Question

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

Short answer

0 - \frac{4i}{3}

Step-by-step solution

Understand Hyperbolic Cosine

Recall the definition of hyperbolic cosine: \ \[\cosh(z) = \frac{e^z + e^{-z}}{2}\].

Substitute the Given Expression

Substitute the given expression \(\frac{i \pi}{2} - \ln 3\) into the definition of hyperbolic cosine: \[\cosh\left(\frac{i \pi}{2} - \ln 3\right) = \frac{e^{\frac{i \pi}{2} - \ln 3} + e^{-\left(\frac{i \pi}{2} - \ln 3\right)}}{2}\].

Simplify Exponents

Simplify the exponents separately: \[e^{\frac{i \pi}{2} - \ln 3} = e^{\frac{i \pi}{2}} \cdot e^{-\ln 3} = i \cdot \frac{1}{3} = \frac{i}{3}\]. Similarly, simplify the other term: \[e^{-\left(\frac{i \pi}{2} - \ln 3\right)} = e^{-\frac{i \pi}{2}} \cdot e^{\ln 3} = -i \cdot 3 = -3i\].

Combine Simplified Exponents

Combine the results from the simplified exponents: \[\cosh\left(\frac{i \pi}{2} - \ln 3\right) = \frac{\frac{i}{3} + (-3i)}{2} = \frac{\frac{i}{3} - 3i}{2}\].

Simplify the Final Expression

Simplify the final expression further: \[\cosh\left(\frac{i \pi}{2} - \ln 3\right) = \frac{-\frac{8i}{3}}{2} = -\frac{4i}{3}\]. Therefore, the expression in the form \(x + i y\) is \(0 - \frac{4i}{3}\).

Key concepts

hyperbolic functions

To tackle problems with hyperbolic functions, it's crucial to first understand what they are. These functions are analogs of trigonometric functions but for a hyperbola rather than a circle. They are defined using the exponential function. For example:

• The hyperbolic cosine, denoted \(\text{cosh}(z)\), is given by \(\text{cosh}(z) = \frac{e^z + e^{-z}}{2}\).
• The hyperbolic sine, denoted \(\text{sinh}(z)\), is given by \(\text{sinh}(z) = \frac{e^z - e^{-z}}{2}\).

These functions appear frequently in various mathematical fields because they simplify certain complex integral and differential equations.
Understanding these definitions will help you substitute and simplify complex expressions involving hyperbolic functions.

Euler's formula

Euler's formula is a remarkable equation that relates complex exponentials to trigonometric functions. It states: \(e^{ix} = \text{cos}(x) + i \text{sin}(x)\). This equation bridges the gap between exponential functions and trigonometry, making it easier to handle complex numbers in exponential form.
For instance, in the exercise, we encounter terms like \(e^{i\frac{\text{\pi}}{2}}\), which can be directly converted using Euler's formula. This term simplifies to \(\text{i}\), because \(e^{i\frac{\text{\pi}}{2}} = \text{cos}(\frac{\text{\pi}}{2}) + i \text{sin}(\frac{\text{\pi}}{2}) = i\). Learning this formula provides foundational knowledge for solving complex exponentiation problems.

complex exponentiation

Complex exponentiation involves raising numbers to the power of complex exponents. It is deeply linked with Euler's formula and hyperbolic functions.
In this exercise, we deal with terms like \( e^{\frac{i \pi}{2} - \ln 3}\). The key steps are:

• Separate and handle real and imaginary parts individually.
• Convert the imaginary exponential using Euler's formula.
• Combine the results.

Remember, for any complex number \(z \): \( e^z = e^{a+bi} = e^a \times e^{bi}\). Applying these principles allows you to solve complex exponentiation problems systematically.

mathematical simplification

Mathematical simplification means making an equation or expression as simple as possible. This often involves:

• Combining like terms.
• Breaking down complex expressions.
• Using identities (like Euler’s) to simplify terms.

In the exercise, after substituting and simplifying individual terms, the expression \( \frac{i}{3} - 3i \) was simplified to \( \frac{-\frac{8i}{3}}{2} = -\frac{4i}{3} \). This simplification is what eventually gives us the final \ x + iy \ form.
By mastering these simplification techniques, complex expressions become manageable, leading to clearer and more accurate solutions.