Question

Find each of the following in the \(x+i y\) form and check your answers by computer. $$\cosh \left(\frac{i \pi}{2}-\ln 3\right)$$

Short answer

\(0 + i0\)

Step-by-step solution

Understanding the Hyperbolic Cosine Function

The hyperbolic cosine function, denoted as \(\text{cosh}(z)\), can be expressed in terms of exponential functions: \(\text{cosh}(z) = \frac{e^z + e^{-z}}{2}\).

Substitute the Given Expression

We need to find \(\text{cosh}\bigg(\frac{i \pi}{2} - \ln 3\bigg)\). Substitute \(\frac{i \pi}{2} - \ln 3\) into the hyperbolic cosine formula: \[ \text{cosh} \bigg(\frac{i \pi}{2} - \ln 3\bigg) = \frac{e^{\frac{i \pi}{2} - \ln 3} + e^{-\big(\frac{i \pi}{2} - \ln 3\big)}}{2} \]

Simplify the Exponential Terms

First simplify each exponential term separately. \(\begin{align*} e^{\frac{i \pi}{2} - \ln 3} &= e^{\frac{i \pi}{2}} \cdot e^{-\ln 3} \cdot 3}\ && = (i)\cdot\frac{1}{3} \end{align*}\)

Combine Exponential Terms

Combine the simplified exponential terms: \[ \text{cosh} \bigg(\frac{i \pi}{2} - \ln 3\bigg) = \frac{i/3 + (-i/3)}{2}\bigg) = \frac{\frac{i}{3} - \frac{i}{3}}{2} = 0 \]

Convert to \(x + iy\) Form

Since the simplified term is zero, we can express it in the \(x + iy\) form as \(0 + i(0)\).

Key concepts

Hyperbolic Cosine

The hyperbolic cosine function, denoted as \(\text{cosh}(z)\), resembles the regular cosine function but applies to hyperbolic geometry. It is defined using exponential functions. In fact, the formula is as follows:

<\[ \text{cosh}(z) = \frac{e^z + e^{-z}}{2} \]

This formula tells us that we can compute the hyperbolic cosine of any complex number or real number by taking the average of its exponential and its reciprocal exponential. Understanding this formula is key to solving the exercise problem.

Hyperbolic functions like \(\text{cosh}(z)\) play an essential role in various fields of mathematics, including solving differential equations and describing hyperbolic geometry.

Complex Numbers

Complex numbers are an extension of the real numbers and include a real part and an imaginary part. The general form of a complex number is written as \(a + bi\), where \(a\) is the real part, \(b\) is the imaginary part and \(i\) is the imaginary unit, defined by \(i^2 = -1\).

When dealing with complex numbers in hyperbolic functions, it's essential to treat the real and imaginary parts separately. For instance, in our exercise, we substitute the complex expression \(\frac{i \pi}{2} - \ln 3\) into the hyperbolic cosine formula.

This involves understanding the behavior of the exponential function when applied to complex numbers. Simplifying these requires properties like Euler's formula, which connects complex exponentials to trigonometric functions.

Exponential Functions

Exponential functions are fundamental in mathematics, represented as \(e^x\) where \(e\) is Euler's number, approximately equal to 2.71828. Exponential functions grow rapidly and have essential applications in various fields including calculus, differential equations, and complex analysis.

The key property we're using here is that the exponential function can handle complex numbers. When we have an expression of the form \(e^{a + bi}\), it can be split using properties of exponents:

<\[ e^{a + bi} = e^a \cdot e^{bi} \]

Euler's formula comes into play here, allowing us to express \(e^{bi}\) in terms of trigonometric functions:

<\[ e^{bi} = \cos(b) + i\sin(b) \]

Applying this split is crucial in simplifying the given expression in our problem.

Euler's Formula

Euler's formula is a stunningly beautiful connection between complex exponentials and trigonometric functions. It is stated as:

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

This formula shows that exponentials with imaginary exponents result in a complex number on the unit circle in the complex plane. The angle \(\theta\) corresponds to the argument of the complex number, giving a direct relationship with sine and cosine functions.

In our exercise, to simplify \( e^{i \pi/2} \), we use Euler's formula:

<\[ e^{i \pi/2} = \cos(\pi/2) + i\sin(\pi/2) = i \]

Using Euler's formula makes the exponential simplifications more manageable, confirming our intermediate steps' accuracy and leading us to the solution.