{2}\, as we used in the exercise.

hyperbolic functions

Hyperbolic functions are analogues of trigonometric functions but for a hyperbola instead of a circle. The key hyperbolic functions are \text{sinh}\text{(x)}\ and \text{cosh}\text{(x)}\:\[ \text{sinh}(x) = \frac{\text{e}^x - \text{e}^{-x}}{2} \] \[ \text{cosh}(x) = \frac{\text{e}^x + \text{e}^{-x}}{2} \]. While they look similar to sine and cosine, these functions are related to exponential growth and decay. In the complex plane, they demonstrate unique properties that are useful in solving various mathematical problems involving complex numbers.

Euler's formula

Euler's formula creates a powerful bridge between trigonometry and exponential functions. The classic formulation is:\[ \text{e}^{ix} = \text{cos}(x) + i\text{sin}(x) \]This reveals that complex exponential functions can be represented using trigonometric functions. For our problem, understanding that \text{i}\ is at \frac{\text{\pi}}{2}\ allowed us to simplify \text{ln}(i)\, helping us handle other operations with ease.

imaginary unit

The imaginary unit \text{i}\ is fundamental to complex numbers. It is defined as:\[ \text{i} = \text{\sqrt{-1}} \]This unit allows us to quantify and handle values involving \text{\sqrt{-1}}, expanding the numeric system we use. For instance, \text{i}^2 = -1\, adds meaningful explanation to the steps in the exercise. The manipulation of \text{i}\ was critical in evaluating the given problem, transforming and simplifying expressions along the way.