Problem 9
Question
$$\cosh (\pi i)=\cos (i(\pi i))=\cos (-\pi)=\cos \pi=-1$$
Step-by-Step Solution
Verified Answer
\( \cosh(\pi i) = -1 \).
1Step 1: Understand the Hyperbolic Cosine Function
The function given is the hyperbolic cosine function, which can be expressed in terms of the exponential function as \( \cosh(x) = \frac{e^x + e^{-x}}{2} \). In this problem, our task is to evaluate \( \cosh(\pi i) \).
2Step 2: Substitute into Hyperbolic Cosine Formula
Substitute \( x = \pi i \) into the hyperbolic cosine formula: \( \cosh(\pi i) = \frac{e^{\pi i} + e^{-\pi i}}{2} \). Next, recall Euler's formula \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \) to help simplify this expression.
3Step 3: Calculate \( e^{\pi i} \) Using Euler's Formula
According to Euler's formula, \( e^{\pi i} = \cos(\pi) + i\sin(\pi) \), which simplifies to \( -1 + 0i = -1 \).
4Step 4: Calculate \( e^{-\pi i} \) Using Euler's Formula
Similarly, \( e^{-\pi i} = \cos(-\pi) + i\sin(-\pi) \), which also simplifies to \( -1 + 0i = -1 \).
5Step 5: Substitute into Formula and Simplify
Substitute the calculated values of \( e^{\pi i} \) and \( e^{-\pi i} \) into the hyperbolic cosine formula: \( \cosh(\pi i) = \frac{-1 + (-1)}{2} = \frac{-2}{2} = -1 \). This shows that \( \cosh(\pi i) = -1 \), confirming the result.
Key Concepts
Hyperbolic FunctionsEuler's FormulaComplex Exponentials
Hyperbolic Functions
Hyperbolic functions are analogues of trigonometric functions for the hyperbola. They are essential in many areas of mathematics, including complex analysis. Two of the most important hyperbolic functions are the hyperbolic sine, \( \sinh(x) \), and the hyperbolic cosine, \( \cosh(x) \). These functions are defined using the exponential function:
Using the formula for the hyperbolic cosine function, \( \cosh(\pi i) \) becomes \( \frac{e^{\pi i} + e^{-\pi i}}{2} \). From here, further calculations simplify the expression using Euler's formula.
- \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
- \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)
Using the formula for the hyperbolic cosine function, \( \cosh(\pi i) \) becomes \( \frac{e^{\pi i} + e^{-\pi i}}{2} \). From here, further calculations simplify the expression using Euler's formula.
Euler's Formula
Euler's formula is a fundamental bridge between the exponential function and trigonometric functions for complex numbers. It states that for any real number \( \theta \),
\[ e^{i\theta} = \cos(\theta) + i\sin(\theta) \]
This formula allows us to express complex exponentials as trigonometric functions. For example, in the given exercise, we know that \( e^{\pi i} \) can be calculated using Euler's formula:
\[ e^{\pi i} = \cos(\pi) + i\sin(\pi) = -1 + 0i = -1 \]
Similarly, \( e^{-\pi i} \) simplifies to the same value because cosine is an even function and sine is an odd function:
\[ e^{-\pi i} = \cos(-\pi) + i\sin(-\pi) = -1 + 0i = -1 \]
This fundamental property helps to maintain symmetry in calculations involving complex numbers and is crucial for solving problems in complex analysis.
\[ e^{i\theta} = \cos(\theta) + i\sin(\theta) \]
This formula allows us to express complex exponentials as trigonometric functions. For example, in the given exercise, we know that \( e^{\pi i} \) can be calculated using Euler's formula:
\[ e^{\pi i} = \cos(\pi) + i\sin(\pi) = -1 + 0i = -1 \]
Similarly, \( e^{-\pi i} \) simplifies to the same value because cosine is an even function and sine is an odd function:
\[ e^{-\pi i} = \cos(-\pi) + i\sin(-\pi) = -1 + 0i = -1 \]
This fundamental property helps to maintain symmetry in calculations involving complex numbers and is crucial for solving problems in complex analysis.
Complex Exponentials
Complex exponentials involve extending the real exponential function to complex numbers. The compatibility with Euler's formula makes them very manageable to work with. Recognizing how exponential functions exhibit repetitive behavior when applied to imaginary units helps solve many complex questions.
This gets particularly interesting when we calculate expressions like \( e^{i\theta} \), as they revolve around a circular path in the complex plane. In the case of \( e^{\pi i} \) and \( e^{-\pi i} \), both fall on the negative x-axis of the unit circle. This repetition leads to the cosine and sine values used in Euler's formula and highlights the elegant nature of complex exponentials.
For our example, substituting these into the hyperbolic cosine function allows us to translate a potentially complex problem into a simple arithmetic one:
This gets particularly interesting when we calculate expressions like \( e^{i\theta} \), as they revolve around a circular path in the complex plane. In the case of \( e^{\pi i} \) and \( e^{-\pi i} \), both fall on the negative x-axis of the unit circle. This repetition leads to the cosine and sine values used in Euler's formula and highlights the elegant nature of complex exponentials.
For our example, substituting these into the hyperbolic cosine function allows us to translate a potentially complex problem into a simple arithmetic one:
- \( e^{\pi i} = -1 \)
- \( e^{-\pi i} = -1 \)
Other exercises in this chapter
Problem 8
\(u=\frac{x}{x^{2}+y^{2}}, v=\frac{y}{x^{2}+y^{2}} ; \quad \frac{\partial u}{\partial x}=\frac{y^{2}-x^{2}}{\left(x^{2}+y^{2}\right)^{2}}, \quad \frac{\partial
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$$4\left(\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}\right)$$
View solution Problem 9
$$\frac{3 \sqrt{2}}{2}\left(\cos \frac{5 \pi}{4}+i \sin \frac{5 \pi}{4}\right)$$
View solution Problem 9
$$11-10 i$$
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