Problem 25
Question
$$ \text { Prove that } \cosh z=\cosh x \cos y+i \sinh x \sin y \text {. } $$
Step-by-Step Solution
Verified Answer
The identity \( \cosh z = \cosh x \cos y + i \sinh x \sin y \) is proven using the definitions of hyperbolic and trigonometric identities.
1Step 1: Recall the definition of hyperbolic cosine
The hyperbolic cosine function is defined as \( \cosh z = \frac{e^z + e^{-z}}{2} \). This identity is central to proving the given equation.
2Step 2: Express \( z \) in terms of real and imaginary parts
In the complex number \( z = x + iy \), \( x \) and \( y \) are real numbers. Express \( e^z \) as \( e^{x+iy} = e^x e^{iy} \).
3Step 3: Simplify \( e^z \) using Euler's formula
Using Euler's formula, \( e^{iy} = \cos y + i \sin y \). Therefore, \( e^{x+iy} = e^x (\cos y + i \sin y) \). Similarly, \( e^{-z} = e^{-x} (\cos y - i \sin y) \) as derived from \( e^{-z} = e^{-(x+iy)} = e^{-x}e^{-iy} \).
4Step 4: Compute \( \cosh z \) with derived components
Substitute back into the definition: \[ \cosh z = \frac{e^x (\cos y + i \sin y) + e^{-x} (\cos y - i \sin y)}{2} \].
5Step 5: Distribute and Combine Terms
Distribute and combine real and imaginary terms: \[ \Rightarrow \cosh z = \frac{e^x \cos y + i e^x \sin y + e^{-x} \cos y - i e^{-x} \sin y}{2} \]. Combine like terms: \[ = \frac{(e^x + e^{-x}) \cos y}{2} + i \frac{(e^x - e^{-x}) \sin y}{2} \].
6Step 6: Simplify using identities for hyperbolic and trigonometric functions
Recall that \( \cosh x = \frac{e^x + e^{-x}}{2} \) and \( \sinh x = \frac{e^x - e^{-x}}{2} \). Substitute these identities back to get: \[ \cosh z = \cosh x \cos y + i \sinh x \sin y \].
7Step 7: Conclusion
We obtained that \( \cosh z \) simplifies exactly to \( \cosh x \cos y + i \sinh x \sin y \), proving the given equation.
Key Concepts
Hyperbolic FunctionsEuler's FormulaComplex Numbers
Hyperbolic Functions
Hyperbolic functions are analogs of trigonometric functions but for a hyperbola rather than a circle. They have important applications in areas such as calculus, complex analysis, and many engineering fields. The two most basic hyperbolic functions are hyperbolic sine (\( \,\sinh z\, \)) and hyperbolic cosine (\( \,\cosh z\, \)).
They are crucial when dealing with hyperbolas, much like how sine and cosine relate to circles. One key relationship to note is the identity:\[ \cosh^2 z - \sinh^2 z = 1 \]which mimics the trigonometric Pythagorean identity \(\cos^2 x + \sin^2 x = 1\).
- Hyperbolic cosine, \(\cosh z\), is defined as \(\cosh z = \frac{e^z + e^{-z}}{2}\).
- Hyperbolic sine, \(\sinh z\), is defined as \(\sinh z = \frac{e^z - e^{-z}}{2}\).
They are crucial when dealing with hyperbolas, much like how sine and cosine relate to circles. One key relationship to note is the identity:\[ \cosh^2 z - \sinh^2 z = 1 \]which mimics the trigonometric Pythagorean identity \(\cos^2 x + \sin^2 x = 1\).
Euler's Formula
Euler's formula is a fundamental bridge between the exponential function and trigonometric functions. It is typically expressed as \( e^{ix} = \cos x + i \sin x \).
This relationship elegantly shows how complex numbers interact with rotations in the plane. It's deeply linked to the complex exponential function and facilitates expressing periodic wave functions in physics and engineering.
In the given problem, Euler's formula allows us to break down complex exponents:
thereby simplifying calculations and understanding of phenomena such as oscillations and waves.
This relationship elegantly shows how complex numbers interact with rotations in the plane. It's deeply linked to the complex exponential function and facilitates expressing periodic wave functions in physics and engineering.
In the given problem, Euler's formula allows us to break down complex exponents:
- For complex number \(z = x + iy\), we use \(e^{iy} = \cos y + i \sin y\).
- Consequently, any expression involving \(e^{z}\) or \(e^{-z}\) can be decomposed using Euler's insights.
thereby simplifying calculations and understanding of phenomena such as oscillations and waves.
Complex Numbers
Complex numbers extend the real numbers with a unique component: \(i\), the imaginary unit, satisfying \(i^2 = -1\). Each complex number has the form \(z = x + iy\), where \(x\) is the real part and \(y\) is the imaginary part.
The capability to write real and imaginary components allows complex numbers to be manipulated like vectors. Operations such as addition, multiplication, and exponentiation can be performed under rules extending those for real numbers.
The capability to write real and imaginary components allows complex numbers to be manipulated like vectors. Operations such as addition, multiplication, and exponentiation can be performed under rules extending those for real numbers.
- Adding or subtracting involves managing real and imaginary parts separately — \((x_1 + iy_1) + (x_2 + iy_2) = (x_1 + x_2) + i(y_1 + y_2)\).
- Multiplication is done through distributive law: \((x + iy)(u + iv) = (xu - yv) + i(xv + yu)\).
Other exercises in this chapter
Problem 25
Express \(\ln z\) in the form \(a+i b\). \(z=-2+2 i\)
View solution Problem 25
Verify that the given function \(u\) is harmonic. Find \(v\), the harmonic conjugate function of \(u\). Form the corresponding analytic function \(f(z)=u+i v .\
View solution Problem 25
In Problems 23-28, express \(\ln z\) in the form \(a+i b\). $$ z=-2+2 i $$
View solution Problem 25
In Problems 23-28, verify that the given function \(u\) is harmonic. Find \(v\), the harmonic conjugate function of \(u\). Form the corresponding analytic funct
View solution